Question

How many litres of detonating gas may be produced at $0^\circ C$ and $1atm$ from the decomposition of $0.1mole$ of water, by an electric current?
A) $2.241$
B) $1.121$
C) $3.361$
D) $4.481$

Answer 1

The mixture of hydrogen and oxygen gas is called the detonating gas. To calculate the litres of detonating gas produced at $0^\circ C$ and $1atm$ from the decomposition of water, we first need to figure out the balanced chemical reaction and then based on the stoichiometric ratio of the equation, the calculation can be done.

Complete answer:
The balanced chemical reaction for the generation of detonating gas will be - $2{H_2}O(g) \to 2{H_2}(g) + {O_2}(g)$
Now, as the balanced chemical equation suggests – two moles of water produce one mole of oxygen and two moles of hydrogen. Here we can see that the number of moles of oxygen gas produced is half as the number of moles of water and the number of moles of hydrogen gas is similar to the number of moles of water. Hence, the total number of moles of detonating gas produced in a standard reaction will be - $2 + 1 = 3mol$ .
As we are given with $0.1mole$ of water, the number of moles of hydrogen will be the same, i.e. $0.1mole$ and the number of moles of oxygen will be half of it, i.e. $0.1mole/2 = 0.05mole$ . This makes the total number of moles to $0.1mole + 0.05mole = 0.15mole$ .
Now converting the number of moles of gas into litres –
$0.15mole \times (22.4) = 3.36litre$
Hence, option C) $3.361$ is the correct answer.

Note:
One mole of gas at STP ( $0^\circ C$ and $1atm$ ) occupies $22.4$ litres of volume so for calculating the volume of a given number of moles of the sample of gas, we can multiply it by $22.4$ .