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Question: If \(5g{H_2}\) is mixed with \(14g\) of nitrogen gas for the following reaction \({N_2} + 3{H_2} \to...

If 5gH25g{H_2} is mixed with 14g14g of nitrogen gas for the following reaction N2+3H22NH3{N_2} + 3{H_2} \to 2N{H_3} . At the end, mass (in grams) of H2{H_2} left unreacted is?

Explanation

Solution

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 N2+3H22NH3{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, 3×2=6gH23 \times 2 = 6g{H_2} ,
1×28=28gN21 \times 28 = 28g{N_2} and
2×17=34gNH32 \times 17 = 34gN{H_3}
It is given to us that
N2=14g{N_2} = 14g
H2=5g{H_2} = 5g
So, we know that,
28g N26g H228g{\text{ }}{N_2} \to 6g{\text{ }}{H_2}
1g N2628g H2\Rightarrow 1g{\text{ }}{N_2} \to \dfrac{6}{{28}}g{\text{ }}{H_2}
So, 14g N2628×14g H214g{\text{ }}{{\text{N}}_2} \to \dfrac{6}{{28}} \times 14g{\text{ }}{{\text{H}}_2}
14g N23g H2\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 14gN214g{N_2} that is not sufficient for complete reaction with 5gH25g{H_2} and thus, N2{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 H2{H_2} left is 53=2g5 - 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 N234g NH228g{\text{ }}{{\text{N}}_2} \to 34g{\text{ N}}{{\text{H}}_2}
1g N23428g NH3\Rightarrow 1g{\text{ }}{N_2} \to \dfrac{{34}}{{28}}g{\text{ }}N{H_3}
14g N23428×14g NH3\Rightarrow 14g{\text{ }}{N_2} \to \dfrac{{34}}{{28}} \times 14g{\text{ }}N{H_3}
Thus, 14gN214g{N_2} produces 17gNH317g N{H_3} and the unreacted H2{H_2} amount is 2g2g

Note:
The production of ammonia NH3N{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.