Question

How many liters of water vapor can be produced if 8.9 liters of methane gas (CH4) are combusted, if all measurements are taken at the same temperature and pressure?

Answer 1

Task:

How many liters of water vapor can be produced if 8.9 liters of methane gas (CH4) are combusted, if all measurements are taken at the same temperature and pressure?

Solution:

The chemical equation for the combustion reaction is

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

If the measurements are taken at STP (T=273 K, p = 1 atm), the number of moles of $\mathrm{CH_4}$ is

$$n(\mathrm{mol}) = \frac{V(L)}{V_0(L)}$$

n- number of moles of methane

V – volume of $\mathrm{CH_4}$, (L)

$V_0$ – molar volume of the gas at STP, (L)

$$n(\mathrm{CH_4}) = \frac{V(\mathrm{CH_4})}{V_0} = \frac{8.9}{22.4} = 0.397 \mathrm{mol}$$

According to the chemical equation the number of moles of water vapor is twice the number of moles of methane.

$$n(\mathrm{H_2O}) = 2n(\mathrm{CH_4}) = 2 \cdot 0.397 = 0.794 \mathrm{mol}$$

The volume of water vapor is

$$V(\mathrm{H_2O}) = n(\mathrm{mol}) \cdot V_0(L) = 0.794 \cdot 22.4 = 17.8 \mathrm{L}$$

Answer: $V(\mathrm{H_2O}) = 17.8 \mathrm{L}$