the haber process is 3h2(g) + n2(g) --> 2nh3(g). if 34.5 g of nitrogen gas is reacted with 34.5 g of…

Explanation:

Step1: Calculate moles of reactants

Molar mass of $N_2$ is $M_{N_2}=2\times14\ g/mol = 28\ g/mol$. Moles of $N_2$, $n_{N_2}=\frac{m_{N_2}}{M_{N_2}}=\frac{34.5\ g}{28\ g/mol}\approx1.23\ mol$. Molar mass of $H_2$ is $M_{H_2}=2\times1\ g/mol = 2\ g/mol$. Moles of $H_2$, $n_{H_2}=\frac{m_{H_2}}{M_{H_2}}=\frac{34.5\ g}{2\ g/mol}=17.25\ mol$.

Step2: Determine limiting - reactant

From the balanced equation $3H_2(g)+N_2(g)\longrightarrow2NH_3(g)$, the mole - ratio of $H_2$ to $N_2$ is 3:1. For 1.23 mol of $N_2$, the moles of $H_2$ required is $n_{H_2\ required}=3\times n_{N_2}=3\times1.23\ mol = 3.69\ mol$. Since we have 17.25 mol of $H_2$ and only 3.69 mol are required for the complete reaction of $N_2$, $N_2$ is the limiting reactant.

Step3: Calculate moles of $NH_3$ produced

The mole - ratio of $N_2$ to $NH_3$ is 1:2. So, moles of $NH_3$ produced, $n_{NH_3}=2\times n_{N_2}=2\times1.23\ mol = 2.46\ mol$.

Step4: Calculate mass of $NH_3$ produced

Molar mass of $NH_3$ is $M_{NH_3}=14 + 3\times1\ g/mol=17\ g/mol$. Mass of $NH_3$, $m_{NH_3}=n_{NH_3}\times M_{NH_3}=2.46\ mol\times17\ g/mol = 41.82\ g$.

Answer:

41.82 g