Volume 1
1.8. MAIN TYPES OF CHEMICAL REACTIONS. CHEMICAL EQUATIONS AND THEIR BALANCING
Are There Many Types of Chemical Reactions? • Combination Reactions • Decomposition Reactions • Substitution (Displacement) Reactions • Exchange (Metathesis) Reactions • Balancing Simple Chemical Equations • Performing Selected Reactions of Different Types at Home • Exercises
Are There Many Types of Chemical Reactions? • Combination Reactions • Decomposition Reactions • Substitution (Displacement) Reactions • Exchange (Metathesis) Reactions • Balancing Simple Chemical Equations • Performing Selected Reactions of Different Types at Home • Exercises
1.8.1. Are There Many Types of Chemical Reactions? There are. Chemists classify chemical reactions by a pretty broad variety of factors. In our course, however, we will deal mostly with four main categories of chemical transformations: (a) combination reactions; (b) decomposition reactions; (c) substitution reactions, also known as displacement reactions; and (d) exchange reactions that are often called metathesis reactions.
1.8.2. Combination Reactions. A combination reaction is a chemical reaction between two or more substances to give a single substance. Chemical equations for a few typical combination reactions are presented in Figure 1-22. Note that all of these reactions give rise to only one product. If a chemical transformation produces more than one substance, it is not a combination reaction. (The non-subscript numbers in the equations will be explained later in this section.)
1.8.2. Combination Reactions. A combination reaction is a chemical reaction between two or more substances to give a single substance. Chemical equations for a few typical combination reactions are presented in Figure 1-22. Note that all of these reactions give rise to only one product. If a chemical transformation produces more than one substance, it is not a combination reaction. (The non-subscript numbers in the equations will be explained later in this section.)
Figure 1-22. Examples of combination reactions.
We are already familiar with the reaction between iron and sulfur. The next example on the list (Figure 1-22) is the reaction of aluminum (Al) with iodine (I2) to give AlI3. Watch this spectacular demonstration (Video 1-16) to see how the heat produced in the reaction causes some of the iodine to sublime in the form of purple vapors. Sublimation is the phase transition of a substance from the solid state directly to the gas phase (vapor) without melting. Examples of substances that sublime include iodine (I2) and dry ice, which is solid carbon dioxide (CO2). Solid CO2 is called dry ice because, while being cold (much colder than ice), it does not melt on heating the way ice does, but rather evaporates to become a gas.
We are already familiar with the reaction between iron and sulfur. The next example on the list (Figure 1-22) is the reaction of aluminum (Al) with iodine (I2) to give AlI3. Watch this spectacular demonstration (Video 1-16) to see how the heat produced in the reaction causes some of the iodine to sublime in the form of purple vapors. Sublimation is the phase transition of a substance from the solid state directly to the gas phase (vapor) without melting. Examples of substances that sublime include iodine (I2) and dry ice, which is solid carbon dioxide (CO2). Solid CO2 is called dry ice because, while being cold (much colder than ice), it does not melt on heating the way ice does, but rather evaporates to become a gas.
Video 1-16. The reaction of aluminum (Al) with iodine (I2) (source).
Digression. Did you notice in Video 1-16 that in order to initiate the reaction of aluminum with iodine, a drop of water had to be added to the mixture? Why and how does the water promote this transformation? No problem if you cannot answer this question right now. Just consider making a note to revisit this conundrum after you have mastered the material of Volume 3 (sections 3.3 and 3.7 in particular). There is also a hint in section 3.1 to help you. When you can explain why water initiates the reaction between aluminum and iodine and write the corresponding chemical equations, you are no longer at the introductory chemistry level.
One more example of a combination reaction is the reaction of calcium oxide, CaO, also known as quicklime, with water (Figure 1-22). The product of this reaction is calcium hydroxide (slaked lime), Ca(OH)2. Note that the formula of calcium hydroxide is presented as Ca(OH)2, not CaO2H2. Parentheses are often used in chemical formulas to indicate subgroups of atoms within a molecule. This provides some basic structural information about the compound on top of its composition (what elements are present and in what ratio). In principle, there is nothing wrong with presenting the formula of calcium hydroxide as CaO2H2, but chemists prefer Ca(OH)2 just because it is more informative. It will become self-explanatory later in the course. It is important to remember, however, that the subscript 2 outside the parenthesis means that there are two O atoms and two H atoms per one Ca atom in the compound. Likewise, in the formula of aluminum hydroxide, Al(OH)3, the subscript 3 indicates that there are three O atoms and three H atoms per one Al atom.
A very important combination reaction is the reaction of hydrogen with oxygen (Figure 1-22, bottom). This reaction, 2 H2 + O2 = 2 H2O, constitutes the foundation of the hydrogen economy for two main reasons. First, the reaction produces a large amount of energy that can be used for a variety of purposes, such as transportation (watch this video). Second, the only product of the reaction is water, an environmentally benign compound.
1.8.3. Decomposition Reactions. In a decomposition reaction, one substance breaks down into two or more other substances. Examples of decomposition reactions are presented in Figure 1-23.
A very important combination reaction is the reaction of hydrogen with oxygen (Figure 1-22, bottom). This reaction, 2 H2 + O2 = 2 H2O, constitutes the foundation of the hydrogen economy for two main reasons. First, the reaction produces a large amount of energy that can be used for a variety of purposes, such as transportation (watch this video). Second, the only product of the reaction is water, an environmentally benign compound.
1.8.3. Decomposition Reactions. In a decomposition reaction, one substance breaks down into two or more other substances. Examples of decomposition reactions are presented in Figure 1-23.
Figure 1-23. Examples of decomposition reactions.
The upper equation in Figure 1-23 is for the decomposition of hydrogen peroxide (H2O2) to water (H2O) and oxygen (O2). The arrow pointing upwards next to the formula of oxygen (O2) is sometimes used in chemical equations to indicate that the substance produced is a gas that evolves from the reaction mixture.
A very important decomposition reaction that has been used on a large scale since the ancient times is the decomposition of naturally abundant calcium carbonate (CaCO3), also known as limestone, to calcium oxide (CaO) and carbon dioxide (CO2). This reaction (Figure 1-23) is conducted in special reactors that are called kilns at a high temperature of approximately 1000 oC. The product of this decomposition reaction, calcium oxide (CaO), also known as quicklime (which was mentioned earlier), finds numerous important applications, such as in the production of steel and cement for making concrete. Interestingly, before the invention of electric lighting, the so-called Drummond light (also known as limelight or calcium light) was used for stage illumination in theaters and music halls. Drummond light, named after its inventor, Thomas Drummond, a Scottish engineer, is based on an intense glow that CaO emits when heated to 2,400 °C (Video 1-17).
