4.7.6-4. Mass percentage concentration is the gram amount of a solute in 100 g of solution. Therefore, to solve the problem, we need to know the total mass of the solution after the reaction and the amount of sodium ethoxide produced in the reaction. The equation is: 2 C2H5OH + 2 Na = 2 C2H5ONa + H2. The hydrogen gas produced bubbles off. For the reaction, we have 1 g of Na and 100 mL of ethanol. The amount of ethanol in grams is 100 mL x 0.79 g/mL = 79 g. The total mass of the solution after the reaction is therefore 79 g (ethanol) plus 1 g (sodium) minus the mass of the H2 gas that was formed and left the reaction mixture. This follows from the law of conservation of matter.
We need to determine which of the two reactants, ethanol or sodium, is the limiting reagent. The atomic weight of sodium is 23 and the molecular weight of ethanol is 46, twice as much. Therefore, 2 g of ethanol are needed for the reaction with 1 g of sodium metal. We have 79 g of ethanol, vastly more than 2 g that is needed for the reaction. Hence, sodium is our limiting reagent.
It is clear from the equation that 2 moles of Na (2 x 23 g = 46 g) react with ethanol to give 2 moles of sodium ethoxide (2 x 68 g = 136 g) and 1 mole of H2 (2 g). Solving the proportion for x, the amount of sodium ethoxide formed from 1 g of Na,
46 g Na --- gives 136 g C2H5ONa
1 g Na --- gives x g C2H5ONa
we get: x = (136 x 1)/46 = 2.96 g. This is the amount of sodium ethoxide produced.
Solving the proportion for y, the amount of H2 formed from 1 g of Na,
46 g Na --- gives 2 g H2
1 g Na --- gives y H2
we get: y = (2 x 1)/46 = 0.04 g. This is the amount of H2 produced.
The total weight of the solution after the reaction has gone to completion and the hydrogen gas produced has escaped, is 79 g (ethanol) + 1 g (Na) - 0.04 g (H2 formed and lost) = 79.96 g.
Knowing the weight of the solution (79.96 g) and the amount of sodium ethoxide in it (2.96 g), we determine the percentage concentration:
79.96 g of solution ------ contains 2.96 g of sodium ethoxide
100 g of such solution ------ would contain z g of sodium ethoxide;
z = (100 x 2.96)/79.96 = 3.7 g. Thus the concentration of C2H5ONa in the solution produced is 3.7%.