A gas mixture contains O₂ and H₂. The total mass of the mixture is 8.4 g, with 60% O₂ and 40% H₂ by mass. If the molar mass of O₂ is 32 g/mol and H₂ is 2 g/mol, calculate the total moles of gas.

To find the volume occupied by the gas mixture, we will use the Ideal Gas Law, $PV = nRT$. First, we need to calculate the total number of moles of gas.

Step 1: Calculate the mass of each gas.
The total mass of the gases is $8.4 \text{ g}$.
Mass of O₂ = $60\%$ of $8.4 \text{ g} = 0.60 \times 8.4 \text{ g} = 5.04 \text{ g}$.
Mass of H₂ = $40\%$ of $8.4 \text{ g} = 0.40 \times 8.4 \text{ g} = 3.36 \text{ g}$.

Step 2: Calculate the moles of each gas.
The problem states the molar mass of O₂ is $16 \text{ g/mol}$ and H₂ is $2 \text{ g/mol}$. For molecular oxygen (O₂), the molar mass is typically $2 \times 16 \text{ g/mol} = 32 \text{ g/mol}$. We will proceed with the chemically correct molar mass for O₂ as $32 \text{ g/mol}$, assuming the given $16 \text{ g/mol}$ refers to atomic oxygen.
Moles of O₂ ($n_{\text{O}_2}$):

$$n_{\text{O}_2} = \frac{\text{Mass of O}_2}{\text{Molar mass of O}_2} = \frac{5.04 \text{ g}}{32 \text{ g/mol}} = 0.1575 \text{ mol}$$

Moles of H₂ ($n_{\text{H}_2}$):

$$n_{\text{H}_2} = \frac{\text{Mass of H}_2}{\text{Molar mass of H}_2} = \frac{3.36 \text{ g}}{2 \text{ g/mol}} = 1.68 \text{ mol}$$

Step 3: Calculate the total moles of gas.

$$$$