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Chemistry Question on Stoichiometry and Stoichiometric Calculations

The number of moles of methane required to produce 11g CO2CO_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

0.25

Explanation

Solution

The general combustion reaction of alkanes is:
CnH2n+2+3n+12O2→nCO2+(n+1)H2O.\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 (CH4\text{CH}_4), the specific balanced equation is:
CH4+2O2→CO2+2H2O.\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}.
Step 1: Molar mass of CO2\text{CO}_2
The molar mass of CO2\text{CO}_2 is:
Molar mass of CO2=12+(2×16)=44 g/mol.\text{Molar mass of } \text{CO}_2 = 12 + (2 \times 16) = 44 \, \text{g/mol}.
Step 2: Calculate moles of CO2\text{CO}_2
The number of moles of CO2\text{CO}_2 in 11 g11 \, \text{g} is:
Moles of CO2=Mass of CO2Molar mass of CO2=1144=0.25 mol.\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: Relate CH4\text{CH}_4 to CO2\text{CO}_2
From the balanced reaction:
1 mole of CH4 produces 1 mole of CO2.1 \, \text{mole of } \text{CH}_4 \, \text{produces } 1 \, \text{mole of } \text{CO}_2.
Therefore, to produce 0.25 moles of CO20.25 \, \text{moles of } \text{CO}_2, the moles of CH4\text{CH}_4 required are:
Moles of CH4=0.25 mol.\text{Moles of } \text{CH}_4 = 0.25 \, \text{mol}.Step 4: Mass of CH4\text{CH}_4
The mass of 0.25 moles of CH40.25 \, \text{moles of } \text{CH}_4 can be verified as:
Mass of CH4=Moles of CH4×Molar mass of CH4=0.25×16=4 g.\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:
The number of moles of methane required to produce 11 g of CO211 \, \text{g of } \text{CO}_2 is 0.25 mol0.25 \, \text{mol}.
Final Answer: (2).