If the percent yield for the following reaction is 75.0%, 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)

The amount of nitric acid produced is 30.81 grams

Explanation:

To calculate the number of moles, we use the equation:

    .....(1)

Given mass of nitrogen dioxide = 45.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

For the given chemical equation:

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 moles of nitric acid

So, 0.978 moles of nitrogen dioxide will produce = of nitric acid

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.652 moles

Putting values in equation 1, we get:

Percentage yield of nitric acid = 75 %

Theoretical yield of nitric acid = 41.08 g

Putting values in above equation, we get:

Hence, the amount of nitric acid produced is 30.81 grams

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm