What volume of ethylene is required for the synthesis of ethyl alcohol weighing 115 g?

We compose the reaction equation:
Н2С = СН2 + Н2О = С2Н5 – ОН – the reaction of ethylene hydration, ethyl alcohol was obtained;
Let’s make calculations: M (C2H5OH) = 12 * 2 = 6 = 16 = 46 g / mol;
Let us determine the number of moles of C2H5OH, if the mass is known: m (C2H5OH) = 115 g;
Y (C2H5OH) = m (C2H5OH) / M (C2H5OH); Y (C2H5OH) = 115/46 = 2.5 mol;
We make the proportion based on the data of the reaction equation:
X mol (C2H4) – 2.5 mol (C2H5OH);
-1 mol – 1 mol hence, X mol (C2H4) = 2.5 mol;
We calculate the volume of ethylene according to Avogadro’s law:
V (C2H4) = 2.5 * 22.4 = 56 l.
Answer: for the synthesis of ethyl alcohol, ethylene with a volume of 56 liters is required.