Define relative atomic mass. (a) Balance the equation: X₂CO₃(s) + HCl(aq) → XCl(aq) + H₂O(l) + CO₂(g). (b) If 23.6 cm³ of 0.15 mol/dm³ HCl reacts completely with X₂CO₃, calculate the moles of HCl used

To balance the X atoms, place a coefficient of 2 in front of XCl.

$$\text{X}_2\text{CO}_3\text{(s)} + \text{HCl(aq)} \rightarrow 2\text{XCl(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$$

To balance the H and Cl atoms, place a coefficient of 2 in front of HCl.

$$\boxed{\text{X}_2\text{CO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow 2\text{XCl(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}}$$

$$\text{Moles of HCl} = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of HCl} = 0.15 \text{ mol/dm}^3 \times 0.0236 \text{ dm}^3$$ $$\text{Moles of HCl} = 0.00354 \text{ mol}$$ $$\boxed{0.00354 \text{ mol}}$$

From the balanced equation: $1 \text{ mole of X}_2\text{CO}_3$ reacts with $2 \text{ moles of HCl}$.

$$\text{Moles of X}_2\text{CO}_3 = \frac{\text{Moles of HCl}}{2}$$ $$\text{Moles of X}_2\text{CO}_3 = \frac{0.00354 \text{ mol}}{2}$$ $$\text{Moles of X}_2\text{CO}_3 = 0.00177 \text{ mol}$$ $$\boxed{0.00177 \text{ mol}}$$

$$\text{M}_r(\text{X}_2\text{CO}_3) = \frac{\text{Mass of X}_2\text{CO}_3}{\text{Moles of X}_2\text{CO}_3}$$ $$\text{M}_r(\text{X}_2\text{CO}_3) = \frac{0.245 \text{ g}}{0.00177 \text{ mol}}$$ $$\text{M}_r(\text{X}_2\text{CO}_3) \approx 138.418 \text{ g/mol}$$ $$\boxed{138.4 \text{ g/mol}}$$

Assume RAM(C) = 12.0 g/mol and RAM(O) = 16.0 g/mol.

$$\text{M}_r(\text{X}_2\text{CO}_3) = 2 \times \text{RAM(X)} + \text{RAM(C)} + 3 \times \text{RAM(O)}$$ $$138.418 = 2 \times \text{RAM(X)} + 12.0 + 3 \times 16.0$$ $$138.418 = 2 \times \text{RAM(X)} + 12.0 + 48.0$$ $$138.418 = 2 \times \text{RAM(X)} + 60.0$$

$$2 \times \text{RAM(X)} = 138.418 - 60.0$$ $$2 \times \text{RAM(X)} = 78.418$$ $$\text{RAM(X)} = \frac{78.418}{2}$$ $$\text{RAM(X)} = 39.209 \text{ g/mol}$$ $$\boxed{39.2 \text{ g/mol}}$$