30.9 g of a mixture of acetic acid, ethyl alcohol and acetaldehyde were sequentially treated with an aqueous

30.9 g of a mixture of acetic acid, ethyl alcohol and acetaldehyde were sequentially treated with an aqueous solution of sodium bicarbonate, while 11.2 liters of gas were evolved. Sequential treatment with an ammonia solution of silver oxide formed 2.16 g of a precipitate. Determine the content (in% by weight) of ethyl alcohol and acetaldehyde in the initial mixture.

Given:
m mixture = 30.9 g
V (gas) = 11.2 l
m (sediment) = 2.16 g

To find:
ω (C2H5OH) -?
ω (CH3CHO) -?

Decision:
1) CH3COOH + NaHCO3 => CH3COONa + H2O + CO2 ↑;
CH3CHO + Ag2O => CH3COOH + 2Ag ↓;
2) n (CO2) = V (CO2) / Vm = 11.2 / 22.4 = 0.5 mol;
3) n (CH3COOH) = n (CO2) = 0.5 mol;
4) m (CH3COOH) = n (CH3COOH) * M (CH3COOH) = 0.5 * 60 = 30 g;
5) n (Ag) = m (Ag) / M (Ag) = 2.16 / 108 = 0.02 mol;
6) n (CH3CHO) = n (Ag) / 2 = 0.02 / 2 = 0.01 mol;
7) m (CH3CHO) = n (CH3CHO) * M (CH3CHO) = 0.01 * 44 = 0.44 g;
8) m (C2H5OH) = m mixture – m (CH3COOH) – m (CH3CHO) = 30.9 – 30 – 0.44 = 0.46 g;
9) ω (C2H5OH) = m (C2H5OH) * 100% / m mixture = 0.46 * 100% / 30.9 = 1.49%;
10) ω (CH3CHO) = m (CH3CHO) * 100% / m mixture = 0.44 * 100% / 30.9 = 1.42%.

Answer: The mass fraction of C2H5OH is 1.49%; CH3CHO 1.42%.