The upper equation in Figure 1-23 is for the decomposition of hydrogen peroxide (H2O2) to water (H2O) and oxygen (O2). The arrow pointing upwards next to the formula of oxygen (O2) is sometimes used in chemical equations to indicate that the substance produced is a gas that evolves from the reaction mixture.
A very important decomposition reaction that has been used on a large scale since the ancient times is the decomposition of naturally abundant calcium carbonate (CaCO3), also known as limestone, to calcium oxide (CaO) and carbon dioxide (CO2). This reaction (Figure 1-23) is conducted in special reactors that are called kilns at a high temperature of approximately 1000 oC. The product of this decomposition reaction, calcium oxide (CaO), also known as quicklime (which was mentioned earlier), finds numerous important applications, such as in the production of steel and cement for making concrete. Interestingly, before the invention of electric lighting, the so-called Drummond light (also known as limelight or calcium light) was used for stage illumination in theaters and music halls. Drummond light, named after its inventor, Thomas Drummond, a Scottish engineer, is based on an intense glow that CaO emits when heated to 2,400 °C (Video 1-17).
Video 1-17. Drummond light (source).
The bottom equation in Figure 1-23 is for the decomposition of malachite, a beautiful mineral that is broadly used to make jewelry (Figure 1-24). Malachite is a copper compound that has the formula Cu2(OH)2CO3.
The bottom equation in Figure 1-23 is for the decomposition of malachite, a beautiful mineral that is broadly used to make jewelry (Figure 1-24). Malachite is a copper compound that has the formula Cu2(OH)2CO3.
Figure 1-24. Native malachite, Cu2(OH)2CO3 (source).
Malachite jewelry should not be exposed to high temperatures because on heating to approximately 200 oC, this green mineral decomposes to carbon dioxide (CO2), water (H2O), and black copper oxide (CuO), according to the bottom equation in Figure 1-23. Watch the decomposition of a sample of green malachite to black copper oxide in the first 1 min 20 sec of Video 1-18.
Malachite jewelry should not be exposed to high temperatures because on heating to approximately 200 oC, this green mineral decomposes to carbon dioxide (CO2), water (H2O), and black copper oxide (CuO), according to the bottom equation in Figure 1-23. Watch the decomposition of a sample of green malachite to black copper oxide in the first 1 min 20 sec of Video 1-18.
Video 1-18. Thermal decomposition of malachite, Cu2(OH)2CO3 (source).
1.8.4. Substitution (Displacement) Reactions. In a substitution reaction, one atom or group of atoms of a molecule is replaced with another atom or group of atoms. A few examples of substitution reactions are presented in Figure 1-25.
1.8.4. Substitution (Displacement) Reactions. In a substitution reaction, one atom or group of atoms of a molecule is replaced with another atom or group of atoms. A few examples of substitution reactions are presented in Figure 1-25.
Figure 1-25. Examples of substitution (displacement) reactions.
The upper equation in Figure 1-25 is for the substitution reaction of mercury chloride (HgCl2) with copper (Cu). In this reaction, the mercury atom of HgCl2 is substituted (displaced) by a Cu atom of copper metal. As a result, mercury metal is produced. Since copper is red and mercury is silver in color, this reaction is used in the old demonstration experiment where a copper coin is turned into a "silver" one (Video 1-19). Of course, there is no silver involved in this experiment, just the formation of mercury metal that imitates silver. Mercury is highly toxic, so do not try this experiment at home.
The upper equation in Figure 1-25 is for the substitution reaction of mercury chloride (HgCl2) with copper (Cu). In this reaction, the mercury atom of HgCl2 is substituted (displaced) by a Cu atom of copper metal. As a result, mercury metal is produced. Since copper is red and mercury is silver in color, this reaction is used in the old demonstration experiment where a copper coin is turned into a "silver" one (Video 1-19). Of course, there is no silver involved in this experiment, just the formation of mercury metal that imitates silver. Mercury is highly toxic, so do not try this experiment at home.
Video 1-19. The substitution reaction of mercury chloride (HgCl2) with copper (Cu) of a copper coin (source).
The middle equation in Figure 1-25 is for the so-called thermite reaction. Thermite is a mixture of aluminum powder (Al) and iron oxide (Fe2O3). If ignited, the mixture is engaged in a spectacular reaction that produces aluminum oxide (Al2O3) and iron metal (Fe). The reaction is highly energetic, generating dazzling light and so much heat that the iron produced melts. Watch a stunning thermite reaction experiment in Video 1-20 below. The thermite reaction is also a typical substitution reaction where the Fe atoms of Fe2O3 are displaced by Al atoms.
The middle equation in Figure 1-25 is for the so-called thermite reaction. Thermite is a mixture of aluminum powder (Al) and iron oxide (Fe2O3). If ignited, the mixture is engaged in a spectacular reaction that produces aluminum oxide (Al2O3) and iron metal (Fe). The reaction is highly energetic, generating dazzling light and so much heat that the iron produced melts. Watch a stunning thermite reaction experiment in Video 1-20 below. The thermite reaction is also a typical substitution reaction where the Fe atoms of Fe2O3 are displaced by Al atoms.
Video 1-20. Thermite reaction (source).
I would not recommend an inexperienced person to perform the thermite reaction because of its violent nature and serious dangers associated with it. Neither can I recommend trying another truly spectacular substitution reaction (Video 1-21), the reaction of sodium metal with water that occurs according to the bottom equation in Figure 1-25. In this reaction, a sodium atom (Na) displaces one of the hydrogen atoms on a molecule of water (H2O). The reaction of sodium metal with water generates so much heat that the hydrogen (H2) produced is ignited and can even explode due to its combination reaction with oxygen in the air, 2 H2 + O2 = 2 H2O (Figure 1-22, bottom).
I would not recommend an inexperienced person to perform the thermite reaction because of its violent nature and serious dangers associated with it. Neither can I recommend trying another truly spectacular substitution reaction (Video 1-21), the reaction of sodium metal with water that occurs according to the bottom equation in Figure 1-25. In this reaction, a sodium atom (Na) displaces one of the hydrogen atoms on a molecule of water (H2O). The reaction of sodium metal with water generates so much heat that the hydrogen (H2) produced is ignited and can even explode due to its combination reaction with oxygen in the air, 2 H2 + O2 = 2 H2O (Figure 1-22, bottom).
Video 1-21. The substitution reaction of sodium (Na) with water (H2O) performed on a pound scale (source).
1.8.5. Exchange (Metathesis) Reactions. In an exchange reaction (Figure 1-26), two different compounds interchange their parts. For example, if we mix a solution of table salt (NaCl) and silver nitrate (AgNO3), an exchange reaction will occur, resulting in the formation of AgCl that is insoluble in water and precipitates out. The downward arrow drawn after a formula in a chemical equation indicates that the compound precipitates out of solution. Before the reaction, the sodium atom was bonded to the chlorine atom (NaCl) and the silver atom was attached to the NO3 group. During the reaction, the sodium (Na) and silver (Ag) atoms swapped their original partners (Cl for Na and NO3 for Ag), as illustrated by the round arrows in the equation (Figure 1-26, top).
