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Chemistry Question on Equilibrium

28 g of N2{{N}_{2}} and 6 g6\,g of H2{{H}_{2}} were mixed. At equilibrium, 17 g17\,g NH3N{{H}_{3}} was produced. The weights of N2{{N}_{2}} and H2{{H}_{2}} at equilibrium are respectively

A

11 g, 0 g

B

1 g, 3 g

C

14 g, 3 g

D

11 g, 3 g

Answer

14 g, 3 g

Explanation

Solution

N2+3H22NH3N _{2}+3 H _{2} 2 NH _{3}
28 g 6 g 34 g28 \,g \,6 \,g \,34\, g
∵34 g NH3\because 34 \, g \, NH _{3}
was produced from =28gN2=28 g N _{2}
∴17 g NH3\therefore 17 \, g \, NH _{3}
will be produced from =28×1734=\frac{28 \times 17}{34}
=14 g N2=14 \, g\, N_{2}
Similarly, 17 g17 \, g of NH3NH _{3}
will be produced from =6×1734=\frac{6 \times 17}{34}
=3 g H2=3 \, g\, H _{2}