Question

56g of nitrogen and 8g of hydrogen gas are heated in a closed vessel. At equilibrium 34g ammonia is present. The equilibrium number of moles of nitrogen, hydrogen and ammonia are respectively:
A.1, 2, 2
B.2, 2, 1
C.1, 1, 2
D.2, 1, 2

Answer 1

Hint: We can solve this question by keeping two things in mind, first is that the reactants react and give products in accordance to the stoichiometry of the reaction and the other is the formula for finding number of moles of any compound, which is as follows:
Number of moles= \[\dfrac{given mass}{molar mass}\]

Complete step by step answer:
\[{{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}}\]
Molar mass of \[{{N}_{2}}\] is 28g and Molar mass of \[{{H}_{2}}\] is 2g
Number of moles= \[\dfrac{given mass}{molar mass}\]
Number of Moles of \[{{N}_{2}}\] initially present = \[\dfrac{56}{28}\] = 2
Number of Moles of \[{{H}_{2}}\] initially present = \[\dfrac{8}{2}\]= 4
Initially we have 2 moles of nitrogen gas\[{{N}_{2}}\], 4 moles of hydrogen gas \[{{H}_{2}}\] and 0 moles of ammonia .
At equilibrium say x moles of reacted then, at equilibrium
Number of Moles of \[{{N}_{2}}\]= 2-x
Number of Moles of \[{{H}_{2}}\]= 4-3x
Number of Moles of \[N{{H}_{3}}\]= 2x
Molar mass of \[N{{H}_{3}}\] is 17g.
Number of moles= \[\frac{given mass}{molar mass}\]= \[\dfrac{34}{17}\] = 2 moles of \[N{{H}_{3}}\]
As, given in the question we get 2 moles of \[N{{H}_{3}}\] at equilibrium so
2x = 2
Hence, we get x=1
So substituting the value of x in above equations we get,
Number of Moles of \[{{N}_{2}}\]= 2-x= 2-1= 1
Number of Moles of \[{{H}_{2}}\] = 4-3x= 4 - 3(1)=1
Number of Moles of \[N{{H}_{3}}\] =2x = 2

So the answer is option C

Note:
In a chemical reaction, chemical equilibrium is defined as the state in which both reactants and products are present in concentrations which have no further tendency to change with time and hence there is no observable change in the properties of the system.