Question

If the percent yield for the following reaction is 80.37%, and 63.21 g of NO2 are used in the reaction mixture, how many grams of nitric acid, HNO3(aq), are produced in the experiment? 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) Molar mass NO2 = 46.01 g/mol and HNO3 = 63.01 g/mol.

Answer 1

Explanation:

It is given that mass of $NO_{2}$ is 46.01 g/mol and mass of $HNO_{3}$ is 63.01 g/mol.

So, in 1 mole calculate the number of moles present in 63.21 g/mol of $NO_{2}$ as follows.

                 No. of moles = $\frac{mass}{\text{molar mass}}$

                                       = $\frac{63.21 g}{46.01 g/mol}$

                                       = 1.37 mol

Now, according to the reaction equation we require 3 moles of $NO_{2}$ to make 2 moles of $HNO_{3}$.

        $1.37 mol NO_{3} \times \frac{\text{2 moles} HNO_{3}}{\text{3 moles} NO_{2}}$

                          = 0.91 mol

Hence, calculate the mass of $HNO_{3}$ as follows.

                  No. of moles = $\frac{mass}{\text{molar mass}}$

                           0.91 mol = $\frac{mass}{63 g/mol}$

                          mass = 57.33 g

Now, mass of $HNO_{3}$ actually produced by 80.37% yield is calculated as follows.

                     $0.8037 \times 57.33 g$

                       = 46.07 g

Thus, we can conclude that 46.07 g of nitric acid, $HNO_{3}(aq)$, are produced in the experiment.