A mixture of methane $(\mathrm{CH} 4)$ and ethane $(\mathrm{C} 2 \mathrm{H} 6)$ is stored in a container at $294 \mathrm{~mm} \mathrm{Hg}$. The gases are burned in air to form $\mathrm{CO} 2$ and $\mathrm{H} 2 \mathrm{O}$. If the pressure of $\mathrm{CO} 2$ is $349 \mathrm{~mm}$ $\mathrm{Hg}$ measured at the same temperature and volume as the original mixture, calculate the mole fraction of the gases.

Mole fraction of methane:

Mole fraction of ethane:

Question

A mixture of methane $(\mathrm{CH} 4)$ and ethane $(\mathrm{C} 2 \mathrm{H} 6)$ is stored in a container at $294 \mathrm{~mm} \mathrm{Hg}$. The gases are burned in air to form $\mathrm{CO} 2$ and $\mathrm{H} 2 \mathrm{O}$. If the pressure of $\mathrm{CO} 2$ is $349 \mathrm{~mm}$ $\mathrm{Hg}$ measured at the same temperature and volume as the original mixture, calculate the mole fraction of the gases.

Mole fraction of methane:

Mole fraction of ethane:

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Write the balanced chemical equations for the combustion of methane and ethane**: The combustion of methane and ethane can be represented by the following balanced chemical equations: For methane: $CH_4 + 2O_2 ightarrow CO_2 + 2H_2O$ and for ethane: $C_2H_6 + \frac{7}{2}O_2 ightarrow 2CO_2 + 3H_2O$. ‖ ‖ 2 ⋮ **Use stoichiometry to relate the moles of CO2 produced to the moles of CH4 and C2H6 burned**: From the balanced equations, we see that 1 mole of $CH_4$ produces 1 mole of $CO_2$ and 1 mole of $C_2H_6$ produces 2 moles of $CO_2$. ‖ ‖ 3 ⋮ **Apply Dalton's Law of Partial Pressures**: The total pressure of the gas mixture before combustion is the sum of the partial pressures of methane and ethane. After combustion, the pressure of $CO_2$ alone is given. Since the conditions of temperature and volume are constant, the pressure of $CO_2$ produced is directly proportional to the moles of $CO_2$ produced. ‖ ‖ 4 ⋮ **Set up the equations to solve for the mole fractions**: Let $x$ be the mole fraction of $CH_4$ and $(1-x)$ be the mole fraction of $C_2H_6$. The total moles of $CO_2$ produced is equal to the sum of the moles of $CO_2$ from $CH_4$ and $C_2H_6$. Therefore, $x \cdot 294 ext{ mm Hg} + 2(1-x) \cdot 294 ext{ mm Hg} = 349 ext{ mm Hg}$. ‖ ‖ 5 ⋮ **Solve for x**: Solving the equation from step 4 gives us the mole fraction of methane. $x \cdot 294 + 2(1-x) \cdot 294 = 349$. Simplifying this equation will give us the value of $x$. ‖ ‖ 6 ⋮ **Calculate the mole fraction of ethane**: The mole fraction of ethane is simply $1 - x$. ‖ ‖ 7 ⋮ **Perform the calculations**: Solving the equation from step 5, we get $x \cdot 294 + 588 - 2x \cdot 294 = 349$. This simplifies to $-x \cdot 294 = 349 - 588$, which gives $x = \frac{239}{294}$. Therefore, the mole fraction of methane is $x$ and the mole fraction of ethane is $1 - x$. ‖ ∻Answer∻ ⚹ Mole fraction of methane: $x = \frac{239}{294}$ ⚹ ⚹ Mole fraction of ethane: $1 - x = \frac{55}{294}$ ⚹ ∻Key Concept∻ ⚹ Dalton's Law of Partial Pressures and stoichiometry in chemical reactions are used to calculate mole fractions in a mixture. ⚹ ∻Explanation∻ ⚹ The pressure of CO2 produced after combustion is used to determine the mole fractions of methane and ethane in the original mixture, taking into account the stoichiometry of the combustion reactions. ⚹◊What is the formula to calculate the mole fraction of a gas in a mixture?⍭ Generate me a similar question◊

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