A steel cylinder contains 5.00 mole of graphite (pure carbon) and 5.00 moles of $\mathrm{O}_{2}$. The mixture is ignited and all the graphite reacts. Combustion produces a mixture of $C O$ gas and $\mathrm{CO}_{2}$ gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by $17.0 \%$. Calculate the mole fractions of $\mathrm{CO}, \mathrm{CO}_{2}$, and $\mathrm{O}_{2}$ in the final gaseous mixture.

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∻Solution∻ ‖ 1 ⋮ **Write the balanced chemical equations**: The combustion of carbon (graphite) can produce either carbon monoxide ($CO$) or carbon dioxide ($CO_2$). The two possible reactions are: $C + O_2 ightarrow CO_2$ and $2C + O_2 ightarrow 2CO$. ‖ ‖ 2 ⋮ **Determine the limiting reactant**: Since we have equal moles of $C$ and $O_2$, and the reaction to form $CO_2$ requires a 1:1 mole ratio, while the reaction to form $CO$ requires a 2:1 mole ratio of $C$ to $O_2$, $O_2$ is the limiting reactant for both reactions. ‖ ‖ 3 ⋮ **Calculate the moles of gases after reaction**: If all $O_2$ is used to form $CO_2$, we would have 5 moles of $CO_2$. If all $O_2$ is used to form $CO$, we would have 10 moles of $CO$. Since the pressure increased by 17%, we know that the total moles of gas increased. This means that some $O_2$ must have been used to form $CO$. Let $x$ be the moles of $O_2$ that reacted to form $CO_2$, then $5 - x$ is the moles of $O_2$ that reacted to form $CO$. The total moles of gas after reaction is $x + 2(5 - x) + (5 - x) = 10 + x$. ‖ ‖ 4 ⋮ **Relate the change in moles to change in pressure**: Assuming ideal gas behavior and constant temperature and volume, the pressure is proportional to the number of moles of gas. Let $P_0$ be the original pressure, then the final pressure is $P_0 imes 1.17$. The initial moles of gas is 10 (5 moles of $C$ and 5 moles of $O_2$), so $10 imes 1.17 = 10 + x$. Solving for $x$ gives $x = 1.7$ moles. ‖ ‖ 5 ⋮ **Calculate the mole fractions**: The mole fraction of a component is the number of moles of that component divided by the total number of moles. The mole fraction of $CO_2$ is $\frac{x}{10 + x}$, the mole fraction of $CO$ is $\frac{2(5 - x)}{10 + x}$, and the mole fraction of $O_2$ is $\frac{5 - x}{10 + x}$. Plugging in $x = 1.7$ gives the mole fractions. ‖ ∻Answer∻ ⚹ Mole fraction of $CO_2$: $\frac{1.7}{11.7} \approx 0.145$ ⚹ Mole fraction of $CO$: $\frac{2(5 - 1.7)}{11.7} \approx 0.568$ ⚹ Mole fraction of $O_2$: $\frac{5 - 1.7}{11.7} \approx 0.282$ ⚹ ∻Key Concept∻ ⚹ The mole fraction of a component in a mixture is the ratio of the number of moles of that component to the total number of moles of all components in the mixture. ⚹ ∻Explanation∻ ⚹ The mole fractions were calculated based on the stoichiometry of the combustion reactions and the given increase in pressure, which indicates the total moles of gas after the reaction. By assuming ideal gas behavior, the change in pressure directly relates to the change in the number of moles of gas. ⚹◊What is the balanced chemical equation for the combustion reaction of graphite with oxygen to produce $CO$ and $CO_{2}$ gases?⍭ Generate me a similar question◊

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