Question

If $2L$ of ${{N}_{2}}$ ​ is mixed with $2L$ of ${{H}_{2}}$ ​ at a constant temperature and pressure, then what will be the volume of $N{{H}_{3}}$ ​ formed?

Answer 1

The reaction for ammonia formation is, ${{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}}$ From this, by converting mass into moles; we can find the moles of ammonia formed. Further, the limiting reagent is the reactant which is present in smaller quantities and the quantity of product is determined from it. So, the reactant which will decide the formation of the product will be the limiting reactant.

Complete step by step solution:
We know the reaction for formation of ammonia can be written as; ${{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}}$
This means one mole of Nitrogen reacts with three moles of hydrogen to give ammonia gas. We have the volume of ${{H}_{2}}$ which is ${{V}_{{{H}_{2}}}}=2L$ and the volume of ${{N}_{2}}$ which is ${{V}_{{{N}_{2}}}}=2L$
So, the amount of ammonia formation will be decided by hydrogen. We have that the three moles of hydrogen give two moles of ammonia.
Now from the reaction, ${{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}}$ and by their equivalent is given by $1V+3V\to 2V$
Thus it can be rewritten as; ${{H}_{2}}\to \dfrac{{{V}_{{{H}_{2}}}}}{3}=\dfrac{2}{3}=0.667$ similarly for ${{N}_{2}}\to \dfrac{{{V}_{{{N}_{2}}}}}{1}=\dfrac{2}{1}=2$
Since we know that $2>0.667$ , since ${{H}_{2}}$ is the limiting agent.
Now, we know that $N{{H}_{3}}={{H}_{2}}$ ,
$\dfrac{x}{2V}=\dfrac{2L}{3V}$ from here we have get the value of x;
$\Rightarrow x=\dfrac{2L\times 2V}{3V}=\dfrac{4}{3}L=1.333L$
Thus the volume of $N{{H}_{3}}$ formed is $1.333L$ .

Note:
It must be noted that the number of moles of a molecule can be found by dividing the given mass of the molecule with the molar mass of the molecule. The molar mass of nitrogen is $28$ and not $14$ because it exists as ${{N}_{2}}$ . It has two atoms. Similarly, the molar mass of hydrogen is $2$ and not $1$ because it exists as ${{H}_{2}}$ .