1.8.5. Exchange (Metathesis) Reactions. In an exchange reaction (Figure 1-26), two different compounds interchange their parts. For example, if we mix a solution of table salt (NaCl) and silver nitrate (AgNO3), an exchange reaction will occur, resulting in the formation of AgCl that is insoluble in water and precipitates out. The downward arrow drawn after a formula in a chemical equation indicates that the compound precipitates out of solution. Before the reaction, the sodium atom was bonded to the chlorine atom (NaCl) and the silver atom was attached to the NO3 group. During the reaction, the sodium (Na) and silver (Ag) atoms swapped their original partners (Cl for Na and NO3 for Ag), as illustrated by the round arrows in the equation (Figure 1-26, top).
Figure 1-26. Examples of exchange reactions.
The reaction between magnesium sulfate (MgSO4) and barium chloride (BaCl2) shown at the bottom of Figure 1-26 is another example of an exchange reaction. Both MgSO4 and BaCl2 are soluble in water. On mixing aqueous solutions of these two compounds, a white precipitate of barium sulfate (BaSO4) is formed. The other product of the exchange, magnesium chloride (MgCl2) is water-soluble and, consequently, remains in solution.
The reaction between magnesium sulfate (MgSO4) and barium chloride (BaCl2) shown at the bottom of Figure 1-26 is another example of an exchange reaction. Both MgSO4 and BaCl2 are soluble in water. On mixing aqueous solutions of these two compounds, a white precipitate of barium sulfate (BaSO4) is formed. The other product of the exchange, magnesium chloride (MgCl2) is water-soluble and, consequently, remains in solution.
Digression. Interestingly, both sulfates, MgSO4 and BaSO4, have applications in medicine. Magnesium sulfate, also known as Epsom salt, is a laxative. Barium sulfate is widely used as a radiocontrast agent for X-ray imaging and diagnostics in order to enhance the visibility of gastro-intestinal tract structures. Prior to the test, a patient is given the so-called barium meal, a thick suspension of BaSO4 in water with added flavors. Although water-soluble barium salts such as barium chloride, BaCl2, and barium nitrate, Ba(NO3)2, are toxic, BaSO4 is not and is perfectly safe to take because it is insoluble in water and is excreted from the body unchanged.
1.8.6. Balancing Simple Chemical Equations. In most of the equations presented in Figures 1-22, 1-23, and 1-25, we see non-subscript numbers in front of certain chemical formulas. Some of these equations are repeated below with those numbers in red.
2 Al + 3 I2 = 2 AlI3
2 H2 + O2 = 2 H2O
2 Na + 2 H2O = 2 NaOH + H2
The emphasized non-subscript numbers in the three equations above are called the coefficients. Coefficients indicate the stoichiometry of a chemical reaction, the ratio of reagents and products involved in a chemical transformation. The equation 2 Al + 3 I2 = 2 AlI3 tells us that two aluminum atoms (Al) react with three molecules of iodine (I2) to give two molecules of aluminum iodide (AlI3).
From the next equation, 2 H2 + O2 = 2 H2O, we gather that two molecules of hydrogen (H2) react with one molecule of oxygen (O2) to produce two molecules of water (H2O). Note that if the coefficient is 1, no number is placed before the corresponding formula in the equation.
A correct chemical equation is balanced, which means that the total number of atoms of the same type on the left side and on the right side of the equation must be the same. This requirement follows from the law of conservation of mass, which states that in chemical reactions atoms are neither created nor destroyed, but are only redistributed to form new molecules.
Let us begin to learn how to balance chemical equations using a very simple reaction as an example, the reaction between oxygen (O2) and hydrogen (H2). When balancing a reaction of O2, it is always a good idea to start by balancing the number of oxygen atoms.
Step 1. We know that H2 and O2 react to give water, H2O. We express that in the form of a chemical equation. The equation is not balanced though.
H2 + O2 = H2O (H balanced; O not balanced; overall not balanced)
Step 2. The originally written equation is not balanced. It is balanced for hydrogen (two H atoms on the left and two on the right), but not for oxygen. The number of oxygen atoms on the left is two, but on the right is only one. Important: the number of oxygen atoms on the left is even and will always be even because a molecule of oxygen is diatomic (consists of two atoms). No matter what coefficient we place before the O2 in the equation, the number of O atoms on the left will be even. Consequently, the number of O atoms on the right side of the final balanced equation must be even, too. Currently the number is odd because there is one molecule of H2O on the right, and a molecule of H2O contains one O atom, an odd number. We can make the number of O atoms on the right even by placing a coefficient 2 in front of the H2O. Let us do that. As a result, the oxygen is now balanced. But the hydrogen is not.
H2 + O2 = 2 H2O (H not balanced; O balanced; overall not balanced)
Step 3. Now we count the number of H atoms on the left and on the right of the equation just above. There are two H atoms on the left and four on the right, twice as many. To balance the hydrogen, we just have to put the coefficient 2 in front of the H2 on the left.
2 H2 + O2 = 2 H2O (H balanced; O balanced; overall balanced)
Finally, the equation is fully balanced. To double-check, we should compare the number of H atoms and O atoms on the left and on the right of the equation. There are four H atoms on the left and four H atoms on the right. There are two O atoms on the left and two on the right. All is well, the equation is balanced.
Important: When balancing chemical equations we can change only the coefficients, the numbers before chemical formulas in the equation. The subscripts in the formulas must never be changed for balancing chemical equations because the subscripts specify the number of atoms in a molecule or, in other words, its composition, which is constant for a particular substance.
In a correctly balanced equation, the coefficients should be minimal. For example, although the equation 4 H2 + 2 O2 = 4 H2O is balanced, it can and should be simplified by dividing all of the coefficients by two, which gives: 2 H2 + O2 = 2 H2O.
Let us practice a bit more by balancing the reaction of sodium (Na) with water (H2O) to give sodium hydroxide (NaOH) and hydrogen (H2).
Step 1. Knowing that Na reacts with H2O to give NaOH and H2, we write the equation.
Na + H2O = NaOH + H2 (Na balanced; O balanced; H not balanced; overall not balanced)
Step 2. The original equation (Step 1 above) is balanced for Na and O, but not for H. We notice that the number of H atoms on the left will always be even because a molecule of H2O contains two H atoms. Therefore, in the balanced equation the number of H atoms on the right must also be even. There are two hydrogen-containing substances on the right. One is H2 with an even number of H atoms. The other is NaOH with just one H atom, which is an odd number. Therefore, we should make the number of molecules of NaOH even. Let us make it 2 and see what happens.
