Question

50kg of N2 and 10kg of H2 are mixed to produce NH3 . Calculate the moles of NH3 formed. Identify the limiting reagent in the production of NH3 in this situation.

Answer 1

N2 + 3H2 ---->2NH3
here.
28kg N reacts with 6kg of H
so,
50 kg of M react with 6/28×50kg H = 10.7kg but we have only 10 so H2 is limiting reagent..

6kg H2 gives 34kg NH3
10kg H2 will give 34/6×10=56.67kgNH3

Answer 2

During the reaction, ammonia formed 56.67 kg.

Explanation:

A balanced reaction between nitrogen and hydrogen is as follows:

$N_{2}+3 H_{2} \rightarrow 2 N H_{3}$

From the reaction 28 g of nitrogen is reacted with 6 g of hydrogen to form 34 grams of ammonia.

Therefore, 6 g of hydrogen reacts with 28 g of nitrogen to form 34 grams of $NH_3$.

From the given,

Weight of hydrogen = 1 kg

$10 \ \mathrm{kg} \text { of } \mathrm{H}_{2} \rightarrow \mathrm{X} \ \mathrm{gms} \text { of } \mathrm{N}_{2}$

$x=\frac{10 \times 28}{6}=46.67 \ \mathrm{kg}$

Therefore, 10 kg of hydrogen reacts with 46.67 kg of nitrogen.

Amount of ammonia = 10 kg + 46.67 kg = 56.67 kg

The limiting agent for the formation of ammonia is hydrogen.