Hydration of ethyl alcohol weighing 36.8 g according to Lebedev’s method yielded butadiene 1.3, V = 5.6 l.

Hydration of ethyl alcohol weighing 36.8 g according to Lebedev’s method yielded butadiene 1.3, V = 5.6 l. What is the mass fraction of the product yield?

Given:
m (C2H5OH) = 36.8 g
V practical. (C4H6) = 5.6 L

To find:
η (C4H6) -?

Decision:
1) 2C2H5OH => C4H6 + H2O + H2;
2) M (C2H5OH) = Mr (C2H5OH) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (O) * N (O) + Ar (H) * N (H ) = 12 * 2 + 1 * 5 + 16 * 1 + 1 * 1 = 46 g / mol;
3) n (C2H5OH) = m (C2H5OH) / M (C2H5OH) = 36.8 / 46 = 0.8 mol;
4) n theory. (C4H6) = n (C2H5OH) / 2 = 0.8 / 2 = 0.4 mol;
5) V theor. (C4H6) = n theory. (C4H6) * Vm = 0.4 * 22.4 = 8.96 L;
6) η (C4H6) = V practical. (C4H6) * 100% / V theor. (C4H6) = 5.6 * 100% / 8.96 = 62.5%.

Answer: The product yield is 62.5%.