Na + H2O = 2 NaOH + H2 (Na not balanced; O not balanced; H not balanced; overall not balanced)
Step 3. In the equation modified in Step 2, not a single element is balanced. No sweat. Sometimes things get worse before they get better. Just think logically. What is important is that there are even numbers of H atoms on both sides of the equation, four on the right and two on the left. To balance the hydrogen, we put 2 in front of the formula of water. Interestingly, this balances not only the hydrogen but also the oxygen.
Na + 2 H2O = 2 NaOH + H2 (Na not balanced; O balanced; H balanced; overall not balanced)
Step 4. The only remaining problem is that there are two Na atoms on the right but only one on the left. All we need to do is just put 2 in front of the Na in the left side of the equation. The equation is now fully balanced.
2 Na + 2 H2O = 2 NaOH + H2 (Na balanced; O balanced; H balanced; overall balanced)
Finally, let us balance one more equation, the one for the reaction between aluminum and iodine.
Step 1. Write the unbalanced original equation.
Al + I2 = AlI3 (Al balanced; I not balanced; overall not balanced)
Step 2. We note that the number of iodine atoms on the left will always be even because a molecule of I2 is diatomic. On the right side, however, the number of I atoms is odd (three). We make it even by placing 2 before the formula of AlI3.
Al + I2 = 2 AlI3 (Al not balanced; I not balanced; overall not balanced)
Step 3. Now that the number of I atoms on both sides is even, we can balance the iodine. There are two I atoms on the left and 6 on the right. By placing 3 before the I2 formula, we get the iodine balanced.
Al + 3 I2 = 2 AlI3 (Al not balanced; I balanced; overall not balanced)
Step 4. All we have to do is balance the aluminum. There are two Al atoms on the right but only one on the left. We put 2 before the Al on the left to have the equation fully balanced.
2 Al + 3 I2 = 2 AlI3 (Al balanced; I balanced; overall balanced)
Done.
1.8.7. Performing Selected Reactions of Different Types at Home. If you wish to try some fun reactions of different types at home, here are some experiments for you to consider trying. First, however, please carefully read and understand the following:
DISCLAIMER: Although most of the experiments described in this subsection and elsewhere in this website are regarded as low hazard, I expressly disclaim all liability for any occurrence, including, but not limited to, damage, injury or death which might arise as consequences of the use of any experiment(s) discussed, listed, described, or otherwise mentioned in the free online course Chemistry from Scratch. Therefore, you assume all the liability and use these experiments at your own risk (see Terms of Use).
If you decide not to do the experiments, please still read this subsection.
Experiment 5. Decomposition of Hydrogen Peroxide (H2O2) in the Presence of Baker's Yeast. This reaction is used in a cool experiment titled "Elephant Toothpaste", (Video 1-22). There are many more videos of various modifications of this experiment, which can be easily found on the Internet.
2 Al + 3 I2 = 2 AlI3
2 H2 + O2 = 2 H2O
2 Na + 2 H2O = 2 NaOH + H2
The emphasized non-subscript numbers in the three equations above are called the coefficients. Coefficients indicate the stoichiometry of a chemical reaction, the ratio of reagents and products involved in a chemical transformation. The equation 2 Al + 3 I2 = 2 AlI3 tells us that two aluminum atoms (Al) react with three molecules of iodine (I2) to give two molecules of aluminum iodide (AlI3).
From the next equation, 2 H2 + O2 = 2 H2O, we gather that two molecules of hydrogen (H2) react with one molecule of oxygen (O2) to produce two molecules of water (H2O). Note that if the coefficient is 1, no number is placed before the corresponding formula in the equation.
A correct chemical equation is balanced, which means that the total number of atoms of the same type on the left side and on the right side of the equation must be the same. This requirement follows from the law of conservation of mass, which states that in chemical reactions atoms are neither created nor destroyed, but are only redistributed to form new molecules.
Let us begin to learn how to balance chemical equations using a very simple reaction as an example, the reaction between oxygen (O2) and hydrogen (H2). When balancing a reaction of O2, it is always a good idea to start by balancing the number of oxygen atoms.
Step 1. We know that H2 and O2 react to give water, H2O. We express that in the form of a chemical equation. The equation is not balanced though.
H2 + O2 = H2O (H balanced; O not balanced; overall not balanced)
Step 2. The originally written equation is not balanced. It is balanced for hydrogen (two H atoms on the left and two on the right), but not for oxygen. The number of oxygen atoms on the left is two, but on the right is only one. Important: the number of oxygen atoms on the left is even and will always be even because a molecule of oxygen is diatomic (consists of two atoms). No matter what coefficient we place before the O2 in the equation, the number of O atoms on the left will be even. Consequently, the number of O atoms on the right side of the final balanced equation must be even, too. Currently the number is odd because there is one molecule of H2O on the right, and a molecule of H2O contains one O atom, an odd number. We can make the number of O atoms on the right even by placing a coefficient 2 in front of the H2O. Let us do that. As a result, the oxygen is now balanced. But the hydrogen is not.
H2 + O2 = 2 H2O (H not balanced; O balanced; overall not balanced)
Step 3. Now we count the number of H atoms on the left and on the right of the equation just above. There are two H atoms on the left and four on the right, twice as many. To balance the hydrogen, we just have to put the coefficient 2 in front of the H2 on the left.
2 H2 + O2 = 2 H2O (H balanced; O balanced; overall balanced)
Finally, the equation is fully balanced. To double-check, we should compare the number of H atoms and O atoms on the left and on the right of the equation. There are four H atoms on the left and four H atoms on the right. There are two O atoms on the left and two on the right. All is well, the equation is balanced.
Important: When balancing chemical equations we can change only the coefficients, the numbers before chemical formulas in the equation. The subscripts in the formulas must never be changed for balancing chemical equations because the subscripts specify the number of atoms in a molecule or, in other words, its composition, which is constant for a particular substance.
In a correctly balanced equation, the coefficients should be minimal. For example, although the equation 4 H2 + 2 O2 = 4 H2O is balanced, it can and should be simplified by dividing all of the coefficients by two, which gives: 2 H2 + O2 = 2 H2O.
Let us practice a bit more by balancing the reaction of sodium (Na) with water (H2O) to give sodium hydroxide (NaOH) and hydrogen (H2).
Step 1. Knowing that Na reacts with H2O to give NaOH and H2, we write the equation.
Na + H2O = NaOH + H2 (Na balanced; O balanced; H not balanced; overall not balanced)
Step 2. The original equation (Step 1 above) is balanced for Na and O, but not for H. We notice that the number of H atoms on the left will always be even because a molecule of H2O contains two H atoms. Therefore, in the balanced equation the number of H atoms on the right must also be even. There are two hydrogen-containing substances on the right. One is H2 with an even number of H atoms. The other is NaOH with just one H atom, which is an odd number. Therefore, we should make the number of molecules of NaOH even. Let us make it 2 and see what happens.
