Question

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)

Answer 1

Answer: The amount of nitric acid produced is 30.81 grams

Step-by-step explanation:

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of nitrogen dioxide = 45.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

`\text{Moles of nitrogen dioxide}=(45g)/(46g/mol)=0.978mol`

For the given chemical equation:

`3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)`

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 =
`(2)/(3)* 0.978=0.652mol` of nitric acid

• Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.652 moles

Putting values in equation 1, we get:

`0.652mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.652mol* 63g/mol)=41.08g`

• To calculate the experimental yield of nitric acid, we use the equation:

`\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

Percentage yield of nitric acid = 75 %

Theoretical yield of nitric acid = 41.08 g

Putting values in above equation, we get:

`75\%=\frac{\text{Experimental yield of nitric acid}}{41.08g}* 100\\\\\text{Experimental yield of nitric acid}=(75* 41.08)/(100)=30.81g`

Hence, the amount of nitric acid produced is 30.81 grams