A mixture of carbon dioxide and oxygen has volume \(8310\) cm\(^3\), temperature \(300\) K, pressure \(100\) kPa and mass \(13.2\) g. The number of moles of carbon dioxide and oxygen gases in the mixture respectively are _______.
(Assume both carbon dioxide and oxygen gases behave like ideal gases) \([R = 8.31\) J/mol.K\(]\)

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 \(0.12\)