Calculate the total volume of hydrogen gas that was liberated at S.T.P conditions (Zn = 65.4, molar gas volume = 22.4 litres at S.T.P) assuming one mole of zinc reacted.

A common reaction is zinc reacting with hydrochloric acid:

$$\text{Zn(s)} + 2\text{HCl(aq)} \rightarrow \text{ZnCl}_2\text{(aq)} + \text{H}_2\text{(g)}$$

From the stoichiometry, 1 mole of $\text{Zn}$ produces 1 mole of $\text{H}_2$.

Assuming 1 mole of $\text{Zn}$ reacts, then 1 mole of $\text{H}_2$ is produced.

$$\text{Moles of H}_2 = 1 \text{ mol}$$

At S.T.P., 1 mole of any gas occupies 22.4 litres.

$$\text{Volume of H}_2 = \text{Moles of H}_2 \times \text{Molar gas volume at S.T.P.}$$ $$\text{Volume of H}_2 = 1 \text{ mol} \times 22.4 \text{ litres/mol}$$ $$\text{Volume of H}_2 = 22.4 \text{ litres}$$

The total volume of hydrogen gas liberated, assuming one mole of zinc reacted, is $\boxed{\text{22.4 litres}}$.

2. Calculate the mass of nitrogen (IV) oxide gas that would occupy the same volume as 10g of hydrogen gas at the same temperature and pressure. (H = 1.0, N = 14.0, O = 16.0)

$$\text{Molar mass of H}_2 = 2 \times \text{Atomic mass of H}$$ $$\text{Molar mass of H}_2 = 2 \times 1.0 \text{ g/mol} = 2.0 \text{ g/mol}$$

$$\text{Moles of H}_2 = \frac{\text{Mass of H}_2}{\text{Molar mass of H}_2}$$ $$\text{Moles of H}_2 = \frac{10 \text{ g}}{2.0 \text{ g/mol}} = 5.0 \text{ mol}$$

According to Avogadro's Law, at the same temperature and pressure, equal volumes of all gases contain the same number of moles. Since nitrogen (IV) oxide gas occupies the same volume as hydrogen gas at the same temperature and pressure, they must have the same number of moles.

$$\text{Moles of NO}_2 = \text{Moles of H}_2 = 5.0 \text{ mol}$$

$$\text{Molar mass of NO}_2 = \text{Atomic mass of N} + (2 \times \text{Atomic mass of O})$$ $$\text{Molar mass of NO}_2 = 14.0 \text{ g/mol} + (2 \times 16.0 \text{ g/mol})$$ $$\text{Molar mass of NO}_2 = 14.0 \text{ g/mol} + 32.0 \text{ g/mol} = 46.0 \text{ g/mol}$$

$$\text{Mass of NO}_2 = \text{Moles of NO}_2 \times \text{Molar mass of NO}_2$$ $$\text{Mass of NO}_2 = 5.0 \text{ mol} \times 46.0 \text{ g/mol}$$ $$\text{Mass of NO}_2 = 230 \text{ g}$$

The mass of nitrogen (IV) oxide gas is $\boxed{\text{230 g}}$.