Question

Methane undergoes a combustion reaction according to the reaction $C{H}_4\left(g\right)+2O_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(l\right)$. How many grams of methane gas were burned if 67.2 $L$ of carbon dioxide gas are produced in the reaction? (assume $STP$)

• A. 16 grams
• B. 48 grams
• C. 3 grams
• D. 132 grams
• E. 22.4 grams

Answer 1

The correct option is B 48 grams

At $STP$, 1 mole of any gas occupies a volume of 22.4 $L$.
The volume of 67.2 $L$ corresponds to $\frac{67.2}{22.4}=3$ moles.
1 mole of $C{O}_2$ is obtained from 1 mole of $C{H}_4$.
3 moles of $C{H}_4$ will be obtained from 3 moles of $C{O}_2$.
The molar mass of methane is 16 $g/mol$.
The mass of 3 moles of methane will be $3\times 16=48g$