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.
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