A student heats 8.70 g of manganese dioxide (87.0 g/mol) with 40.0 mL of hydrochloric acid of…

Step 1. Moles. MnO2: $\dfrac{8.70\ \text{g}}{87.0\ \text{g/mol}} = 0.100\ \text{mol}$. HCl: $0.0400\ \text{L} \times 6.00\ \text{mol/L} = 0.240\ \text{mol}$.

Step 2. The limiting reagent. $0.100\ \text{mol}$ of MnO2 needs $0.400\ \text{mol}$ of HCl. There is only $0.240\ \text{mol}$, so the acid limits.

Step 3. Chlorine. One mole of Cl2 comes from 4 mol of HCl: $\dfrac{0.240}{4} = 0.0600\ \text{mol}$. The volume is $0.0600\ \text{mol} \times 22.4\ \text{L/mol} = 1.34\ \text{L}$.

Step 4. Manganese dioxide left. $0.0600\ \text{mol}$ reacted, so $0.0400\ \text{mol}$ is left, which is $0.0400 \times 87.0\ \text{g/mol} = 3.48\ \text{g}$.