Question

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

Answer 1

Hint: 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 ${N_2}$ is 50 kg.
Mass of ${H_2}$ is 10 kg.
The reaction between ${N_2}$ and ${H_2}$ is shown below.
${N_2}(g) + 3{H_2}(g) \to 2N{H_3}$
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 ${N_2}$ is 28 g/mol
The molecular mass of ${H_2}$ is 2 g/mol
The formula for calculating the moles is shown below.
$n = \dfrac{m}{M}$
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 ${N_2}$, substitute the values in the above equation.
$n = \dfrac{{50000g}}{{28g/mol}}$
$\Rightarrow n = 1785.71mol$
To calculate the moles of ${H_2}$, substitute the values in above expression.
$n = \dfrac{{10000g}}{{2g/mol}}$
$\Rightarrow n = 5000mol$
As one mole of ${N_2}$ is reacting with three moles of ${H_2}$ to give two moles of $N{H_3}$.
So 1785.71 mol of ${N_2}$ will react with $3 \times 1785.71 = 5357.13$ mol of ${H_2}$ but 5000 mol is present.
Thus, the limiting reagent is hydrogen.
The amount of ammonia form when 5000 mol of ${H_2}$ react $= \dfrac{2}{3} \times 5000 = 3333.33$ mol
The molar mass of ammonia is 17 g/mol


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