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Question: 50 kg of \({N_2}\)(g) and 10 kg of \({H_2}\)(g) are mixed to produce \(N{H_3}\) (g). Calculate the a...

50 kg of N2{N_2}(g) and 10 kg of H2{H_2}(g) are mixed to produce NH3N{H_3} (g). Calculate the amount of NH3N{H_3} formed. Identify the limiting reagent in the production of NH3N{H_3}.

Explanation

Solution

Limiting reagent is defined as the reactant in the chemical reaction which is consumed completely in the chemical reaction and stops any further reaction to take place. It is present in a limited amount.

Complete step by step answer:
Given,
Mass of N2{N_2} is 50 kg.
Mass of H2{H_2} is 10 kg.
The reaction between N2{N_2} and H2{H_2} is shown below.
N2(g)+3H2(g)2NH3{N_2}(g) + 3{H_2}(g) \to 2N{H_3}
In this reaction, nitrogen gas reacts with hydrogen gas to form ammonia.
Convert the mass in kg to gm.
50 kg = 50000 gm
10 kg = 10000 gm
The molecular mass of N2{N_2} is 28 g/mol
The molecular mass of H2{H_2} is 2 g/mol
The formula for calculating the moles is shown below.
n=mMn = \dfrac{m}{M}
Where,
*n is the moles of the compound
*m is the mass of the compound
*M is the molar mass of the compound
To calculate the moles of N2{N_2}, substitute the values in the above equation.
n=50000g28g/moln = \dfrac{{50000g}}{{28g/mol}}
n=1785.71mol\Rightarrow n = 1785.71mol
To calculate the moles of H2{H_2}, substitute the values in above expression.
n=10000g2g/moln = \dfrac{{10000g}}{{2g/mol}}
n=5000mol\Rightarrow n = 5000mol
As one mole of N2{N_2} is reacting with three moles of H2{H_2} to give two moles of NH3N{H_3}.
So 1785.71 mol of N2{N_2} will react with 3×1785.71=5357.133 \times 1785.71 = 5357.13 mol of H2{H_2} but 5000 mol is present.
Thus, the limiting reagent is hydrogen.
The amount of ammonia form when 5000 mol of H2{H_2} react =23×5000=3333.33= \dfrac{2}{3} \times 5000 = 3333.33 mol
The molar mass of ammonia is 17 g/mol

The molar mass of NH3N{H_3} is 17.031 g/mol.
To calculate the mass of NH3N{H_3}, substitute the values in the given equation.
3333.33=m17.0313333.33 = \dfrac{m}{{17.031}}
m=3333.33×17.031\Rightarrow m = 3333.33 \times 17.031
m=56769.94\Rightarrow m = 56769.94g
56.76 kg.
Thus, the amount of NH3N{H_3} formed is 56.76 kg.

Note:
The reactant which is available in the excess amount is called as excess reactant or reagent. The excess reagent remains in the reaction even after the reaction is completed.