Question

Sol. [Key Concept: Use $PV = nRT$ for total moles, then solve $n_{1} + n_{2}$ and $44n_{1} + 32n_{2} =$ mass.]

First, we determine the total number of moles ( $n_{\text{total}\text{~}}$ ) in the mixture using the Ideal Gas Law: PV = nRT. Given the values:

Pressure (P): $100kPa = 10^{5}\text{ }Pa$

Volume (V): $8310\text{ }cm3 = 8.31 \times 10^{- 3}{\text{ }m}^{3}$

Temperature (T): 300 K

Gas Constant (R): $8.31\text{ }J/mol \cdot K$

$$\frac{PV}{RT} = \frac{10^{5} \times 8.31 \times 10^{- 3}}{8.31 \times 300} = \frac{100}{300} \approx 0.333\text{ }mole$$

Mole equation: $n_{1} + n_{2} = 0.333$

Mass equation: $44n_{1} + 32n_{2} = 13.2$ (where 44 and 32 are the molar masses of ${CO}_{2}$ and $O_{2}$ respectively)

Solve for individual moles

Using the first equation to express $n_{2}$ as $0.333 - n_{1}$, we substitute this into the mass equation:

$$\begin{matrix} & 44n_{1} + 32\left( 0.333 - n_{1} \right) = 13.2 \\ & 44n_{1} + 10.667 - 32n_{1} = 13.2 \\ & 12n_{1} = 2.533 \\ & n_{1} \approx 0.211\text{ }mol \end{matrix}$$

Then, calculate $n_{2}$ :

$$n_{2} = 0.333 - 0.211 = 0.122\text{ }mol$$