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

Step-by-step 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}}`

=
`(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) * \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 =
`(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 * 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.