Question

How many liters of methane gas (CH4) need to be combusted to produce 8.5 liters of water vapor, if all measurements are taken at the same temperature and pressure? Show all of the work used to solve this problem. CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g)

Answer 1

Answer : The volume of methane gas needed are 4.25 liters.

Explanation :

The balanced chemical reaction is:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

At STP, 1 mole of substance contains 22.4 L volume of gas.

From the given reaction we conclude that,

1 mole of methane react to give 2 moles of water vapor.

As, $2\times 22.4L$ volume of water vapor produced from 22.4 L volume of methane gas

So, 8.5 L volume of water vapor produced from $\frac{8.5}{2\times 22.4}\times 22.4=4.25L$ volume of methane gas

Therefore, the volume of methane gas needed are 4.25 liters.

Answer 2

8.5L H2O (1 mol H2O/22.4L)=.37946 mol H2O
.3796 mol H2O(1 mol CH4/2 mol H2O)=.18973 mol CH4
.18973 mol CH4(22.4L/1mol CH4)=4.25L CH4 gas