Na + H2O = 2 NaOH + H2 (Na not balanced; O not balanced; H not balanced; overall not balanced)
Step 3. In the equation modified in Step 2, not a single element is balanced. No sweat. Sometimes things get worse before they get better. Just think logically. What is important is that there are even numbers of H atoms on both sides of the equation, four on the right and two on the left. To balance the hydrogen, we put 2 in front of the formula of water. Interestingly, this balances not only the hydrogen but also the oxygen.
Na + 2 H2O = 2 NaOH + H2 (Na not balanced; O balanced; H balanced; overall not balanced)
Step 4. The only remaining problem is that there are two Na atoms on the right but only one on the left. All we need to do is just put 2 in front of the Na in the left side of the equation. The equation is now fully balanced.
2 Na + 2 H2O = 2 NaOH + H2 (Na balanced; O balanced; H balanced; overall balanced)
Finally, let us balance one more equation, the one for the reaction between aluminum and iodine.
Step 1. Write the unbalanced original equation.
Al + I2 = AlI3 (Al balanced; I not balanced; overall not balanced)
Step 2. We note that the number of iodine atoms on the left will always be even because a molecule of I2 is diatomic. On the right side, however, the number of I atoms is odd (three). We make it even by placing 2 before the formula of AlI3.
Al + I2 = 2 AlI3 (Al not balanced; I not balanced; overall not balanced)
Step 3. Now that the number of I atoms on both sides is even, we can balance the iodine. There are two I atoms on the left and 6 on the right. By placing 3 before the I2 formula, we get the iodine balanced.
Al + 3 I2 = 2 AlI3 (Al not balanced; I balanced; overall not balanced)
Step 4. All we have to do is balance the aluminum. There are two Al atoms on the right but only one on the left. We put 2 before the Al on the left to have the equation fully balanced.
2 Al + 3 I2 = 2 AlI3 (Al balanced; I balanced; overall balanced)
Done.
1.8.7. Performing Selected Reactions of Different Types at Home. If you wish to try some fun reactions of different types at home, here are some experiments for you to consider trying. First, however, please carefully read and understand the following:
DISCLAIMER: Although most of the experiments described in this subsection and elsewhere in this website are regarded as low hazard, I expressly disclaim all liability for any occurrence, including, but not limited to, damage, injury or death which might arise as consequences of the use of any experiment(s) discussed, listed, described, or otherwise mentioned in the free online course Chemistry from Scratch. Therefore, you assume all the liability and use these experiments at your own risk (see Terms of Use).
If you decide not to do the experiments, please still read this subsection.
Experiment 5. Decomposition of Hydrogen Peroxide (H2O2) in the Presence of Baker's Yeast. This reaction is used in a cool experiment titled "Elephant Toothpaste", (Video 1-22). There are many more videos of various modifications of this experiment, which can be easily found on the Internet.
Video 1-22. "Elephant toothpaste" (decomposition of hydrogen peroxide, H2O2) (source).
We will do a similar experiment using a 3% solution of hydrogen peroxide in water, which is sold in pharmacies and grocery stores. As the decomposition occurs, oxygen gas (O2) bubbles off (Figure 1-23, top equation). At room temperature, the decomposition of H2O2, especially in such a dilute solution is extremely slow, but we can speed it up by using a catalyst. As briefly mentioned earlier, a catalyst is a substance that accelerates a chemical reaction. An excellent catalyst for the decomposition of hydrogen peroxide is manganese dioxide (MnO2). If you happen to have a tiny bit of MnO2, you can use it. Alternatively, you may get this compound from a zinc-carbon battery, as shown in this video. Just keep in mind that (a) to get MnO2 you need a zinc-carbon battery, one of those marked "Heavy Duty" or "Super Heavy Duty" and (b) the manganese dioxide that you get from such batteries is not very pure, but still can be used as a catalyst to decompose H2O2.
There is, however, another excellent catalyst for decomposition of H2O2, and that catalyst is likely available from your kitchen. It is… baker's yeast! Baker's yeast contains the enzyme catalase, a huge and very complex biomolecule that accelerates the decomposition reaction of hydrogen peroxide very efficiently. It is baker's yeast that we will use in our experiment described below.
Place about ¼ teaspoon of dry baker's yeast at the bottom of a highball glass. Slowly pour 3% hydrogen peroxide onto the yeast and observe the formation of bubbles as the two get in contact. The gas that bubbles off is pure oxygen, the same gas that is present in air and that is vital to life. We breathe in air to get oxygen, which is needed for biological processes constantly occurring in our bodies. Also, it is oxygen that is needed for combustion. Combustion in pure oxygen is much more efficient than in air, because air contains only about 20% O2.
How do we know that the gas produced in the decomposition of hydrogen peroxide is oxygen? A good simple test for pure O2 is a smoldering wooden toothpick or coffee stirrer. When exposed to pure oxygen, the sluggishly smoldering areas immediately flare up and the wood burns rapidly and brightly, as demonstrated in Video 1-23.
We will do a similar experiment using a 3% solution of hydrogen peroxide in water, which is sold in pharmacies and grocery stores. As the decomposition occurs, oxygen gas (O2) bubbles off (Figure 1-23, top equation). At room temperature, the decomposition of H2O2, especially in such a dilute solution is extremely slow, but we can speed it up by using a catalyst. As briefly mentioned earlier, a catalyst is a substance that accelerates a chemical reaction. An excellent catalyst for the decomposition of hydrogen peroxide is manganese dioxide (MnO2). If you happen to have a tiny bit of MnO2, you can use it. Alternatively, you may get this compound from a zinc-carbon battery, as shown in this video. Just keep in mind that (a) to get MnO2 you need a zinc-carbon battery, one of those marked "Heavy Duty" or "Super Heavy Duty" and (b) the manganese dioxide that you get from such batteries is not very pure, but still can be used as a catalyst to decompose H2O2.
There is, however, another excellent catalyst for decomposition of H2O2, and that catalyst is likely available from your kitchen. It is… baker's yeast! Baker's yeast contains the enzyme catalase, a huge and very complex biomolecule that accelerates the decomposition reaction of hydrogen peroxide very efficiently. It is baker's yeast that we will use in our experiment described below.
Place about ¼ teaspoon of dry baker's yeast at the bottom of a highball glass. Slowly pour 3% hydrogen peroxide onto the yeast and observe the formation of bubbles as the two get in contact. The gas that bubbles off is pure oxygen, the same gas that is present in air and that is vital to life. We breathe in air to get oxygen, which is needed for biological processes constantly occurring in our bodies. Also, it is oxygen that is needed for combustion. Combustion in pure oxygen is much more efficient than in air, because air contains only about 20% O2.
