Question

methane was burned in an incorrectly adjusted burner. The methane was converted into a mixture of CO2 and CO in the ratio of 99:1, togetherwith water vapour. What will b the volume of oxygen consumed when y dm3 of methane is burned?

Answer 1

The methane conversation into $\mathrm{CO}_{2}$ and CO can be described by next reactions:

$$\mathrm{CH_4} + 2\mathrm{O_2} = 2\mathrm{H_2O} + \mathrm{CO_2}$$ $$2\mathrm{CH_4} + 3\mathrm{O_2} = 4\mathrm{H_2O} + 2\mathrm{CO}$$

If the ratio between $\mathrm{CO_2}$ and CO is 99:1, total volume of methane is y dm³

In both reactions methane is converted in equivalent amount/volume of carbon containing gas. So volumes of $\mathrm{CO_2}$ and CO are 0.99 y and 0.01 y. Now it is possible to find volume of oxygen:

X1 0.99 y

$$\mathrm{CH_4} + 2\mathrm{O_2} = 2\mathrm{H_2O} + \mathrm{CO_2}$$

2 1

$$\mathrm{X1} = 2 \times 0.99\mathrm{y} = 1.98\mathrm{y}$$

For next reaction:

X2 0.01 y

$$2\mathrm{CH_4} + 3\mathrm{O_2} = 4\mathrm{H_2O} + 2\mathrm{CO}$$

3 2

$$\mathrm{X2} = 3 \times 0.01\mathrm{y} / 2 = 0.015\mathrm{y}$$

Total volume is $\mathrm{X1 + X2 = 1.98y + 0.015y = 1.995y}$