Calculate the volume in dm³ of hydrogen produced at s.t.p. during the electrolysis of dilute tetraoxosulphate (VI) acid, if 1.0 A was used for 15 minutes. [1F = 96,500 C, Volume of gas at s.t.p. = 22.4 dm³]

First, convert the time from minutes to seconds.

$$t = 15 \text{ min} \times 60 \text{ s/min} = 900 \text{ s}$$

Now, calculate the charge using the formula $Q = I \times t$.

$$Q = 1.0 \text{ A} \times 900 \text{ s} = 900 \text{ C}$$

During the electrolysis of dilute tetraoxosulphate (VI) acid ($H_2SO_4$), hydrogen ions ($H^+$) are reduced at the cathode.

$$2H^+(aq) + 2e^- \to H_2(g)$$

This reaction shows that 2 moles of electrons are required to produce 1 mole of hydrogen gas.

Use Faraday's constant ($F = 96,500 \text{ C/mol}$) to find the moles of electrons ($n_e$).

$$n_e = \frac{Q}{F} = \frac{900 \text{ C}}{96,500 \text{ C/mol}}$$

From the cathode reaction, 2 moles of electrons produce 1 mole of $H_2$.

$$\text{Moles of } H_2 = \frac{1}{2} \times n_e = \frac{1}{2} \times \frac{900}{96,500} \text{ mol}$$ $$\text{Moles of } H_2 = \frac{450}{96,500} \text{ mol} \approx 0.0046632 \text{ mol}$$

At s.t.p., 1 mole of any gas occupies $22.4 \text{ dm}^3$.

$$\text{Volume of } H_2 = \text{Moles of } H_2 \times \text{Molar volume at s.t.p.}$$ $$\text{Volume of } H_2 = \frac{450}{96,500} \text{ mol} \times 22.4 \text{ dm}^3/\text{mol}$$ $$\text{Volume of } H_2 = \frac{10080}{96500} \text{ dm}^3$$ $$\text{Volume of } H_2 \approx 0.10445 \text{ dm}^3$$

Rounding to three significant figures:

$$\text{Volume of } H_2 \approx \boxed{\text{0.104 dm}^3}$$