In an electrolysis, a current of 200 A was passed through molten oxide of metal Q for 58 minutes and 64.8 g of the metal deposited. Determine:

First, convert time from minutes to seconds:

$$t = 58 \text{ min} \times 60 \text{ s/min} = 3480 \text{ s}$$

Now, calculate the charge:

$$Q_{total} = I \times t$$ $$Q_{total} = 200 \text{ A} \times 3480 \text{ s} = 696000 \text{ C}$$

$$\text{Moles of Q} = \frac{\text{Mass of Q}}{\text{Molar mass of Q}}$$ $$\text{Moles of Q} = \frac{64.8 \text{ g}}{27 \text{ g/mol}} = 2.4 \text{ mol}$$

$$\text{Moles of electrons} = \frac{Q_{total}}{\text{Faraday constant}}$$ $$\text{Moles of electrons} = \frac{696000 \text{ C}}{96500 \text{ C/mol}} \approx 7.2124 \text{ mol}$$

The deposition of metal Q from its oxide involves the reduction of $\text{Q}^{n+}$ ions:

$$\text{Q}^{n+} + n\text{e}^- \to \text{Q}(s)$$

From this equation, $n$ moles of electrons are required to deposit 1 mole of Q.
Therefore, the ratio of moles of electrons to moles of Q is $n$:

$$n = \frac{\text{Moles of electrons}}{\text{Moles of Q}}$$ $$n = \frac{7.2124 \text{ mol}}{2.4 \text{ mol}} \approx 3.005$$

Rounding to the nearest whole number, the charge on metal Q is 3.

The charge on metal Q is $\boxed{3}$

Since molten oxide of metal Q is electrolyzed, oxide ions ($\text{O}^{2-}$) are oxidized at the anode to produce oxygen gas.

$$2\text{O}^{2-} \to \text{O}_2(g) + 4\text{e}^-$$

This equation shows that 4 moles of electrons are required to produce 1 mole of oxygen gas.

$$\text{Moles of O}_2 = \frac{\text{Moles of electrons}}{4}$$ $$\text{Moles of O}_2 = \frac{7.2124 \text{ mol}}{4} \approx 1.8031 \text{ mol}$$

$$\text{Volume of O}_2 = \text{Moles of O}_2 \times \text{Molar gas volume at STP}$$ $$\text{Volume of O}_2 = 1.8031 \text{ mol} \times 22.4 \text{ dm}^3\text{/mol}$$ $$\text{Volume of O}_2 \approx 40.389 \text{ dm}^3$$

Rounding to three significant figures:

The volume of oxygen gas produced is $\boxed{40.4 \text{ dm}^3}$