Question

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

Answer 1

Hint: Start by writing the reaction of hydrogen gas with oxygen gas to form water. Make a note of the number of moles of hydrogen gas and oxygen gas required to form one mole of water. Think about the relationship between the number of moles and volume of a gas. Equate the two parameters to get the answer.

Complete step by step solution:
- Let’s begin by writing the balanced chemical reaction of formation of water.
\[2{{H}_{2(g)}}+{{O}_{2\left( g \right)}}\to 2{{H}_{2}}{{O}_{\left( g \right)}}\]
- Two moles of dihydrogen gas also known as hydrogen gas, ${{H}_{2}}$ reacts with oxygen gas, ${{O}_{2}}$ to form two moles of water vapour.
- The ratio of number of moles of hydrogen to oxygen is 2:1.
- According to Avogadro’s law, the number of moles of a gas is directly proportional to the volume occupied by gas.
- Therefore, for the reaction $2{{H}_{2\left( g \right)}}+{{O}_{2\left( g \right)}}\to 2{{H}_{2}}{{O}_{\left( g \right)}}$, two volumes of hydrogen gas reacts with one volume of oxygen gas to give two volumes of water vapour.
- Therefore, the ratio of volumes of hydrogen gas to oxygen gas is equal to 2:1 and volumes of hydrogen is equal to volumes of water vapour.
- Now, let’s look at the question. Ten volumes of dihydrogen gas reacts with five volumes of oxygen gas, so the ratio of volumes of hydrogen gas to oxygen gas is 10:5 that is, 2:1.
- Therefore, volumes of water vapour produced will be equal to volumes of hydrogen gas present that is, 10volumes.

- Therefore, ten volumes of dihydrogen gas reacts with five volumes of oxygen gas to produce ten volumes of water vapour.

Note: Remember the Avogadro’s law which describes the relationship between the volume and number of moles of a gas. According to Avogadro’s law, the number of moles of a gas is directly proportional to its volume. Also, remember the ratio of hydrogen to oxygen in production of water is 2:1.