Let’s write down the reaction of butane combustion and arrange the coefficients:
2С4Н10 + 13О2 = 8СО2 + 10Н2О.
It is known that all gases under normal conditions occupy the same volume, called molar – Vm = 22.4 l / mol. In this case, the volume of CO2 will be calculated by the formula: V (CO2) = n (CO2) * Vm, where n (CO2) is the amount of CO2. The reaction equation shows that n (CO2) = 4n (C4H10), where n (C4H10) is the amount of butane substance.
n (C4H10) = m (C4H10) / M (C4H10), where:
m (C4H10) – mass of butane,
M (C4H10) = 58 g / mol is the molar mass of butane.
n (C4H10) = 8/58 = 0.138 mol,
n (CO2) = 0.138 * 4 = 0.552 mol.
Then V (CO2) = 0.552 * 22.4 = 12.36 L
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