Chapter 11, Problem 46A

When solid copper is added to nitric acid, copper (ll) nitrate, nitrogen dioxide, and water are produced. Write the balanced chemical equation for the reaction. List six mole ratios for the reaction.

Interpretation Introduction

Interpretation:

The reaction equation for the reaction of solid copper with nitric acid to form copper (II) nitrate, nitrogen dioxide and water with six mole rations for the reaction needs to be determined.

Concept introduction:

A chemical reaction involves the conversion of certain molecules (reactant) to new substance (product) by bond breaking and making. One of the reactants, which is present in limited amount and determines the amount of product is known as limiting reactant. Whereas another reactant is known as excess reactant.

Answer to Problem 46A

$\begin{array}{l}{\text{Cu + 4 HNO}}_{\text{3 }}\stackrel{}{\to }{\text{ Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ + 2 NO}}_{\text{2}}{\text{ + 2 H}}_{\text{2}}^{}\text{O }\\ \frac{\text{1 mole Cu}}{{\text{4 moles HNO}}_{\text{3 }}}\text{ and }\frac{\text{1 mole Cu}}{{\text{1 mole Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\\ \frac{{\text{4 moles HNO}}_{\text{3 }}}{{\text{2 moles NO}}_{\text{2}}}\text{ and }\frac{{\text{4 moles HNO}}_{\text{3 }}}{{\text{1 mole Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\\ \frac{\text{1 mole Cu}}{{\text{2 moles NO}}_{\text{2}}}\text{and}\frac{\text{1 mole Cu}}{{\text{2 moles H}}_{\text{2}}^{}\text{O }}\end{array}$

Explanation of Solution

Given information:

Solid copper reacts with nitric acid to form copper (II) nitrate, nitrogen dioxide and water.

The balance chemical equation for solid copper with nitric acid to form copper (II) nitrate, nitrogen dioxide and water can be writen as:

${\text{Cu + 4 HNO}}_{\text{3 }}\stackrel{}{\to }{\text{ Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ + 2 NO}}_{\text{2}}{\text{ + 2 H}}_{\text{2}}^{}\text{O }$

The mole ratio relates the moles of reactant with other reactant and product molecules in the balance chemical equation. Thus, one mole ratio relates with another mole ratio. The mole ratio in the given balance reaction must be:

$\begin{array}{l}{\text{Cu + 4 HNO}}_{\text{3 }}\stackrel{}{\to }{\text{ Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ + 2 NO}}_{\text{2}}{\text{ + 2 H}}_{\text{2}}^{}\text{O }\\ \frac{\text{1 mole Cu}}{{\text{4 moles HNO}}_{\text{3 }}}\text{ and }\frac{\text{1 mole Cu}}{{\text{1 mole Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\\ \frac{{\text{4 moles HNO}}_{\text{3 }}}{{\text{2 moles NO}}_{\text{2}}}\text{ and }\frac{{\text{4 moles HNO}}_{\text{3 }}}{{\text{1 mole Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\\ \frac{\text{1 mole Cu}}{{\text{2 moles NO}}_{\text{2}}}\text{and}\frac{\text{1 mole Cu}}{{\text{2 moles H}}_{\text{2}}^{}\text{O }}\end{array}$

Conclusion

Mole ratio can calculate with the help of balance chemical reaction equation.