Question

If 3.78 L of nitrogen gas and 28.73 L of hydrogen gas were allowed to react, how many litres of ammonia gas could form?

Answer 1

Answer:

Volume of ammonia formed is 7.56 L

Explanation:

Volume of 1 mol of gas is 24 L.

No. of mol in 3.78 L of nitrogen = $\frac{3.78}{24}=0.1575\ mol$

No. of mol in 3.78 L of hydrogen =$\frac{28.73 }{24}=1.197\ mol$

$N_2 +3H_2 \rightarrow 2NH_3$

I mol of nitrogen required 3 moles of hhydrogen.

So, 0.1575 mol of nitrogen will require 0.4725 moles of hydrogen.

As moles of hydrogen is 1.197, therefore, nitrogen will be the limiting reagent.

So,

1 mol of nitrogen forms 2 moles of ammonia

0.1575 mol of nitrogen will form 0.1575 mol × 2 = 0.315 mol of ammonia

Volume of ammonia = 0.315 × 24

                               = 7.56 L

Volume of ammonia formed is 7.56 L