Question

The number of moles of methane required to produce 11g \(CO_2\) (g) after complete combustion is:
(Given molar mass of methane in g mol^–1 : 16)

• A. 0.75
• B. 0.25
• C. 0.35
• D. 0.5

Answer 1

The Correct Option is B

The general combustion reaction for alkanes is: \[\text{C}_n\text{H}_{2n+2} + \frac{3n+1}{2}\text{O}_2 \rightarrow n\text{CO}_2 + (n+1)\text{H}_2\text{O}.\] For methane (\(\text{CH}_4\)), the specific balanced equation is: \[\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}.\] Step 1: Molar mass of \(\text{CO}_2\). The molar mass of \(\text{CO}_2\) is calculated as: \[\text{Molar mass of } \text{CO}_2 = 12 + (2 \times 16) = 44 \, \text{g/mol}.\] Step 2: Calculate moles of \(\text{CO}_2\). The number of moles of \(\text{CO}_2\) in \(11 \, \text{g}\) is: \[\text{Moles of } \text{CO}_2 = \frac{\text{Mass of } \text{CO}_2}{\text{Molar mass of } \text{CO}_2} = \frac{11}{44} = 0.25 \, \text{mol}.\] Step 3: Stoichiometric relationship between \(\text{CH}_4\) and \(\text{CO}_2\). The balanced reaction indicates that: \[1 \, \text{mole of } \text{CH}_4 \, \text{produces } 1 \, \text{mole of } \text{CO}_2.\] Consequently, to produce \(0.25 \, \text{moles of } \text{CO}_2\), the required moles of \(\text{CH}_4\) are: \[\text{Moles of } \text{CH}_4 = 0.25 \, \text{mol}.\] Step 4: Mass of \(\text{CH}_4\). The mass of \(0.25 \, \text{moles of } \text{CH}_4\) is: \[\text{Mass of } \text{CH}_4 = \text{Moles of } \text{CH}_4 \times \text{Molar mass of } \text{CH}_4 = 0.25 \times 16 = 4 \, \text{g}.\] Conclusion: To produce \(11 \, \text{g of } \text{CO}_2\), \(0.25 \, \text{mol}\) of methane is required. Final Answer: (2).