A $10.0 ext{ L}$ gaseous mixture of ethane ($\mathrm{C}_2\mathrm{H}_6$) and butane ($\mathrm{C}_4\

Question

A $10.0 	ext{ L}$ gaseous mixture of ethane ($\mathrm{C}_2\mathrm{H}_6$) and butane ($\mathrm{C}_4\mathrm{H}_{10}$) is completely combusted at $0^{\circ} 	ext{C}$ and $1 	ext{ atm}$ pressure, consuming $45.0 	ext{ L}$ of oxygen gas measured under identical conditions. Assuming the standard enthalpies of combustion are $\Delta H_{	ext{comb}}(\mathrm{C}_2\mathrm{H}_6) = -1560 	ext{ kJ mol}^{-1}$ and $\Delta H_{	ext{comb}}(\mathrm{C}_4\mathrm{H}_{10}) = -2880 	ext{ kJ mol}^{-1}$, the total net heat energy liberated from this combustion process is roughly

Professor PC Thomas Classes · Answer

1. Combustion Equations:
$	ext{C}_2	ext{H}_6 + 3.5	ext{O}_2 ightarrow 2	ext{CO}_2 + 3	ext{H}_2	ext{O}$
$	ext{C}_4	ext{H}_{10} + 6.5	ext{O}_2 ightarrow 4	ext{CO}_2 + 5	ext{H}_2	ext{O}$

2. Volume Calculation (Avogadro's Law):
Let $v = 	ext{volume of } 	ext{C}_2	ext{H}_6$. Therefore, $	ext{volume of } 	ext{C}_4	ext{H}_{10} = 10.0 - v$.
$3.5v + 6.5(10.0 - v) = 45.0 	ext{ L}$
$65.0 - 3.0v = 45.0 \implies 3.0v = 20.0$

$v_{	ext{ethane}} = \frac{20}{3} 	ext{ L}$
$v_{	ext{butane}} = 10.0 - \frac{20}{3} = \frac{10}{3} 	ext{ L}$

3. Moles at STP ($0^\circ	ext{C}, 1 	ext{ atm} \approx 22.4 	ext{ L/mol}$):

$n_{	ext{ethane}} = \frac{20/3}{22.4} \approx 0.2976 	ext{ mol}$
$n_{	ext{butane}} = \frac{10/3}{22.4} \approx 0.1488 	ext{ mol}$

4. Total Heat Liberated:
$q = (n_{	ext{ethane}} 	imes 1560) + (n_{	ext{butane}} 	imes 2880)$
$q = (0.2976 	imes 1560) + (0.1488 	imes 2880)$
$q = 464.26 + 428.54 = \mathbf{892.8 	ext{ kJ}}$ (Rounds to $893 	ext{ kJ}$)