1.15.4-1. NaOH is a strong base. Therefore, it will react with acids and acidic and amphoteric oxides and hydroxides.

(a) NaOH + HCl = NaCl + H2O. HCl is a strong acid.

(b) No reaction with other bases, including Ca(OH)2.

(c) No reaction with basic oxides, including MgO.

(d) 2 NaOH + CO2 = Na2CO3 + H2O (NaOH in excess); NaOH + CO2 = NaHCO3 (CO2 in excess). CO2 is an acidic oxide.

(e) 2 NaOH + Zn(OH)2 = Na2ZnO2 + 2 H2O. Zn(OH)2 is amphoteric.

(f) 2 NaOH + H2SiO3 = Na2SiO3 + 2 H2O. H2SiO3 is an acid.










1.15.4-2. SO3 is an acidic oxide. Therefore, it will react with bases and basic species.

(a) No reaction. HCl is an acid.

(b) SO3 + Mg(OH)2 = MgSO4 + H2O. Mg(OH)2 is a base.

(c) SO3 + CaO = CaSO4. CaO is a basic oxide.

(d) No reaction. CO2 is an acidic oxide.

(e) SO3 + Ag2O = Ag2SO4. Ag2O is a basic oxide.










1.15.4-3. (a), (b), (d), (e), (g), (h), (i), (k), (l), (m), (n), (o), and (p).










1.16.4-1. False. Organic acids are, whereas mineral acids are not.










1.16.4-2. False. The reaction of H2SO4 with NaCl is a laboratory method for making HCl in small quantities. In industry, HCl is obtained as a co-product of chlorination reactions of some organic compounds.










1.16.4-3.
(a) H2SO4 + Mg = MgSO4 + H2
(b) SO3 + H2O = H2SO4
(c) SO3 + 2 KOH = K2SO4
(d) CaCO3 + 2 HCl = CaCl2 + H2O + CO2
(e) H3PO4 + 3 NaOH = Na3PO4 + 3 H2O
(f) 2 HNO3 + Cu(OH)2 = Cu(NO3)2 + 2 H2O
(g) H2SO4 + BaCl2 = BaSO4↓ + 2 HCl
(h) P2O5 + 3 H2O = 2 H3PO4










1.17.4-1. False. NaOH and KOH are too reactive toward many substances to exist in nature, especially pure. Furthermore, NaOH and KOH are hygroscopic and deliquescent and, consequently, remain solid in humid air only for a very short period of time (see 1.12.9 and 1.12.13, problem 8).










1.17.4-2. True.










1.17.4-3.

(a) NaOH + HNO3 = NaNO3 + H2O
(b) 2 KOH + CuCl2 = Cu(OH)2↓ + 2 KCl
(c) Ca(OH)2 + CO2 = CaCO3↓ + H2O
(d) Fe(OH)3 + 3 HCl = FeCl3 + 3 H2O
(e) AlCl3 + 3 NaOH = Al(OH)3↓ + 3 NaCl
(f) AlCl3 + 4 NaOH = NaAlO2 + 3 NaCl + 2 H2O
(g) Cu(OH)2 = CuO + H2O
(h) Zn(OH)2 + 2 KOH = K2ZnO2 + 2 H2O
(i) MgCl2 + 2 NaOH = Mg(OH)2↓ + 2 NaCl
(j) Na2O + H2O = 2 NaOH










1.18.3-1. (b) because Hg is more reactive than Ag.










1.18.3-2.
(a) 2 K + 2 HCl = 2 KCl + H2
(b) Ag2SO4 + BaCl2 = BaSO4↓ + 2 AgCl↓
(c) Zn(OH)2 + 2 HCl = ZnCl2 + 2 H2O
(d) Cu(NO3)2 + Ca(OH)2 = Ca(NO3)2 + Cu(OH)2
(e) CuCl2 + Zn = ZnCl2 + Cu
(f) Cu + HCl = no reaction
(g) 2 Al(OH)3 + 3 H2SO4 = Al2(SO4)3 + 6 H2O
(h) Al(OH)3 + KOH = KAlO2 + 2 H2O
(i) Al2O3 + 2 KOH = 2 KAlO2 + H2O
(j) CaBr2 + Na2CO3 = 2 NaBr + CaCO3
(k) SiO2 + Na2O = Na2SiO3
(l) K2CO3 + 2 HCl = 2 KCl + H2O + CO2
(m) Zn + 2 HCl = ZnCl2 + H2
(n) Fe + H2SO4 = FeSO4 + H2










1.19.1-1. hydrogen (simple substance); oxygen (simple substance); copper (II) oxide (oxide); copper (I) oxide (oxide); potassium oxide (oxide); aluminum oxide (oxide); zinc oxide (oxide); barium hydroxide (hydroxide; base); iron (II) hydroxide (hydroxide; base); iron (III) hydroxide (hydroxide; base); sodium carbonate (salt); magnesium sulfate (salt); calcium chloride (salt); phosphoric acid (acid); hydrogen fluoride or hydrofluoric acid (acid); ozone (simple substance); potassium hydrosulfate or potassium bisulfate (hydro salt); strontium nitrate (salt); tin (II) chloride (salt); aluminum sulfate (salt).










