Question

part a if the percent yield for the following reaction is 75.0%, and 40.0 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) 36.5 g 27.4 g 61.6 g 48.7 g

Explanation:

Step1: Calculate molar - mass of NO₂ and HNO₃

The molar - mass of NO₂ ($M_{NO_2}$) is $M_{NO_2}=14 + 2\times16=46\ g/mol$, and the molar - mass of HNO₃ ($M_{HNO_3}$) is $M_{HNO_3}=1 + 14+3\times16 = 63\ g/mol$.

Step2: Determine the theoretical yield of HNO₃

From the balanced chemical equation $3NO_2(g)+H_2O(l)
ightarrow2HNO_3(aq)+NO(g)$, the mole ratio of $NO_2$ to $HNO_3$ is $n_{NO_2}:n_{HNO_3}=3:2$.
The number of moles of $NO_2$ consumed, $n_{NO_2}=\frac{m_{NO_2}}{M_{NO_2}}=\frac{40.0\ g}{46\ g/mol}=\frac{40}{46}\ mol$.
The theoretical number of moles of $HNO_3$, $n_{HNO_3}^{theo}=\frac{2}{3}n_{NO_2}=\frac{2}{3}\times\frac{40}{46}\ mol$.
The theoretical mass of $HNO_3$, $m_{HNO_3}^{theo}=n_{HNO_3}^{theo}\times M_{HNO_3}=\frac{2}{3}\times\frac{40}{46}\ mol\times63\ g/mol=\frac{2\times40\times63}{3\times46}\ g=\frac{5040}{138}\ g\approx36.52\ g$.

Step3: Calculate the actual yield of HNO₃

The percent - yield formula is $\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%$.
We know that the percent yield is $75.0\%$, or $0.75$ in decimal form.
Let the actual mass of $HNO_3$ be $m_{HNO_3}^{act}$. Then $0.75=\frac{m_{HNO_3}^{act}}{36.52\ g}$.
$m_{HNO_3}^{act}=0.75\times36.52\ g = 27.4\ g$.

Answer:

$27.4\ g$