An excess of sodium was added to 16 g of a mixture of ethyl and propyl alcohols, while 3.36 liters

An excess of sodium was added to 16 g of a mixture of ethyl and propyl alcohols, while 3.36 liters of hydrogen were released. Determine the composition of the initial mixture.

Given:
m mixture = 16 g
V (H2) = 3.36 L
To find:
ω (C2H5OH) -?
ω (C3H7OH) -?
Decision:
1) 2C2H5OH + 2Na => 2C2H5ONa + H2 ↑;
2C3H7OH + 2Na => 2C3H7ONa + H2 ↑;
2) Let m (C2H5OH) = (x) g, m (C3H7OH) = (16 – x) g;
3) n (C2H5OH) = m (C2H5OH) / M (C2H5OH) = (x / 46) mol;
4) n1 (H2) = n (C2H5OH) / 2 = (x / 46) / 2 = (x / 92) mol;
5) n (C3H7OH) = m (C3H7OH) / M (C3H7OH) = ((16 – x) / 60) mol;
6) n2 (H2) = n (C3H7OH) / 2 = ((16 – x) / 60) / 2 = ((16 – x) / 120) mol;
7) n total (H2) = 3.36 / 22.4 = 0.15 mol;
8) (x / 92) + ((16 – x) / 120) = 0.15;
x = 6.571
m (C2H5OH) = x = 6.571 g;
9) ω (C2H5OH) = m (C2H5OH) * 100% / m mixture = 6.571 * 100% / 16 = 41.07%;
10) ω (C3H7OH) = 100% – ω (C2H5OH) = 100% – 41.07% = 58.93%.
Answer: Mass fraction of C2H5OH is 41.07%; C3H7OH 58.93%.