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 masses

The molar mass of $NH_3$ is $M_{NH_3}=14 + 3\times1=17\ g/mol$. The molar mass of $O_2$ is $M_{O_2}=2\times16 = 32\ g/mol$, and the molar mass of $NO$ is $M_{NO}=14+16 = 30\ g/mol$.

Step3: Determine the limiting - reactant

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 $O_2$, $n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{150\ g}{32\ g/mol}\approx4.69\ mol$.
From the balanced equation, the mole - ratio of $NH_3$ to $O_2$ is $\frac{n_{NH_3}}{n_{O_2}}=\frac{4}{5}$.
For $n_{O_2} = 4.69\ mol$, the moles of $NH_3$ required for complete reaction of $O_2$ is $n_{NH_3\ required}=\frac{4}{5}\times4.69\ mol = 3.75\ mol$. Since $8.82\ mol>3.75\ mol$, $O_2$ is the limiting reactant.

Step4: 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 from $4.69\ mol$ of $O_2$ is $n_{NO}=\frac{4}{5}\times4.69\ mol = 3.75\ mol$.
The theoretical mass of $NO$, $m_{NO\ theoretical}=n_{NO}\times M_{NO}=3.75\ mol\times30\ g/mol = 112.5\ g$.

Step5: Calculate the percent yield

The percent yield is given by the formula $\text{Percent yield}=\frac{m_{NO\ actual}}{m_{NO\ theoretical}}\times100\%$.
Substituting $m_{NO\ actual} = 87\ g$ and $m_{NO\ theoretical}=112.5\ g$ into the formula, we get $\text{Percent yield}=\frac{87\ g}{112.5\ g}\times100\%\approx77\%$.

Answer:

B. 77%