Determine Moles of Carbon Dioxide and Oxygen Mixture

To find the number of moles of carbon dioxide ($\text{CO}_2$) and oxygen ($\text{O}_2$) in the mixture, we use the ideal gas law and the total mass of the mixture.

Step 1: Calculate the total number of moles in the mixture

The ideal gas equation for the mixture is: $PV = nRT$

Given:

• Pressure, $P = 100\text{ kPa} = 10^5\text{ Pa}$
• Volume, $V = 8310\text{ cm}^3 = 8310 \times 10^{-6}\text{ m}^3 = 8.31 \times 10^{-3}\text{ m}^3$
• Temperature, $T = 300\text{ K}$
• Gas constant, $R = 8.31\text{ J/mol}\cdot\text{K}$

Substituting the given values to calculate total moles ($n$): $n = \frac{PV}{RT} = \frac{10^5 \times 8.31 \times 10^{-3}}{8.31 \times 300}$

$n = \frac{831}{8.31 \times 300} = \frac{100}{300} = \frac{1}{3}\text{ moles}$

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Step 2: Set up equations for individual moles

Let:

• $n_1$ be the number of moles of $\text{CO}_2$ (Molar mass $M_1 = 44\text{ g/mol}$)
• $n_2$ be the number of moles of $\text{O}_2$ (Molar mass $M_2 = 32\text{ g/mol}$)

1. Total Moles Equation: $n_1 + n_2 = \frac{1}{3} \quad \implies \quad n_2 = \frac{1}{3} - n_1 \quad \text{--- (Equation 1)}$

2. Total Mass Equation: The total mass of the mixture is $13.2\text{ g}$: $M_1 n_1 + M_2 n_2 = 13.2$ $44 n_1 + 32 n_2 = 13.2 \quad \text{--- (Equation 2)}$

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Step 3: Solve for $n_1$ and $n_2$

Substitute Equation 1 into Equation 2: $44 n_1 + 32 \left(\frac{1}{3} - n_1\right) = 13.2$

$44 n_1 + \frac{32}{3} - 32 n_1 = 13.2$

$12 n_1 = 13.2 - \frac{32}{3}$

$12 n_1 = \frac{39.6 - 32}{3}$

$12 n_1 = \frac{7.6}{3}$

$n_1 = \frac{7.6}{36} = \frac{1.9}{9} \approx 0.211\text{ moles}$

Now, find $n_2$: $n_2 = \frac{1}{3} - \frac{1.9}{9} = \frac{3 - 1.9}{9} = \frac{1.1}{9} \approx 0.122\text{ moles}$

Rounding to two decimal places:

• Number of moles of $\text{CO}_2$ ($n_1$) $\approx 0.21$
• Number of moles of $\text{O}_2$ ($n_2$) $\approx 0.12$

Thus, the correct option is C ($0.21\text{ and } 0.12$).