A student treats 2.65 g of anhydrous sodium carbonate, Na₂CO₃ (106 g/mol), with an excess of dilute…

Step 1. $n(\text{Na}_2\text{CO}_3) = \dfrac{2.65\ \text{g}}{106\ \text{g/mol}} = 0.0250\ \text{mol}$. The equation $\text{Na}_{2}\text{CO}_{3}\ +\ \text{H}_{2}\text{SO}_{4}\ \rightarrow \ \text{Na}_{2}\text{SO}_{4}\ +\ \text{H}_{2}\text{O}\ +\ \text{CO}_{2}$ gives $n(\text{CO}_2) = 0.0250\ \text{mol}$.

Step 2. $V = 0.0250\ \text{mol} \times 22.4\ \text{L/mol} = 0.560\ \text{L} = 560\ \text{mL}$.

Step 3. Each mole of carbon dioxide gives one mole of $\text{CaCO}_{3}$ ($100\ \text{g/mol}$): $0.0250\ \text{mol} \times 100\ \text{g/mol} = 2.50\ \text{g}$.