What ethylene hydrocarbon and in what volume should be taken in order to obtain from it by catalytic

What ethylene hydrocarbon and in what volume should be taken in order to obtain from it by catalytic hydration 28.75 ml of monohydric alcohol (p = 0.804 g / cm3, which is 0.5 mol? The yield of alcohol in this reaction is 50% of theoretical.

Given:
HC – ethylene hydrocarbon
V (alcohol) = 28.75 ml
ρ solution (alcohol) = 0.804 g / cm ^ 3 = 0.804 g / ml
n (alcohol) = 0.5 mol
φ out. = 50%

To find:
UV -?
V (UV) -?

Decision:
1) Ethylene hydrocarbons have the general formula:
C (n) H (2n)
2) C (n) H (2n) + H2O => C (n) H (2n + 1) OH;
3) m solution (alcohol) = ρ solution (alcohol) * V solution (alcohol) = 0.804 * 28.75 = 23.115 g;
4) M (alcohol) = m (alcohol) / n (alcohol) = 23.115 / 0.5 = 46.23 ≈ 46 g / mol;
5) M (C (n) H (2n + 1) OH) = Mr (C (n) H (2n + 1) OH) = Ar (C) * n + Ar (H) * (2n + 1) + Ar (O) + Ar (H) = 12 * n + 1 * (2n + 1) + 16 + 1 = (14n + 18) g / mol;
6) 14n + 18 = 46;
14n = 28;
n = 2;
ethylene hydrocarbon – C2H4 – ethene;
monohydric alcohol – C2H5OH – ethyl alcohol;
7) n theory. (C2H4) = n (C2H5OH) = 0.5 mol;
8) V theor. (C2H4) = n theory. (C2H4) * Vm = 0.5 * 22.4 = 11.2 L;
9) V practical. (C2H4) = V theor. (C2H4) * 100% / φ out. = 11.2 * 100% / 50% = 22.4 liters.

Answer: Unknown ethylene hydrocarbon – C2H4 – ethene; the practical volume of C2H4 is 22.4 liters.