Question

What mass of oxygen is needed to complete the combustion of 8.80 x 10^-3 g of methane?

Answer 1

Step-by-step explanation:

The chemical reaction that describes the combustion of methane is

`\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}`

We need to convert the mass of methane to moles:

`0.00880\:\text{g} × \frac{1\:\text{mol CH}_4}{16.04\:\text{g}}`

`= 5.49×10^(-4)\:\text{mol CH}_4`

Now use the molar ratios to determine the amount of oxygen used during the combustion:

`5.49×10^(-4)\:\text{mol CH}_4×\left(\frac{2\:\text{mol O}_2}{1\:\text{mol CH}_4}\right)`

`= 0.00110\:\text{mol O}_2`

Converting this to grams, we find that the mass of oxygen is

`0.00110\:\text{mol O}_2×\left(\frac{15.999\:\text{g O}_2}{1\:\text{mol O}_2}\right)`

`= 0.0176\:\text{g O}_2`