Question

A mixture of ${N_2}$ and ${H_2}$ is caused to react in a closed container to form $N{H_3}$. The reaction ceases before any of the reactant has been totally consumed. At this stage, 4 moles each of ${N_2},{H_2}$ and $N{H_3}$ are present. Then the weight of ${N_2}$ and ${H_2}$ present originally were respectively
(A) 224 g and 16 g
(B) 168 g and 20 g
(C) 168 g and 16 g
(D) 224 g and 20 g

Answer 1

Hint: First write the balanced reaction of nitrogen and hydrogen gas to obtain ammonia gas. Then assume the number of moles of all three gases at initial state and at the equilibrium state to find the number of moles of the gases present.

Complete step by step solution:
It is given that the reaction attains an equilibrium state. At equilibrium state, four moles of each gas is present in the reaction mixture. Now, assume that A moles of ${N_2}$, B moles of ${H_2}$ and zero moles of ammonia were present at the start of the reaction.
The reaction can be given as below.
\[{N_2} + 3{H_2} \to 2N{H_3}\]
Now, we can say that in equilibrium state, moles of ${N_2}$ will be A-x and moles of ${H_2}$ will be B-x. From the reaction, we can say that moles of $N{H_3}$ will be 2x.

	${N_2}$	${H_2}$	$N{H_3}$
Initial state	A	B	0
Equilibrium state	A-x	B-3x	2x

Now, we are given that at equilibrium state, moles of $N{H_3}$ are 4.
Thus, we can say that 2x = 4. Thus, x =2
Now, it is given that moles of A at equilibrium state is 4.
Thus, $4 = A- x$
So, $4 = A – 2$
Therefore, $A = 4+2 = 6 $moles
Moles of B at equilibrium state is also given 4.
Thus, $4 = B – 3x$
So, $B = 4 +3(2) = 10 moles$.
We know that the molecular weight of ${N_2}$ is 28 $gmo{l^{ - 1}}$.
So, 6 moles of ${N_2}$ will weigh 28$ \times $6 = 168 g
Molecular weight of ${H_2}$ is 2$gmo{l^{ - 1}}$.
So, 10 moles of ${H_2}$ will weighs 10$ \times $2 = 20 g
Thus, we can say that the weight of nitrogen gas and hydrogen gas is 168g and 20g respectively.

Therefore, the correct answer is (B).

Note: Note that the reaction needs to be balanced in order to find the weights or number of moles of gases present in the initial state of the reaction. Remember that atomic weight if nitrogen atom is $14gmo{l^{ - 1}}$ .