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: i) Charge on metal Q ii) The volume of oxygen gas produce

The current $I = 200 \text{ A}$ and time $t = 58 \text{ minutes}$.
Convert time to seconds:

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

The total charge $Q_{total}$ is given by $Q_{total} = I \times t$.

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

The molar mass of metal Q is given as $27 \text{ g/mol}$.
The mass of metal Q deposited is $64.8 \text{ g}$.

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

According to Faraday's laws of electrolysis, the charge required to deposit one mole of a metal is $n \times F$, where $n$ is the charge on the metal ion and $F$ is Faraday's constant ($96500 \text{ C/mol}$).
The total charge passed is $696000 \text{ C}$ and $2.4 \text{ mol}$ of metal Q was deposited.
So, $2.4 \text{ mol} \times n \times F = Q_{total}$.

$$2.4 \text{ mol} \times n \times 96500 \text{ C/mol} = 696000 \text{ C}$$ $$n = \frac{696000 \text{ C}}{2.4 \text{ mol} \times 96500 \text{ C/mol}} = \frac{696000}{231600} \approx 3.005$$

The charge on metal Q is approximately 3.
Therefore, the metal ion is $\text{Q}^{3+}$.

i) Charge on metal Q:
The charge on metal Q is $\boxed{3}$

In molten oxide electrolysis, the oxide ions ($\text{O}^{2-}$) are oxidized at the anode to produce oxygen gas.

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

From Step 3, we know that $2.4 \text{ mol}$ of $\text{Q}^{3+}$ ions required 3 moles of electrons per mole of Q.

$$\text{Moles of electrons} = 2.4 \text{ mol Q} \times \frac{3 \text{ mol e}^-}{1 \text{ mol Q}} = 7.2 \text{ mol e}^-$$

Alternatively, using the total charge:

$$\text{Moles of electrons} = \frac{Q_{total}}{F} = \frac{696000 \text{ C}}{96500 \text{ C/mol e}^-} \approx 7.21 \text{ mol e}^-$$

We will use $7.2 \text{ mol e}^-$ for consistency with the calculated charge on Q.

From the half-reaction for oxygen production, $4 \text{ mol}$ of electrons produce $1 \text{ mol}$ of $\text{O}_2$.

$$\text{Moles of O}_2 = 7.2 \text{ mol e}^- \times \frac{1 \text{ mol O}_2}{4 \text{ mol e}^-} = 1.8 \text{ mol O}_2$$

At STP, $1 \text{ mol}$ of any gas occupies $22.4 \text{ dm}^3$.

$$\text{Volume of O}_2 = 1.8 \text{ mol} \times 22.4 \text{ dm}^3/\text{mol} = 40.32 \text{ dm}^3$$

ii) The volume of oxygen gas produced at standard temperature and pressure (STP):
The volume of oxygen gas produced is $\boxed{40.32 \text{ dm}^3}$