the reaction between nitrogen and hydrogen to form ammonia is carried out in a flask containing 30.5 g of n₂…

Explanation:

Step1: Calculate moles of reactants

Molar mass of $N_2$ is $M_{N_2}=28\ g/mol$, moles of $N_2$, $n_{N_2}=\frac{30.5\ g}{28\ g/mol}\approx1.09\ mol$. Molar mass of $H_2$ is $M_{H_2} = 2\ g/mol$, moles of $H_2$, $n_{H_2}=\frac{8.65\ g}{2\ g/mol}=4.325\ mol$.

Step2: Determine limiting reactant

From the balanced equation $N_2(g)+3H_2(g)\rightarrow2NH_3(g)$, the mole - ratio of $N_2$ to $H_2$ is 1:3. For 1.09 mol of $N_2$, we need $n_{H_2\ required}=3\times1.09\ mol = 3.27\ mol$ of $H_2$. Since we have 4.325 mol of $H_2$, $N_2$ 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, $n_{NH_3}=2\times n_{N_2}=2\times1.09\ mol = 2.18\ mol$.

Step4: Calculate theoretical yield of $NH_3$

Molar mass of $NH_3$ is $M_{NH_3}=17\ g/mol$. Theoretical yield of $NH_3$, $m_{NH_3}=n_{NH_3}\times M_{NH_3}=2.18\ mol\times17\ g/mol = 37.06\ g\approx37.1\ g$.

Step5: Calculate percent yield for Part B

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100%$. Given actual yield = 32.4 g and theoretical yield = 37.06 g. Percent yield=$\frac{32.4\ g}{37.06\ g}\times100%\approx87.3%$.

Answer:

For Part A: 37.1 g $NH_3$ For Part B: 87.3%