How do we know that the gas produced in the decomposition of hydrogen peroxide is oxygen? A good simple test for pure O2 is a smoldering wooden toothpick or coffee stirrer. When exposed to pure oxygen, the sluggishly smoldering areas immediately flare up and the wood burns rapidly and brightly, as demonstrated in Video 1-23.
Video 1-23. Decomposition of hydrogen peroxide (H2O2) in the presence of yeast as a catalyst (source).
If you would like to do the Elephant Toothpaste version of the H2O2 decomposition experiment and are not afraid of getting in trouble for the mess created in your kitchen or bathroom, go right ahead and have fun! You will just need to modify the conditions a bit by adding a small amount of liquid soap and a food color to the yeast and stirring the mixture prior to the addition of hydrogen peroxide. Note that usually in the Elephant Toothpaste experiment, a more concentrated than 3% solution of H2O2 is employed. So, do not get too disappointed if the Elephant Toothpaste is produced less vigorously in your experiment than in the ones that are shown in the videos on the Internet.
Experiment 6. Decomposition of Ammonium Dichromate, (NH4)2Cr2O7: The Chemical Volcano. Watch the demonstration of this truly spectacular decomposition reaction in Video 1-24.
If you would like to do the Elephant Toothpaste version of the H2O2 decomposition experiment and are not afraid of getting in trouble for the mess created in your kitchen or bathroom, go right ahead and have fun! You will just need to modify the conditions a bit by adding a small amount of liquid soap and a food color to the yeast and stirring the mixture prior to the addition of hydrogen peroxide. Note that usually in the Elephant Toothpaste experiment, a more concentrated than 3% solution of H2O2 is employed. So, do not get too disappointed if the Elephant Toothpaste is produced less vigorously in your experiment than in the ones that are shown in the videos on the Internet.
Experiment 6. Decomposition of Ammonium Dichromate, (NH4)2Cr2O7: The Chemical Volcano. Watch the demonstration of this truly spectacular decomposition reaction in Video 1-24.
Video 1-24. The chemical volcano: decomposition of ammonium dichromate, (NH4)2Cr2O7 (source).
The reaction is the self-sustaining decomposition of ammonium dichromate, (NH4)2Cr2O7, to chromium oxide (Cr2O3), nitrogen gas (N2) and water (H2O), according to the equation:
(NH4)2Cr2O7 = Cr2O3 + N2 + 4 H2O
If you watched the video demonstration, you understand why this experiment is often called the chemical volcano. The dark-green fluffy solid formed in the decomposition of the bright-orange ammonium dichromate is chromium oxide, Cr2O3. The reaction also produces nitrogen gas and water in the form of steam, which give the process the striking "volcano" effect. If you would like to do this experiment yourself, you need to obtain ammonium dichromate, which is sold in some online stores.
As always, we need to know well the chemicals we work with. None of the three reaction products (N2, H2O, Cr2O3) pose danger. Nitrogen constitutes about 80% of the air that we breathe. Water is water. The chromium product (Cr2O3) is a stable green solid that is widely used as a pigment for paints and ink and as a polishing material conventionally referred to as "Green Polishing Compound".
Now, read carefully. In contrast with the reaction products, the starting material, ammonium dichromate, is a carcinogen and an irritant. I remind you that, as clearly stated in the Disclaimer above, you use any of the experiments discussed, listed, described, or otherwise mentioned in this course at your own risk. Among all materials used for the experiments described in this course, ammonium dichromate is the chemical that requires particularly careful handling. Avoid inhalation and any contact with this chemical. Wear latex or rubber gloves when doing the experiment.
I recommend that the chemical volcano experiment be performed outdoors. While you should decide on how much ammonium dichromate you would like to use for the reaction, I would suggest a quantity between 20 and 50 g. Pour ammonium dichromate onto a large ceramic tile or a metal sheet or a sand tray. Do not use a tray made of combustible materials such as cardboard, wood or plastic. It is also a good idea to lay out a tarp or sheets of grocery plastic bags around the tile. In this way, it will be easy to remove and dispose of the fluffy Cr2O3 "volcano ash" produced in the reaction. And there will be a lot of it produced! Make a cone pile of (NH4)2Cr2O7 on the tray. Ignite the tip of the pile with a kitchen lighter or a long reach match. Step back and watch the volcano erupt. After the reaction has gone to completion give the hot dark green "ashes" a few minutes to cool down before you remove them.
Experiment 7. Substitution Reaction of Silver Nitrate (AgNO3) with Copper (Cu): The Silver Tree. Look at the photo in Figure 1-27 and watch a demonstration of this beautiful experiment in Video 1-25. If after watching the video you would like to grow your own silver tree, you will need silver nitrate, a piece of copper wire, and distilled water for that.
The reaction is the self-sustaining decomposition of ammonium dichromate, (NH4)2Cr2O7, to chromium oxide (Cr2O3), nitrogen gas (N2) and water (H2O), according to the equation:
(NH4)2Cr2O7 = Cr2O3 + N2 + 4 H2O
If you watched the video demonstration, you understand why this experiment is often called the chemical volcano. The dark-green fluffy solid formed in the decomposition of the bright-orange ammonium dichromate is chromium oxide, Cr2O3. The reaction also produces nitrogen gas and water in the form of steam, which give the process the striking "volcano" effect. If you would like to do this experiment yourself, you need to obtain ammonium dichromate, which is sold in some online stores.
As always, we need to know well the chemicals we work with. None of the three reaction products (N2, H2O, Cr2O3) pose danger. Nitrogen constitutes about 80% of the air that we breathe. Water is water. The chromium product (Cr2O3) is a stable green solid that is widely used as a pigment for paints and ink and as a polishing material conventionally referred to as "Green Polishing Compound".
Now, read carefully. In contrast with the reaction products, the starting material, ammonium dichromate, is a carcinogen and an irritant. I remind you that, as clearly stated in the Disclaimer above, you use any of the experiments discussed, listed, described, or otherwise mentioned in this course at your own risk. Among all materials used for the experiments described in this course, ammonium dichromate is the chemical that requires particularly careful handling. Avoid inhalation and any contact with this chemical. Wear latex or rubber gloves when doing the experiment.
I recommend that the chemical volcano experiment be performed outdoors. While you should decide on how much ammonium dichromate you would like to use for the reaction, I would suggest a quantity between 20 and 50 g. Pour ammonium dichromate onto a large ceramic tile or a metal sheet or a sand tray. Do not use a tray made of combustible materials such as cardboard, wood or plastic. It is also a good idea to lay out a tarp or sheets of grocery plastic bags around the tile. In this way, it will be easy to remove and dispose of the fluffy Cr2O3 "volcano ash" produced in the reaction. And there will be a lot of it produced! Make a cone pile of (NH4)2Cr2O7 on the tray. Ignite the tip of the pile with a kitchen lighter or a long reach match. Step back and watch the volcano erupt. After the reaction has gone to completion give the hot dark green "ashes" a few minutes to cool down before you remove them.
