What is the normal volume of carbon dioxide produced by the combustion of 5.6 liters of butane?

univerkov | July 3, 2021 | education

Reaction equation: 2C4H10 + 13O2 = 8CO2 + 10H2O
Let’s find the amount of butane: n = V / Vm, where Vm = 22.4 l / mol – molar volume, constant for all gases at n. at. n (butane) = 5.6 L / 22.4 L / mol = 0.25 mol.
According to the reaction from 2 mol of butane, we get 8 mol of gas, we will make the proportion 0.25 / 2 = n (CO2) / 8. Hence n (CO2) = 0.25 * 8/2 = 1 mol. Carbon dioxide volume: V = n * Vm = 1 mol * 22.4 l / mol = 22.4 l.

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