Chapter 9, Problem 125A

A single-replacement reaction occurs between copper and silver nitrate. When 63.5 g of copper reacts with 339.8 g of silver nitrate, 215.8 g of silver is produced. Write a balanced chemical equation for this reaction. What other product formed? What is the mass of the second product?

Interpretation Introduction

Interpretation:

The balanced chemical reaction for the reaction between copper and silver nitrate needs to be determined. The other product and its mass also need to be calculated.

Concept introduction:

For balancing an equation first write the skeleton equation. After that count the atoms of both side, count each polyatomic ion as a single element. Find the correct coefficient and placed them in front of the atoms and checked both side element equal or not.

Answer to Problem 125A

The balance equation is ${\text{Cu+2AgNO}}_{\text{3}}\to 2\text{Ag+Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$.

Mass of the $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $187.5\text{g}$.

Explanation of Solution

Now, the skeleton equations for copper and silver nitrate is as follows:

${\text{Cu+AgNO}}_{\text{3}}\to \text{Ag+Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

In reactant side, number of Cu is 1, Ag is 1, N is 1 and oxygen atom is 3 while in product side number of Ag is 1, Cu is 1, N is 2 and oxygen is 6. To balance N and O in both sides give coefficient 2 to ${\text{AgNO}}_{\text{3}}$ and coefficient 2 to Ag in the product side. Therefore, the balance equation is as follows:

${\text{Cu+2AgNO}}_{\text{3}}\to 2\text{Ag+Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

Calculate mass of the $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ as follows:

Given,

$\begin{array}{l}\text{Mass}\text{of}\text{Cu=63}\text{.5g}\\ {\text{Mass of AgNO}}_{\text{3}}=339.8g\\ \text{Mass of}\text{Ag=215}\text{.8}\text{g}\\ \text{Mass of}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=?\end{array}$

$\begin{array}{c}\text{Mass}\text{of}{\text{Cu+Mass of AgNO}}_{\text{3}}\text{=Mass of}\text{Ag}+\text{Mass of}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\\ \text{63}\text{.5g+}339.8\text{g=215}\text{.8}\text{g}+\text{Mass of}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\\ \text{Mass of}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{=403}\text{.3g}-\text{215}\text{.8}\text{g}\\ =187.5\text{g}\end{array}$

Therefore, mass of the $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $187.5\text{g}$.

Conclusion

The balance equation is ${\text{Cu+2AgNO}}_{\text{3}}\to \text{Ag+Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$.

Mass of the $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $187.5\text{g}$.