If the percent yield for the following reaction is 60.0%,and 45.0 g of NO₂ are consumed in the reaction,how many grams of nitric acid,HNO₃(aq) ,are produced?
3 NO₂(g) + H₂O(l) → 2 HNO₃(aq) + NO(g)

Question

Quizplus · Accepted Answer

First, calculate the molar mass of $NO_2$ (NO₂) which is $14.01 + 2(16.00) = 46.01 \, \text{g/mol}$. Then, find the moles of $NO_2$ consumed: $45.0 \, \text{g} \div 46.01 \, \text{g/mol} = 0.978 \, \text{moles}$. According to the balanced equation, 3 moles of $NO_2$ produce 2 moles of $HNO_3$, so $0.978 \, \text{moles} \times \frac{2}{3} = 0.652 \, \text{moles of} \, HNO_3$. The molar mass of $HNO_3$ is $1.01 + 14.01 + 3(16.00) = 63.01 \, \text{g/mol}$, so the theoretical yield of $HNO_3$ is $0.652 \, \text{moles} \times 63.01 \, \text{g/mol} = 41.1 \, \text{g}$. With a percent yield of 60.0%, the actual yield is $41.1 \, \text{g} \times 0.60 = 24.6 \, \text{g}$.

If the Percent Yield for the Following Reaction Is 60