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Question: A mixture of \({N_2}\) and \({H_2}\) is caused to react in a closed container to form \(N{H_3}\). Th...

A mixture of N2{N_2} and H2{H_2} is caused to react in a closed container to form NH3N{H_3}. The reaction ceases before any of the reactant has been totally consumed. At this stage, 4 moles each of N2,H2{N_2},{H_2} and NH3N{H_3} are present. Then the weight of N2{N_2} and H2{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

Explanation

Solution

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 N2{N_2}, B moles of H2{H_2} and zero moles of ammonia were present at the start of the reaction.
The reaction can be given as below.
N2+3H2→2NH3{N_2} + 3{H_2} \to 2N{H_3}
Now, we can say that in equilibrium state, moles of N2{N_2} will be A-x and moles of H2{H_2} will be B-x. From the reaction, we can say that moles of NH3N{H_3} will be 2x.

| N2{N_2} | H2{H_2} | NH3N{H_3}
---|---|---|---
Initial state| A| B| 0
Equilibrium state| A-x| B-3x| 2x

Now, we are given that at equilibrium state, moles of NH3N{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−x4 = A- x
So, 4=A–24 = A – 2
Therefore, A=4+2=6A = 4+2 = 6 moles
Moles of B at equilibrium state is also given 4.
Thus, 4=B–3x4 = B – 3x
So, B=4+3(2)=10molesB = 4 +3(2) = 10 moles.
We know that the molecular weight of N2{N_2} is 28 gmol−1gmo{l^{ - 1}}.
So, 6 moles of N2{N_2} will weigh 28×\times6 = 168 g
Molecular weight of H2{H_2} is 2gmol−1gmo{l^{ - 1}}.
So, 10 moles of H2{H_2} will weighs 10×\times2 = 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 14gmol−114gmo{l^{ - 1}} .