Question

Nitrogen dioxide reacts with water to produce nitric acid and nitrogen monoxide. How many grams of nitric acid can be prepared from 95.52 grams of nitrogen dioxide?

Answer 1

Answer:

87.22grams

Explanations:

The chemical reaction between Nitrogen dioxide and water to produce nitric acid and nitrogen monoxide is expressed as:

$3NO_2+H_2O\rightarrow2HNO_3+NO$

Determine the moles of nitrogen dioxide(NO2)

$\begin{gathered} mole=\frac{mass}{molar\text{ mass}} \\ mole=\frac{95.52}{46.0055} \\ mole\text{ of NO}_2=2.076moles \end{gathered}$

According to stoichiometric ratio, 3 moles of nitrogen dioxide produces 2moles of nitric acid, the moles of nitric acid required is expressed as;

$\begin{gathered} moles\text{ of }HNO_3=\frac{2}{3}\times2.076 \\ moles\text{ of }HNO_3=1.384moles \end{gathered}$

Determine the mass of nitric acid

$\begin{gathered} mass\text{ of HNO}_3=mole\times molar\text{ mass} \\ mass\text{ of HNO}_3=1.384\times63.01 \\ mass\text{ of HNO}_3=87.22grams \\ \end{gathered}$

Hence the mass of nitric acid required is 87.22grams