To solve the problem, it is necessary to compose the reaction equation:
С2Н5ОН + 3О2 = 2СО2 + 3Н2О – the ethanol combustion reaction is accompanied by the release of carbon monoxide (4) and water;
M (O2) = 32 g / mol; M (CO2) = 44 g / mol;
Determine the amount of moles of carbon monoxide, oxygen:
1 mol of gas at n. y – 22.4 l;
X mol (CO2) – 0.3 liters from here, X mol (CO2) = 1 * 0.3 / 22.4 = 0.013 mol;
0.013 mol (CO2) – X mol (O2);
-2 mol -3 mol from here, X mol (O2) = 0.013 * 3/2 = 0.019 mol;
Let’s calculate the volume of oxygen:
Y (O2) = 0.019 * 22.4 = 0.425 l.
Answer: the reaction requires oxygen with a volume of 0.425 liters.
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