A mixture of carbon dioxide and oxygen has volume 8310 cm ^3 ,...

Using the ideal gas equation, the total number of moles of the mixture is given by:

$n = \dfrac{PV}{RT}$

Substituting the given values:
$P = 100 \text{ kPa} = 10^5 \text{ Pa}$
$V = 8310 \text{ cm}^3 = 8.31 \times 10^{-3} \text{ m}^3$
$T = 300 \text{ K}$
$R = 8.31 \text{ J/(mol}\cdot\text{K)}$

$n = \dfrac{10^5 \times 8.31 \times 10^{-3}}{8.31 \times 300} = \dfrac{100}{300} = \dfrac{1}{3} \text{ mol}$

Let $n_1$ and $n_2$ be the number of moles of $CO_2$ and $O_2$ respectively.

$n_1 + n_2 = \dfrac{1}{3}$

The total mass of the mixture is $13.2 \text{ g}$. Using the molar masses of $CO_2$ ($44 \text{ g/mol}$)$and O_2$ ($32 \text{ g/mol}$):

$44n_1 + 32n_2 = 13.2$

Substituting $n_2 = \dfrac{1}{3} - n_1$ into the mass equation:

$44n_1 + 32\left(\dfrac{1}{3} - n_1\right) = 13.2$

$12n_1 + \dfrac{32}{3} = 13.2$

$12n_1 = \dfrac{39.6 - 32}{3} = \dfrac{7.6}{3}$

$n_1 = \dfrac{7.6}{36} = \dfrac{19}{90} \approx 0.21 \text{ mol}$

The number of moles of $O_2$ is:

$n_2 = \dfrac{1}{3} - \dfrac{19}{90} = \dfrac{11}{90} \approx 0.12 \text{ mol}$

The number of moles of $CO_2$ and $O_2$ are $0.21$ and $0.12$ respectively.

Answer: $0.21$ and $0.12$