The enthalpy of combustion of butane is -2877 kJ/mol:
$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4C O_{2} (g) + 5 H_{2} O (l)$ What volume of CO₂ at STP (22.4 L/mol) is produced along with the release of 1438.5 kJ of heat?

Question

The enthalpy of combustion of butane is -2877 kJ/mol:

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4C O_{2} (g) + 5 H_{2} O (l)$ What volume of CO₂ at STP (22.4 L/mol) is produced along with the release of 1438.5 kJ of heat?

TestHub · Accepted Answer

Moles of butane burnt = 1438.5/2877 = 0.5 mol. Each mole of butane gives 4 mol CO 2 , so CO 2 = 2 mol. Volume = 2 x 22.4 = 44.8 L. Answer: (B).

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