9.45 g of liquid hexane (C₆H₁₄) is introduced into a 10.0 L vessel containing 13.15 atm of oxygen gas at 21ºC and ignited, yielding carbon dioxide and water. If the vessel is then cooled to -10.ºC, what will be the gas pressure inside the vessel A) 1.42 atm B) 3.09 atm C) 10.9 atm D) 12.6 atm E) 13.15 atm

Question

Quizplus · Accepted Answer

First, determine the moles of hexane (C6H14) using its molar mass (86.18 g/mol): $9.45 \, \text{g} \times \frac{1 \, \text{mol}}{86.18 \, \text{g}} = 0.1096 \, \text{mol}$. The combustion reaction is $2C_6H_{14} + 19O_2 \rightarrow 12CO_2 + 14H_2O$. From the stoichiometry, 0.1096 mol of hexane will consume $0.1096 \, \text{mol} \times \frac{19}{2} = 1.042 \, \text{mol}$ of $O_2$ and produce $0.1096 \, \text{mol} \times 6 = 0.6576 \, \text{mol}$ of $CO_2$ and $0.1096 \, \text{mol} \times 7 = 0.7672 \, \text{mol}$ of $H_2O$. Initially, there are $13.15 \, \text{atm} \times \frac{10.0 \, \text{L}}{0.0821 \, \text{L atm/mol K} \times 294 \, \text{K}} = 5.45 \, \text{mol}$ of $O_2$. After reaction, $5.45 \, \text{mol} - 1.042 \, \text{mol} = 4.408 \, \text{mol}$ of $O_2$ remain. Total moles of gas after reaction: $4.408 + 0.6576 + 0.7672 = 5.8328 \, \text{mol}$. Using PV=nRT for final conditions (263 K), $P = \frac{nRT}{V} = \frac{5.8328 \, \text{mol} \times 0.0821 \, \text{L atm/mol K} \times 263 \, \text{K}}{10.0 \, \text{L}} = 12.6 \, \text{atm}$.

945 G of Liquid Hexane (C6H14) Is Introduced into a 10.0

945 G of Liquid Hexane (C₆H₁₄) Is Introduced into a 10.0