Explanation:
The chemical reaction that describes the combustion of methane is
[tex]\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}[/tex]
We need to convert the mass of methane to moles:
[tex]0.00880\:\text{g} × \dfrac{1\:\text{mol CH}_4}{16.04\:\text{g}}[/tex]
[tex]= 5.49×10^{-4}\:\text{mol CH}_4[/tex]
Now use the molar ratios to determine the amount of oxygen used during the combustion:
[tex]5.49×10^{-4}\:\text{mol CH}_4×\left(\dfrac{2\:\text{mol O}_2}{1\:\text{mol CH}_4}\right)[/tex]
[tex] = 0.00110\:\text{mol O}_2[/tex]
Converting this to grams, we find that the mass of oxygen is
[tex]0.00110\:\text{mol O}_2×\left(\dfrac{15.999\:\text{g O}_2}{1\:\text{mol O}_2}\right)[/tex]
[tex] = 0.0176\:\text{g O}_2[/tex]
