Question

$28{\text{g}}$ of ${{\text{N}}_2}$and $6{\text{g}}$ of \[{{\text{H}}_{\text{2}}}\] were mixed. At equilibrium \[17{\text{g}}\]of ${\text{N}}{{\text{H}}_{\text{3}}}$ was formed. The weight of ${{\text{N}}_2}$ and \[{{\text{H}}_{\text{2}}}\] at equilibrium are _________________ respectively.

Answer 1

Hint: Knowledge of the Gay-Lussac’s Law of combining volumes is important to solve this question. The law states that whenever gases combine with each other to form gaseous products, they do so in amounts that bear a simple whole number ratio to each other.

Complete step by step answer:
Consider the reaction:
\[{{\text{N}}_2} + 3{{\text{H}}_2} \rightleftarrows 2{\text{N}}{{\text{H}}_3}\]
1 mole of${{\text{N}}_2}$ reacts with 3 moles of \[{{\text{H}}_{\text{2}}}\]to form 2 moles of ${\text{N}}{{\text{H}}_{\text{3}}}$.
Therefore to form one mole of ${\text{N}}{{\text{H}}_{\text{3}}}$, $\dfrac{1}{2}$ mole of ${{\text{N}}_2}$ would react with $\dfrac{3}{2}$ moles of \[{{\text{H}}_{\text{2}}}\].
As per the definition of a mole, one mole of particles is equal to the molecular weight of a compound or the atomic weight of an element.
The molecular weight of ${\text{N}}{{\text{H}}_{\text{3}}}$=$\left[ {14 + \left( {3 \times 1} \right)} \right] = 17$g.
At equilibrium also 17 g of ${\text{N}}{{\text{H}}_{\text{3}}}$ was present so total one mole of ${\text{N}}{{\text{H}}_{\text{3}}}$ is formed by the reaction.
Hence the weight of ${{\text{N}}_2}$ at equilibrium = mass of ${{\text{N}}_2}$ = 14 g
And that for $\dfrac{3}{2}$ moles of \[{{\text{H}}_{\text{2}}}\] = 3 g
So, the amount of ${{\text{N}}_2}$and\[{{\text{H}}_{\text{2}}}\] at equilibrium were 14 g and 3 g respectively.

Note:
1. If in any chemical reaction, there are solid or liquid reactants present, along with gaseous reactants, then their volume is considered to be constant or negligible in comparison to gases.
2. This law is mainly applicable to ideal gases reacting with each other under similar conditions of temperature and pressure.