Question

Nitrogen reacts with hydrogen to give ammonia. Calculate the volume of the ammonia gas formed when nitrogen reacts with $6\,litres$ of hydrogen. All volumes measured at STP.

Answer 1

Since both of the reactants and the product are given try to write down a balanced chemical equation corresponding to the reaction. Next calculate the number of moles of the reactants. Check out for the limiting reactant and from that calculate the number of moles of ammonia formed and hence find the volume of ammonia.

Complete step-by-step solution: The balanced chemical equation for the given reaction where ${N_2}$ and ${H_2}$ are the reactants and $N{H_3}$ is the product formed is given by:
${N_2}\, + \,3{H_2}\, \to \,2N{H_3}$
From the equation we can say $1\,mol$ of ${N_2}$ reacts with $3\,mol$ of ${H_2}$to form $2\,mol$of $N{H_3}$.
We will now find out the number of moles of ${N_2}$ and ${H_2}$, and look for the limiting reagent.
Now, number moles of a compound $= \,\dfrac{{Given\,Volume}}{{Volume\,occupied\,at\,STP}}$
For ${H_2}$,
Given Volume $=$ $6\,lit$
We know $1\,mol$of any gas occupies $22.4\,lit$at STP.
$\therefore$ number of moles of ${H_2}$ $= \,\dfrac{{6\,lit}}{{22.4\,lit}}\, = \,0.267\,mol$
Now for ${N_2}$,
Number of moles of ${N_2}\, = \,\dfrac{1}{3} \times \,No.\,of\,moles\,of\,{N_2}\,\, = \,\,\dfrac{1}{3} \times 0.267\,mol\,\, = \,\,0.089\,mol$
So, $0.267\,mol$ of ${H_2}$ and $0.089\,mol$of ${N_2}$ is present. As the number of moles of${H_2}$is more than that of${N_2}$,${H_2}$ will be present in excess after the reaction is over.
Hence ${N_2}$ is the limiting reagent and it will decide the number of moles of $N{H_3}$ formed.
Therefore, we can say that $0.089\,mol$of ${N_2}$ reacts with $0.089\,mol$of ${H_2}$ to form$\left( {2 \times 0.089} \right)\,mol$of $N{H_3}\, = \,0.178\,mol$of $N{H_3}$.
Now since number moles of a compound $= \,\dfrac{{Given\,Volume}}{{Volume\,occupied\,at\,STP}}$
$\therefore$ Given Volume of a gas $=$ $($Number of moles of the gas $\times$ Volume occupied by the gas at STP $)$
$\therefore$ Volume of $N{H_3}$ formed $=$ $\left( {0.178\, \times \,22.4\,} \right)\,lit\, = \,3.98\,lit$

Note: Write a proper balanced equation for the reaction given in the question if there is some error in the equation it will lead to error in the calculation. Finding out the limiting reagent is an important step and hence should be done properly. Also take care of the units. Try to write down the units in each and every step so that error may be avoided.