Question

ammonia can also be synthesized by the reaction: 3h₂(g) + n₂(g) → 2nh₃(g). what is the theoretical yield of ammonia, in kg, that we can synthesize from 5.22 kg of h₂ and 31.5 kg of n₂? express the mass in kilograms to three significant figures. theoretical yield of nh₃ =

Explanation:

Step1: Calculate moles of reactants

The molar mass of $H_2$ is $M_{H_2}=2.016\ g/mol$, and the mass of $H_2$ is $m_{H_2} = 5.22\times10^{3}\ g$. The number of moles of $H_2$, $n_{H_2}=\frac{m_{H_2}}{M_{H_2}}=\frac{5.22\times 10^{3}\ g}{2.016\ g/mol}\approx2590\ mol$. The molar mass of $N_2$ is $M_{N_2}=28.02\ g/mol$, and the mass of $N_2$ is $m_{N_2}=31.5\times 10^{3}\ g$. The number of moles of $N_2$, $n_{N_2}=\frac{m_{N_2}}{M_{N_2}}=\frac{31.5\times 10^{3}\ g}{28.02\ g/mol}\approx1124\ mol$.

Step2: Determine the limiting reactant

From the balanced - chemical equation $3H_2(g)+N_2(g)
ightarrow2NH_3(g)$, the mole - ratio of $H_2$ to $N_2$ is 3:1. For $n_{N_2} = 1124\ mol$ of $N_2$, the amount of $H_2$ required is $n_{H_2\ required}=3\times n_{N_2}=3\times1124\ mol = 3372\ mol$. But we have only $2590\ mol$ of $H_2$. So, $H_2$ is the limiting reactant.

Step3: Calculate moles of $NH_3$ produced

The mole - ratio of $H_2$ to $NH_3$ is 3:2. If $n_{H_2}=2590\ mol$, then the number of moles of $NH_3$ produced, $n_{NH_3}=\frac{2}{3}n_{H_2}=\frac{2}{3}\times2590\ mol\approx1727\ mol$.

Step4: Calculate the mass of $NH_3$ produced

The molar mass of $NH_3$ is $M_{NH_3}=17.04\ g/mol$. The mass of $NH_3$ produced, $m_{NH_3}=n_{NH_3}\times M_{NH_3}=1727\ mol\times17.04\ g/mol\approx29438\ g$. Converting to kg, $m_{NH_3}=29.4\ kg$.

Answer:

$29.4$