Question

Calculate number of moles of NH3 formed by reaction of 2 moles of N2 and
2 moles
of 112
N2(g) + 3112(g) + 2NH3(g).​

Answer 1

Answer:

ANSWER

N2(g)+3H2(g)→2NH3(g) 

In the given reaction, 1 mol of N2 reequires 3 moles of H2 for the formation fo ammonia.

Thus if the number of moles of N2 and H2 should be in ratio 1:3.

We know,

n=molecular weightweight

Thus in option C 56g of N2 means 2 moles of N2 and 10g of H2 means 5 moles of H2.

Thus here H2 acts as limiting reagent.

Answer 2

we have to find the number of moles of NH₃ formed by reaction of 2 moles of N₂ and 2 moles of H₂.

solution : reaction between hydrogen and nitrogen results ammonia as shown below,

N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)

here you see, one mole of nitrogen combines with three moles of hydrogen to form two moles of ammonia.

we have two moles of nitrogen, to completely reduce nitrogen we need 6 moles of hydrogen ( it is clear in above reaction).

but we have given only 2 moles of hydrogen. it means hydrogen is limiting reagent. so the reaction will continue untill hydrogen is there. so we should prefer hydrogen to find the no of moles of ammonia.

     from reaction, it is clear that 3 moles of hydrogen gives 2 moles of ammonia.

∴ 2 moles of hydrogen will give $\frac{2}{3}\times2$ = 1.33 moles of ammonia.

therefore the no of moles of ammonia after the reaction would be 1.33 moles.