We are asked how much heat is emitted

Since the gas are under the same conditions, the volume percent is the same as the mole percent.

Calculate total moles:

$\overline{){\mathbf{PV}}{\mathbf{=}}{\mathbf{}}{\mathbf{nRT}}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{PV}}{\mathbf{RT}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\frac{(751\overline{)\mathrm{mmHg}}\times {\displaystyle \frac{1\overline{)\mathrm{atm}}}{760\overline{)\mathrm{mmHg}}}})(1.53\overline{)L})}{(0.08206{\displaystyle \frac{\overline{)L}-\overline{)\mathrm{atm}}}{\mathrm{mol}-\overline{)K}}})(298\overline{)K})}$

**n = 0.0617 mol**

Calculate moles of each:

$\mathbf{mol}\mathbf{}{\mathbf{CH}}_{\mathbf{4}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{231}(0.0617\mathrm{mol})\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{014}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{CH}}_{\mathbf{4}}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{mol}\mathbf{}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{411}(0.0617\mathrm{mol})\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{025}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathbf{mol}\mathbf{}{\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{8}}\mathbf{}\mathbf{=}(1-0.231-0.411)(0.0617\mathrm{mol})\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{022}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{8}}$

A gaseous fuel mixture contains 23.1% methane (CH_{4}), 41.1% ethane (C_{2}H_{6}) and the rest propane (C_{3}H_{8}) by volume.

When the fuel mixture contained in a 1.53 L tank, stored at 751 mmHg and 298 K, undergoes complete combustion, how much heat is emitted? (Assume that the water produced by the combustion is in the gaseous state.)

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