If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

The correct option is 10.

To find the volume of water vapour produced, we can use Gay-Lussac's Law of Gaseous Volumes. This law states that when gases react together, they do so in volumes which bear a simple whole-number ratio to one another and to the volume of the products formed (if gaseous), provided all gases are at the same temperature and pressure.

First, let us write the balanced chemical equation for the reaction between dihydrogen gas (H₂) and dioxygen gas (O₂) to form water vapour (H₂O):

$$2H_2(g)+O_2(g)\to 2H_{2}O(g)$$

From the balanced equation, we can observe the stoichiometric relationship between the reacting gases and the product:
- 2 volumes of dihydrogen gas react with 1 volume of dioxygen gas to produce 2 volumes of water vapour.

According to the question, we are given:
- Volume of dihydrogen gas = 10 volumes
- Volume of dioxygen gas = 5 volumes

Since the ratio of the reactants given is:
$\frac{10\text{ volumes of }{H}_2}{5\text{ volumes of }{O}_2}=\frac{2}{1}$
This perfectly matches the stoichiometric ratio (2:1) of the reaction. Therefore, both reactants react completely without any reactant being left in excess.

Since the volume ratio of dihydrogen gas reacting to water vapour produced is 2:2 (or 1:1):
$\text{Volume of }{H}_2\text{O produced}=\text{Volume of }{H}_2\text{ reacted}=10\text{ volumes}$
Thus, 10 volumes of water vapour will be produced.