Experiment 7. Substitution Reaction of Silver Nitrate (AgNO3) with Copper (Cu): The Silver Tree. Look at the photo in Figure 1-27 and watch a demonstration of this beautiful experiment in Video 1-25. If after watching the video you would like to grow your own silver tree, you will need silver nitrate, a piece of copper wire, and distilled water for that.
Figure 1-27. Progress of a silver tree experiment (source).
Video 1-25. The reaction of silver nitrate (AgNO3) with copper (Cu): the silver tree (source).
Silver nitrate, AgNO3, is a crystalline solid that is available for purchase from some online stores. Pure AgNO3 is white. But, because AgNO3 is slowly decomposed by light and on contact with some substances, many commercial samples of AgNO3 are off-white or can even be grayish. Such samples are still fine for the experiment. However, if the sample of AgNO3 you purchased or obtained otherwise is black or very dark, do not use it. Also note that if you spill a silver nitrate solution on your skin, rinse the area with water. Do not worry if after working with AgNO3 you find dark spots on your hands. These spots are from metallic silver that is produced in certain reactions of AgNO3 with various organic compounds on the skin. Why are those stains black, not shiny silver in color? That is because the silver metal that is produced in these reactions is in the form of tiny particles that are black. It will take a few days for the stains to come off.
Silver nitrate, AgNO3, is a crystalline solid that is available for purchase from some online stores. Pure AgNO3 is white. But, because AgNO3 is slowly decomposed by light and on contact with some substances, many commercial samples of AgNO3 are off-white or can even be grayish. Such samples are still fine for the experiment. However, if the sample of AgNO3 you purchased or obtained otherwise is black or very dark, do not use it. Also note that if you spill a silver nitrate solution on your skin, rinse the area with water. Do not worry if after working with AgNO3 you find dark spots on your hands. These spots are from metallic silver that is produced in certain reactions of AgNO3 with various organic compounds on the skin. Why are those stains black, not shiny silver in color? That is because the silver metal that is produced in these reactions is in the form of tiny particles that are black. It will take a few days for the stains to come off.
Digression. Silver nitrate and other silver compounds are antiseptics that are used to treat small cuts, scratches, and wounds to kill bacteria and even to remove warts. Dr. Albert C. Barnes (1872-1951) made his fortune from the invention and sales of Argyrol, a very efficient antiseptic that was made with silver nitrate. Being an avid and very knowledgeable art collector, Dr. Barnes used his fortune to acquire many amazing pieces of art, mostly paintings by Cezanne, Renoir, Modigliani, de Chirico, and other outstanding artists. These masterpieces are now on display in the Barnes Foundation, a museum and an educational institution in Philadelphia. Perhaps, we should be thankful for this outstanding art collection not only to the artists and Dr. Barnes, but also to AgNO3.
Before we do the silver tree experiment, we also need to mention that copper wire is available from hardware and other stores. Distilled water can be purchased from grocery stores, pharmacies or some department stores. Can tap water be used in the experiment? It may, but a solution of AgNO3 in tap water is usually opaque (cloudy). This is due to the formation of small quantities of insoluble silver chloride (AgCl) from silver nitrate (AgNO3) and chloride salts such as NaCl that are present in tap water (Figure 1-26).
Place about one quarter of a teaspoon of AgNO3 in a highball glass of the size of a standard cup. Use a plastic teaspoon, not a metal one. Add approximately half a cup of distilled water to the silver nitrate and stir the mixture with a plastic teaspoon until all of the AgNO3 has dissolved. This will take only a few moments because silver nitrate is very easily soluble in water.
Now, shape your copper wire in any form you like. It could be a tree or a flower or a star, or just a simple little spring, or a circular loop. Just make sure that the size of the shaped copper wire can fit inside the glass. Immerse the wire into the solution of AgNO3 and watch the reaction occur. First, the red copper wire will turn black, but just wait and very soon you will see beautiful shiny crystals of pure silver metal growing on the wire. You will also see that, as the silver crystals grow, the originally colorless solution gradually changes its color to blue due to the formation of Cu(NO3)2 that stays in solution and that is blue in color.
After the reaction, you may shake the silver crystals off the wire and wash them with distilled water. This is conveniently done by decantation. Decantation is a simple and convenient separation technique, in which a liquid over a solid material at the bottom of a vessel is gently poured off so as not to disturb the solid. This short video shows how decantation is performed. The silver metal produced in the reaction is heavy (dense) and once shaken off the copper wire, will settle at the bottom nicely. Decant off the blue solution, then add distilled water to the silver and gently swirl the mixture. Decant off the washing. Repeat this procedure 4-5 times and then let the wet silver dry at room temperature. You have made pure silver from AgNO3 in this spectacular reaction!
In the silver tree experiment, silver nitrate (AgNO3) reacts with copper metal (Cu) to give silver metal (Ag) and copper nitrate (Cu(NO3)2), according to the following equation:
2 AgNO3 + Cu = Cu(NO3)2 + 2 Ag
This is a typical substitution reaction, in which the silver atom bonded to the NO3 group is displaced by a copper atom. As a result, silver metal and copper nitrate are formed. Crystals of silver metal grow on the surface of the copper as the copper dissolves in the form of copper nitrate, Cu(NO3)2, which gives the solution the blue color.
You may naturally ask, why is the formula of silver nitrate AgNO3, whereas the formula of copper nitrate is Cu(NO3)2, not CuNO3? A simple answer is that a silver atom has the ability to bind to only one NO3 group of atoms, whereas a copper atom can bind to two of these groups. We will discuss this in more detail in the next section.
Experiment 8. Reaction of Silver Nitrate (AgNO3) with Sodium Chloride (NaCl). In the description of the silver tree experiment above (Experiment 7 above), I told you that it is much better to use distilled rather than tap water for making solutions of AgNO3. A solution of silver nitrate in distilled water is clear, whereas in tap water it is cloudy. There are always small quantities of various mineral salts in tap water, including NaCl that reacts with AgNO3 to give insoluble AgCl, which precipitates out (Figure 1-26). Since tap water contains only small quantities of NaCl, only a small amount of AgCl is produced in the form of tiny solid particles, which make the solution cloudy.
Dissolve a pinch of NaCl (table salt) in a small amount of water. Dissolve a pinch of AgNO3 in a small amount of distilled water. Mix the two solutions. Observe the immediate formation of a white precipitate. This precipitate is silver chloride, AgCl, formed in a typical exchange reaction.
