What would be the mass of the resulting water if 80 liters of ethylene were burned in oxygen?

In accordance with the condition of the problem, we write down the reaction equation:
С2Н4 + 3О2 = 2СО2 + 2Н2О – the reaction of ethylene combustion, carbon dioxide and water are released;
M (C2H4) = 28 g / mol;
M (H2O) = 18 g / mol;
Let’s calculate the number of moles of ethylene:
1 mol of gas at normal level – 22.4 liters;
X mol (C2H4) – -80 L. hence, X mol (C2H4) = 1 * 80 / 22.4 = 3.57 mol;
Let’s make the proportion:
3.57 mol (C2H4) – X mol (H2O);
-1 mol -2 mol hence, X mol (H2O) = 3.57 * 2/1 = 7.142 mol;
We find the mass of water:
m (H2O) = Y * M = 7.142 * 18 = 128.57 g.
Answer: the mass of water is 128.57 g.