1.19.1-2. H2O; HCl; CuO; Cu2O; CuCl; CuCl2; CuCl; CuCl2; H2SO4; H3PO4; HNO3; SnCl2; FeCl3; FeCl2; HBr; KOH; NaHCO3.










1.19.1-3. No. Allotropes are different simple substances formed by the same element.










1.19.1-4. False. See 1.2.2.










1.19.1-5. Amalgams. Homogeneous.










1.19.1-6. False. Carbon, phosphorus, sulfur, iodine, and some other nonmetals are solids.










1.19.1-7. Heterogeneous.










1.19.1-8. NaHCO3 + HCl = NaCl + H2O + CO2










1.19.1-9. There are many ways to solve problems like this one. I prefer to follow the simplest logic. What is mass percentage concentration? By definition, it is the number of grams of a solute in 100 g of solution. As the concentration of HCl is 36%, 100 g of that solution contains 36 g of pure HCl. The rest is 64 g of water: 100 g - 36 g = 64 g.

We need a 20% solution of HCl, which means that 100 g of the desired solution should contain 20 g of pure HCl, with the rest being water (80 g). However, the amount of HCl that we have is not 20 g but 36 g. How much water (x) do we need to add to 36 g of HCl to prepare a 20% solution? Let us solve the simple proportion:

For 20 g HCl – 80 g H2O is needed
For 36 g HCl – x g H2O is needed

x = (80 x 36)/20 = 144 g. This is the total amount of water that is needed to prepare 20% hydrochloric acid from 36 g of pure HCl. However, the solution in the bottle already contains 64 g of water. Therefore, we need to add to this solution 144 – 64 = 80 g of water.

There is another way to solve this problem. Our unknown, x, is the amount of water in grams that we need to dilute our given solution with in order to lower its concentration from 36% to 20%. The desired solution should contain 20 g of pure HCl per each 100 g of the solution. The proportion to be solved for x is:

100 g of desired solution contains – 20 g HCl
(100 g + x g) of desired solution contains – 36 g HCl

100 x 36 = (100 + x) x 20

3600 = 2000 + 20x

20x = 3600 – 2000 = 1600

x = 1600/20 = 80 g

The result is the same: we need to add 80 g of water to lower the original concentration of 36% to 20%.










1.19.1-10. False. Mixtures of gases are always homogeneous.










1.19.1-11. First, heat CaCO3 to decompose it to CaO and CO2: CaCO3 = CaO + CO2. Then treat the CaO produced with water: CaO + H2O = Ca(OH)2.










1.19.1-12. Reaction (a) will occur and reaction (b) will not. See the metal reactivity series (Figure 1-88).










1.19.1-13. First, we need to calculate how much KNO3 and water the starting solution contains. The mass percentage concentration of this solution is 10%, which means that 100 g of an identical solution would contain 90 g of water and 10 g of KNO3. To calculate the quantities of the salt and water in 50 g of our solution, we solve the proportion below, where x is the quantity of KNO3.

100 g of solution contains - 10 g KNO3
50 g of solution contains - x g KNO3

x = (10 x 50)/100 = 5 g

Therefore, 50 g of 10% solution of KNO3 contains 5 g of pure KNO3. The rest is water, 50 – 5 = 45 g.

Our unknown y is the needed quantity of solid KNO3 to be added. The total mass of the desired resultant 20% solution will therefore be (50 + y) g, of which (5 + y) g is KNO3. As the needed concentration is 20%, 100 g of such solution should contain 20 g of KNO3.

100 g of solution contains - 20 g KNO3
(50 + y) g of solution contains - (5 + y) g KNO3

100 x (5 + y) = 20 x (50 + y)

Solving this equation for y, we calculate y = 6.25 g.










1.19.1-14. Add water to dissolve NaCl and sugar. Filter off the insoluble SiO2 and BaSO4. Partially evaporate the filtrate until less soluble NaCl crystallizes out. Separate the crystals of NaCl from the mother liquor containing the dissolved sugar and remaining NaCl.










1.19.1-15. Chemical equations for all of these reactions can be found in Volume 1, except reaction (m): copper metal does not react with acids, see 1.15.3.










1.19.1-16. (b).










1.19.1-17. (a).










1.19.1-18. False. Although this reaction does produce H2, it is too violent to be used safely.










1.19.1-19. False. See 1.12.11.










1.19.1-20. Oxygen gas cannot be called a compound because it is a simple substance, a substance made up of atoms of only one element. Compounds are complex substances whose molecules are composed of atoms of more than one element.