1. **State the problem:** We are given masses of methane (CH₄) and oxygen (O₂) and asked to find the limiting reactant in the reaction: $$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$ 2. **Calculate moles of each reactant:** - Molar mass of CH₄ = 16.05 g/mol - Molar mass of O₂ = 32.00 g/mol 3. **Calculate moles of CH₄:** $$\frac{10.0\text{ g CH}_4}{16.05\text{ g/mol}} = 0.623\text{ moles CH}_4$$ 4. **Calculate moles of O₂:** $$\frac{10.0\text{ g O}_2}{32.00\text{ g/mol}} = 0.3125\text{ moles O}_2$$ 5. **Use stoichiometry to find moles of CO₂ produced from each reactant:** - From CH₄: 1 mole CH₄ produces 1 mole CO₂ $$0.623\text{ moles CH}_4 \times \frac{1\text{ mole CO}_2}{1\text{ mole CH}_4} = 0.623\text{ moles CO}_2$$ - From O₂: 2 moles O₂ produce 1 mole CO₂ $$0.3125\text{ moles O}_2 \times \frac{1\text{ mole CO}_2}{2\text{ moles O}_2} = 0.15625\text{ moles CO}_2$$ 6. **Determine limiting reactant:** Since O₂ produces fewer moles of CO₂ (0.15625) compared to CH₄ (0.623), oxygen is the limiting reactant. 7. **Final answer:** **Oxygen is the limiting reactant.**