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(B) The combustion reactions are:$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$$C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$Let the volume

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(B) The combustion reactions are:$CH_4(g) + 2O_2(g) ightarrow CO_2(g) + 2H_2O(l)$$C_3H_8(g) + 5O_2(g) ightarrow 3CO_2(g) + 4H_2O(l)$Let the volume of $CH_4$ be $x \ L$ and the volume of $C_3H_8$ be $(5-x) \ L$.According to the stoichiometry of the reactions,the volume of $O_2$ consumed is $2x + 5(5-x) = 16$.$2x + 25 - 5x = 16 \implies 3x = 9 \implies x = 3 \ L$.Thus,volume of $CH_4 = 3 \ L$ and volume of $C_3H_8 = 2 \ L$.At $STP$ $(0 \ ^\circ C, 1 \ atm)$,$22.4 \ L$ corresponds to $1 \ mol$. Therefore,moles of $CH_4 = 3/22.4$ and moles of $C_3H_8 = 2/22.4$.Heat released $= (3/22.4) 	imes 890 + (2/22.4) 	imes 2220 = (2670 + 4440) / 22.4 = 7110 / 22.4 \approx 317.4 \ kJ$.

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