Question

If $5g{H_2}$ is mixed with $14g$ of nitrogen gas for the following reaction ${N_2} + 3{H_2} \to 2N{H_3}$ . At the end, mass (in grams) of ${H_2}$ left unreacted is?

Answer 1

Hint:In a reaction there is an excess reagent which is available in more amounts as compared to another reactant and there is limiting reagent which refers to the reactant that is used up completely while the reaction is in process and thus the reaction depends on the limiting reagent because reaction cannot proceed without a reactant.

Complete step by step answer:
The balanced equation of hydrogen reacting with nitrogen gas to give ammonia is ${N_2} + 3{H_2} \to 2N{H_3}$. From this equation we notice that 3 mol of H2 is required with 1 mol of nitrogen gas to produce 2 moles of NH3. Therefore, $3 \times 2 = 6g{H_2}$ ,
$1 \times 28 = 28g{N_2}$ and
$2 \times 17 = 34gN{H_3}$
It is given to us that
${N_2} = 14g$
${H_2} = 5g$
So, we know that,
$28g{\text{ }}{N_2} \to 6g{\text{ }}{H_2}$
 $ \Rightarrow 1g{\text{ }}{N_2} \to \dfrac{6}{{28}}g{\text{ }}{H_2}$
So, $14g{\text{ }}{{\text{N}}_2} \to \dfrac{6}{{28}} \times 14g{\text{ }}{{\text{H}}_2}$
$ \Rightarrow 14g{\text{ }}{{\text{N}}_2} \to 3g{\text{ }}{{\text{H}}_2}$
We are given 5g of H2 in the question thus, we now know that H2 is in excess and is not the limiting reagent. But we also have $14g{N_2}$ that is not sufficient for complete reaction with $5g{H_2}$ and thus, ${N_2}$ is the limiting reagent.
Thus, the given amount of N2 gets used up completely for the reaction but some H2 is left unreacted. So, the unreacted ${H_2}$ left is $5 - 3 = 2g$ .
Now for calculating the amount of ammonia produced by the reaction of the given amount of hydrogen and nitrogen gas will depend on the amount N2 because it is the limiting reagent. From the question we know that
$28g{\text{ }}{{\text{N}}_2} \to 34g{\text{ N}}{{\text{H}}_2}$
$ \Rightarrow 1g{\text{ }}{N_2} \to \dfrac{{34}}{{28}}g{\text{ }}N{H_3}$
$ \Rightarrow 14g{\text{ }}{N_2} \to \dfrac{{34}}{{28}} \times 14g{\text{ }}N{H_3}$
Thus, $14g{N_2}$ produces $17g N{H_3}$ and the unreacted ${H_2}$ amount is $2g$

Note:
The production of ammonia $N{H_3}$ is known as Haber-Bosch process. In this process natural gas or LPG is passed into an ammonia producing plant where gaseous hydrogen is collected and this hydrogen is reacted with nitrogen gas to produce ammonia as a product of their reaction.