Question

1. the first step in the ostwald process for producing nitric acid is nh₃(g) + o₂(g) → no(g) + h₂o(g) (unbalanced reaction) if the reaction of 150. g of ammonia with 150. g of oxygen gas yields 87. g of nitric oxide (no), what is the percent yield of this reaction? a) 62% b) 77% c) 54% d) 84% e) 35%

Explanation:

Step1: Balance the chemical equation

$$4NH_3(g)+5O_2(g) ightarrow4NO(g) + 6H_2O(g)$$

Step2: Calculate the molar - mass of $NH_3$ and $O_2$

The molar - mass of $NH_3$ is $M_{NH_3}=14 + 3\times1=17\ g/mol$, and the molar - mass of $O_2$ is $M_{O_2}=2\times16 = 32\ g/mol$.

Step3: Determine the number of moles of $NH_3$ and $O_2$

The number of moles of $NH_3$, $n_{NH_3}=\frac{m_{NH_3}}{M_{NH_3}}=\frac{150\ g}{17\ g/mol}\approx8.82\ mol$.
The number of moles of $O_2$, $n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{150\ g}{32\ g/mol}\approx4.69\ mol$.

Step4: Identify the limiting reactant

From the balanced equation, the mole ratio of $NH_3$ to $O_2$ is $\frac{n_{NH_3}}{n_{O_2}}=\frac{4}{5}$.
If all $8.82\ mol$ of $NH_3$ react, the moles of $O_2$ required is $n_{O_2\ required}=8.82\ mol\times\frac{5}{4}=11.025\ mol$. But we only have $4.69\ mol$ of $O_2$.
If all $4.69\ mol$ of $O_2$ react, the moles of $NH_3$ required is $n_{NH_3\ required}=4.69\ mol\times\frac{4}{5}=3.752\ mol$. So, $O_2$ is the limiting reactant.

Step5: Calculate the theoretical yield of $NO$

From the balanced equation, the mole ratio of $O_2$ to $NO$ is $\frac{n_{NO}}{n_{O_2}}=\frac{4}{5}$.
The moles of $NO$ produced based on the limiting reactant $O_2$ is $n_{NO}=\frac{4}{5}\times n_{O_2}=\frac{4}{5}\times4.69\ mol = 3.752\ mol$.
The molar - mass of $NO$ is $M_{NO}=14 + 16=30\ g/mol$.
The theoretical yield of $NO$, $m_{NO\ theoretical}=n_{NO}\times M_{NO}=3.752\ mol\times30\ g/mol = 112.56\ g$.

Step6: Calculate the percent yield

The percent yield is given by the formula $\text{Percent yield}=\frac{m_{NO\ actual}}{m_{NO\ theoretical}}\times100\%$.
We know $m_{NO\ actual}=87\ g$ and $m_{NO\ theoretical}=112.56\ g$.
$\text{Percent yield}=\frac{87\ g}{112.56\ g}\times100\%\approx77\%$.

Answer:

B. 77%