Question

the haber process is used to synthesize ammonia (nh3) on a large scale by the reaction of nitrogen gas with hydrogen gas as follows: n2(g) + 3h2(g) → 2nh3(g). consider a reaction between a 4.41 l flask containing nitrogen gas and a 17.3 l flask of hydrogen gas. both gases have a temperature of 290 k and the pressures inside both flasks is 1.15 bar. what mass of ammonia (in g) would you expect to be produced at temperature 290 k and a pressure of 1.15 bar?

Explanation:

Step1: Calculate moles of nitrogen and hydrogen using ideal - gas law

The ideal - gas law is $PV = nRT$. First, convert pressure to SI units: $P=1.15\ bar = 1.15\times10^{5}\ Pa$, $T = 290\ K$, $R=8.314\ J/(mol\cdot K)$.
For nitrogen in the $4.41\ L$ flask ($V_N = 4.41\ L=4.41\times10^{- 3}\ m^{3}$), $n_N=\frac{PV_N}{RT}=\frac{1.15\times10^{5}\times4.41\times10^{-3}}{8.314\times290}$.
For hydrogen in the $17.3\ L$ flask ($V_H = 17.3\ L = 17.3\times10^{-3}\ m^{3}$), $n_H=\frac{PV_H}{RT}=\frac{1.15\times10^{5}\times17.3\times10^{-3}}{8.314\times290}$.

Step2: Determine the limiting reactant

The balanced chemical equation is $N_2(g)+3H_2(g)
ightarrow2NH_3(g)$. The mole - ratio of $N_2$ to $H_2$ is $1:3$. Compare the moles of $N_2$ and $H_2$ to find the limiting reactant.

Step3: Calculate moles of ammonia produced

Based on the mole - ratio of the limiting reactant to ammonia in the balanced equation, calculate the moles of ammonia produced. If the limiting reactant is $N_2$, then $n_{NH_3}=2n_N$ (from the stoichiometry of the reaction where 1 mole of $N_2$ produces 2 moles of $NH_3$). If the limiting reactant is $H_2$, then $n_{NH_3}=\frac{2}{3}n_H$ (since 3 moles of $H_2$ produce 2 moles of $NH_3$).

Step4: Calculate mass of ammonia

The molar mass of ammonia $NH_3$ is $M = 14 + 3\times1=17\ g/mol$. Use the formula $m = nM$ to find the mass of ammonia, where $n$ is the moles of ammonia calculated in Step 3.

$n_N=\frac{1.15\times10^{5}\times4.41\times10^{-3}}{8.314\times290}=\frac{1.15\times4.41\times10^{2}}{8.314\times290}\approx0.21\ mol$
$n_H=\frac{1.15\times10^{5}\times17.3\times10^{-3}}{8.314\times290}=\frac{1.15\times17.3\times10^{2}}{8.314\times290}\approx0.83\ mol$

The mole - ratio of $N_2$ to $H_2$ in the reaction is $1:3$. For $0.21\ mol$ of $N_2$, we need $3\times0.21 = 0.63\ mol$ of $H_2$. Since we have $0.83\ mol$ of $H_2$, $N_2$ is the limiting reactant.

$n_{NH_3}=2n_N = 2\times0.21 = 0.42\ mol$

$m_{NH_3}=n_{NH_3}\times M_{NH_3}=0.42\ mol\times17\ g/mol = 7.14\ g$

Answer:

$7.14\ g$