Question

If the percent yield for the following reaction is 75.0%, and 25.0 g of NO₂ are consumed in the reaction, how many grams of nitric acid, HNO₃(aq) are produced?

Answer 1

Answer:

17.1195 grams of nitric acid are produced.

Explanation:

$3NO_2+H_2O\rightarrow 2HNO_3+NO$

Moles of nitrogen dioxide :

$\frac{25.0 g}{56 g/mol}=0.5434 mol$

According to reaction 3 moles of nitrogen dioxides gives 2 moles of nitric acid.

Then 0.5434 moles of nitrogen dioxides will give:

$\frac{2}{3}\times 0.5434 mol=0.3623 mol$ of nitric acid.

Mass of 0.3623 moles of nitric acid :

$0.3623 mol\times 63 g/mol=22.8260 g$

Theoretical yield = 22.8260 g

Experimental yield = ?

$\%Yield=\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100$

$75\%=\frac{\text{Experimental yield}}{22.8260 g}$

Experimental yield of nitric acid = 17.1195 g