In the atmosphere, the air pollutant nitrogen dioxide (NO₂) reacts with water to produce nitric acid (HNO₃) , a component of acid rain. The reaction for the formation of nitric acid is:
3 NO₂(g) + H₂O (l) $\rightarrow$ 2 HNO₃(aq) + NO (g)
If 8.50 moles of nitrogen dioxide reacts with excess water, what is the actual yield (in grams) of NO if the percent yield is 89.4%?

Question

In the atmosphere, the air pollutant nitrogen dioxide (NO₂) reacts with water to produce nitric acid (HNO₃) , a component of acid rain. The reaction for the formation of nitric acid is:
3 NO₂(g) + H₂O (l)

\rightarrow

2 HNO₃(aq) + NO (g)
If 8.50 moles of nitrogen dioxide reacts with excess water, what is the actual yield (in grams) of NO if the percent yield is 89.4%?

Quizplus · Accepted Answer

First, calculate the theoretical yield of NO based on the stoichiometry of the reaction. For every 3 moles of $NO_2$, 1 mole of NO is produced. Thus, $8.50 \, \text{moles of} \, NO_2$ would theoretically produce $8.50 \times \frac{1}{3} = 2.83 \, \text{moles of} \, NO$. The molar mass of NO is approximately $14.01 \, (\text{N}) + 16.00 \, (\text{O}) = 30.01 \, \text{g/mol}$. Therefore, the theoretical yield in grams is $2.83 \, \text{moles} \times 30.01 \, \text{g/mol} = 85.0 \, \text{g}$. Given the percent yield is 89.4%, the actual yield is $85.0 \, \text{g} \times 0.894 = 76.0 \, \text{g}$.

In the Atmosphere, the Air Pollutant Nitrogen Dioxide (NO2) Reacts →\rightarrow→

In the Atmosphere, the Air Pollutant Nitrogen Dioxide (NO₂) Reacts $\rightarrow$