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青州's Question
Posted 3 months ago

A mixture of methane (CH4)(\mathrm{CH} 4) and ethane (C2H6)(\mathrm{C} 2 \mathrm{H} 6) is stored in a container at 294 mmHg294 \mathrm{~mm} \mathrm{Hg}. The gases are burned in air to form CO2\mathrm{CO} 2 and H2O\mathrm{H} 2 \mathrm{O}. If the pressure of CO2\mathrm{CO} 2 is 349 mm349 \mathrm{~mm} Hg\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:
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Answer from Sia
Posted 3 months ago
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: CH4+2O2CO2+2H2OCH_4 + 2O_2 \rightarrow CO_2 + 2H_2O and for ethane: C2H6+72O22CO2+3H2OC_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O
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 CH4CH_4 produces 1 mole of CO2CO_2 and 1 mole of C2H6C_2H_6 produces 2 moles of CO2CO_2
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 CO2CO_2 alone is given. Since the conditions of temperature and volume are constant, the pressure of CO2CO_2 produced is directly proportional to the moles of CO2CO_2 produced
Set up the equations to solve for the mole fractions: Let xx be the mole fraction of CH4CH_4 and (1x)(1-x) be the mole fraction of C2H6C_2H_6. The total moles of CO2CO_2 produced is equal to the sum of the moles of CO2CO_2 from CH4CH_4 and C2H6C_2H_6. Therefore, x294 mm Hg+2(1x)294 mm Hg=349 mm Hgx \cdot 294 \text{ mm Hg} + 2(1-x) \cdot 294 \text{ mm Hg} = 349 \text{ mm Hg}
Solve for x: Solving the equation from step 4 gives us the mole fraction of methane. x294+2(1x)294=349x \cdot 294 + 2(1-x) \cdot 294 = 349. Simplifying this equation will give us the value of xx
Calculate the mole fraction of ethane: The mole fraction of ethane is simply 1x1 - x
Perform the calculations: Solving the equation from step 5, we get x294+5882x294=349x \cdot 294 + 588 - 2x \cdot 294 = 349. This simplifies to x294=349588-x \cdot 294 = 349 - 588, which gives x=239294x = \frac{239}{294}. Therefore, the mole fraction of methane is xx and the mole fraction of ethane is 1x1 - x
Mole fraction of methane: x=239294x = \frac{239}{294}
Mole fraction of ethane: 1x=552941 - 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.
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.

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