Place about one quarter of a teaspoon of AgNO3 in a highball glass of the size of a standard cup. Use a plastic teaspoon, not a metal one. Add approximately half a cup of distilled water to the silver nitrate and stir the mixture with a plastic teaspoon until all of the AgNO3 has dissolved. This will take only a few moments because silver nitrate is very easily soluble in water.
Now, shape your copper wire in any form you like. It could be a tree or a flower or a star, or just a simple little spring, or a circular loop. Just make sure that the size of the shaped copper wire can fit inside the glass. Immerse the wire into the solution of AgNO3 and watch the reaction occur. First, the red copper wire will turn black, but just wait and very soon you will see beautiful shiny crystals of pure silver metal growing on the wire. You will also see that, as the silver crystals grow, the originally colorless solution gradually changes its color to blue due to the formation of Cu(NO3)2 that stays in solution and that is blue in color.
After the reaction, you may shake the silver crystals off the wire and wash them with distilled water. This is conveniently done by decantation. Decantation is a simple and convenient separation technique, in which a liquid over a solid material at the bottom of a vessel is gently poured off so as not to disturb the solid. This short video shows how decantation is performed. The silver metal produced in the reaction is heavy (dense) and once shaken off the copper wire, will settle at the bottom nicely. Decant off the blue solution, then add distilled water to the silver and gently swirl the mixture. Decant off the washing. Repeat this procedure 4-5 times and then let the wet silver dry at room temperature. You have made pure silver from AgNO3 in this spectacular reaction!
In the silver tree experiment, silver nitrate (AgNO3) reacts with copper metal (Cu) to give silver metal (Ag) and copper nitrate (Cu(NO3)2), according to the following equation:
2 AgNO3 + Cu = Cu(NO3)2 + 2 Ag
This is a typical substitution reaction, in which the silver atom bonded to the NO3 group is displaced by a copper atom. As a result, silver metal and copper nitrate are formed. Crystals of silver metal grow on the surface of the copper as the copper dissolves in the form of copper nitrate, Cu(NO3)2, which gives the solution the blue color.
You may naturally ask, why is the formula of silver nitrate AgNO3, whereas the formula of copper nitrate is Cu(NO3)2, not CuNO3? A simple answer is that a silver atom has the ability to bind to only one NO3 group of atoms, whereas a copper atom can bind to two of these groups. We will discuss this in more detail in the next section.
Experiment 8. Reaction of Silver Nitrate (AgNO3) with Sodium Chloride (NaCl). In the description of the silver tree experiment above (Experiment 7 above), I told you that it is much better to use distilled rather than tap water for making solutions of AgNO3. A solution of silver nitrate in distilled water is clear, whereas in tap water it is cloudy. There are always small quantities of various mineral salts in tap water, including NaCl that reacts with AgNO3 to give insoluble AgCl, which precipitates out (Figure 1-26). Since tap water contains only small quantities of NaCl, only a small amount of AgCl is produced in the form of tiny solid particles, which make the solution cloudy.
Dissolve a pinch of NaCl (table salt) in a small amount of water. Dissolve a pinch of AgNO3 in a small amount of distilled water. Mix the two solutions. Observe the immediate formation of a white precipitate. This precipitate is silver chloride, AgCl, formed in a typical exchange reaction.
Digression. It is often said that AgCl is an insoluble compound. In this course, we consider it insoluble, too. In reality, however, there is no such thing as a totally insoluble substance. Any substance has some finite solubility in any liquid at a certain temperature. To dissolve 1 g of AgCl at room temperature, one would need approximately 500 liters (132 gallons) of water. While this solubility is really low, there are compounds that are trillions of times less soluble than AgCl!
1.8.8. Exercises.
1. Balance equations for the following combination reactions.
(a) Zn + O2 = ZnO
(b) S + O2 = SO2
(c) Al + O2 = Al2O3
(d) BaO + O2 = BaO2
Hint
2. Simplify the following equations by minimizing the coefficients.
(a) 2 Ti + 2 O2 = 2 TiO2
(b) 6 Mg + 3 O2 = 6 MgO
(c) 4 SO2 + 2 O2 = 4 SO3
(d) 14 CO + 7 O2 = 14 CO2
3. Identify the type of each reaction and balance the equations below. What do the arrows written after some of the formulas mean?
(a) HgO = Hg + O2↑
(b) P + O2 = P2O5
(c) CuSO4 + Zn = ZnSO4 + Cu↓
(d) Li + N2 = Li3N
(e) Mg + HCl = MgCl2 + H2↑
(f) KNO3 = KNO2 + O2↑
(g) AgNO3 + KBr = KNO3 + AgBr↓
(h) Zn + Cl2 = ZnCl2
Answer
4. Balance the following equations for substitution reactions.
(a) NaBr + Cl2 = NaCl + Br2
(b) HgCl2 + Cu = CuCl2 + Hg
(c) Al + H2SO4 = Al2(SO4)3 + H2
5. Balance the following equations of decomposition reactions.
(a) HgO = Hg + O2
(b) Ag2O = Ag + O2
(c) H2CO3 = CO2 + H2O
6. Suppose you need 112 tons of CaO to make concrete for your construction project. How much CaCO3 should be decomposed to make the required amount of CaO? Answer
1. Balance equations for the following combination reactions.
(a) Zn + O2 = ZnO
(b) S + O2 = SO2
(c) Al + O2 = Al2O3
(d) BaO + O2 = BaO2
Hint
2. Simplify the following equations by minimizing the coefficients.
(a) 2 Ti + 2 O2 = 2 TiO2
(b) 6 Mg + 3 O2 = 6 MgO
(c) 4 SO2 + 2 O2 = 4 SO3
(d) 14 CO + 7 O2 = 14 CO2
3. Identify the type of each reaction and balance the equations below. What do the arrows written after some of the formulas mean?
(a) HgO = Hg + O2↑
(b) P + O2 = P2O5
(c) CuSO4 + Zn = ZnSO4 + Cu↓
(d) Li + N2 = Li3N
(e) Mg + HCl = MgCl2 + H2↑
(f) KNO3 = KNO2 + O2↑
(g) AgNO3 + KBr = KNO3 + AgBr↓
(h) Zn + Cl2 = ZnCl2
Answer
4. Balance the following equations for substitution reactions.
(a) NaBr + Cl2 = NaCl + Br2
(b) HgCl2 + Cu = CuCl2 + Hg
(c) Al + H2SO4 = Al2(SO4)3 + H2
5. Balance the following equations of decomposition reactions.
(a) HgO = Hg + O2
(b) Ag2O = Ag + O2
(c) H2CO3 = CO2 + H2O
6. Suppose you need 112 tons of CaO to make concrete for your construction project. How much CaCO3 should be decomposed to make the required amount of CaO? Answer