What mass of alcohol can be obtained by hydration of ethylene with a volume of 448 liters?

Let’s write the equation:
С2Н4 + Н2О = С2Н5 – ОН – ethylene hydration reaction, ethyl alcohol was obtained;
Let’s determine the molar masses of ethylene, ethyl alcohol:
M (C2H4) = 28 g / mol;
M (C2H5OH) = 46 g / mol;
Using Avogadro’s law, we calculate the number of moles of ethylene:
1 mol – 22.4 liters.
X mol (C2H4) – 448 L. hence, X mol (C2H4) = 448 * 1 / 22.4 = 20 mol;
Let’s calculate the mass of ethyl alcohol by the formula: m (C2H5OH) = Y (C2H5OH) * M (C2H5OH);
m (C2H5OH) = 20 * 46 = 920 g.
Answer: when ethylene was hydrated, ethyl alcohol weighing 920 g was obtained.