Calculate the volume of O2 required for complete oxidation of 12L ethene.
July 25, 2021 | education
| Let’s implement the solution:
According to the condition of the problem, we write the equation of the process:
С2Н4 + 3О2 = 2СО2 + 2Н2О + Q – combustion, heat, carbon dioxide, water is released;
Calculations by formulas:
M (C2H4) = 28 g / mol;
M (O2) = 32 g / mol.
Proportions:
1 mol of gas at normal level – 22.4 liters;
X mol (C2H4) – 12 liters from here, X mol (C2H4) = 1 * 12 / 22.4 = 0.5 mol.
0.5 mol (C2H4) – X mol (O2);
-1 mol -3 mol from here, X mol (O2) = 0.5 * 3/1 = 1.5 mol.
We find the volume of O2:
V (O2) = 1.5 * 22.4 = 33.6 L
Answer: the volume of oxygen is 33.6 liters
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