Chapter 11.4, Problem 30PP

Interpretation Introduction

Interpretation:

The balanced chemical equation for the reaction between copper wire and silver nitrate to produce silver crystals and copper(II) nitrate should be determined.

Concept introduction:

Chemical equation is written in such a way that the symbolic representation of reaction represents the reaction taking place in the system. The reactants are written on the left-hand side and the products are written on the right-hand side of the equation and are separated by an arrow, two or more reactants and products are separated by “+”. The reactions for those the number of atoms of each element in the reactant and in the product, side are equal, such reactions are said to be a balanced chemical equation.

Answer to Problem 30PP

${\text{Cu(s) + 2AgNO}}_3\text{(aq) }\to {\text{ 2Ag(s) + Cu(NO}}_3{\text{)}}_2\text{(aq)}$

Explanation of Solution

The elemental formula for copper is $\text{Cu}$, for silver is $\text{Ag}$, molecular formula of silver nitrate is ${\text{AgNO}}_3$, and for copper(II) nitrate is ${\text{Cu(NO}}_3{\text{)}}_2$.

The physical state of the reactants and products are shown by writing the symbols after each reactant and product in the reaction. The symbol “l” represents liquid, “aq” represents aqueous, “s” represents solid, and “g” represents gas.

The reaction between copper wire and silver nitrate to produce silver crystals and copper(II) nitrate is written as:

${\text{Cu(s) + AgNO}}_3\text{(aq) }\to {\text{Ag(s) + Cu(NO}}_3{\text{)}}_2\text{(aq)}$

This reaction is not balanced as the number of N and O atoms on the reactant side is 1 and 3 whereas in the product side is 2 and 6 respectively. So, in order to balance the reaction, coefficient 2 is written before ${\text{AgNO}}_3$ in the reactant side and 2 before $\text{Ag}$ on the product side. Hence, the balanced reaction is:

${\text{Cu(s) + 2AgNO}}_3\text{(aq) }\to {\text{ 2Ag(s) + Cu(NO}}_3{\text{)}}_2\text{(aq)}$

Interpretation Introduction

Interpretation:

The theoretical yield of silver should be calculated when 20.0 g of copper is used.

Concept introduction:

The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

The ratio of amount of any two compounds involved in a chemical reaction in terms of mole is said to be mole ratio, and used as a conversion factor between reactants and products.

Answer to Problem 30PP

The theoretical yield of silver is $\text{67}\text{.96 g}$.

Explanation of Solution

The complete balanced reaction is:

${\text{Cu(s) + 2AgNO}}_3\text{(aq) }\to {\text{ 2Ag(s) + Cu(NO}}_3{\text{)}}_2\text{(aq)}$

The molar mass of copper is 63.55 g/mol. The number of moles of copper is calculated as shown:

$\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Number of moles}=\frac{20.0\text{ g}}{\text{63}\text{.55 g/mol}}\\ \text{Number of moles}=0.315\text{ mol}\end{array}$

From the balanced chemical reaction, the mole ratio of copper and silver is 1:2 that means 1 mole of copper produces 2 moles of silver. So, the number of moles of silver produced from 0.315 mol of copper is:

$0.315\text{ mol of Cu×}\frac{\text{2 mol of Ag}}{\text{1 mol of Cu}}\text{ = }0.63\text{ mol of Ag}$

Now, the mass of silver produced from 20.0 g of copper is calculated as:

The molar mass of silver is 107.87 g/mol. So,

$\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ 0.63\text{ mol}=\frac{\text{Mass}}{\text{107}\text{.87 g/mol}}\\ \text{Mass = 0}\text{.63 mol}\times 107.87\text{ g/mol}\\ \text{Mass = 67}\text{.96 g}\end{array}$

Hence, the theoretical yield of silver is $\text{67}\text{.96 g}$.

Interpretation Introduction

Interpretation:

The percent yield for the given reaction needs to be calculated when 60.0 g of silver is recovered from the reaction.

Concept introduction:

In a chemical reaction, the amount of product formed in the relation to the amount of reactant consumed is term as yield. Yield is generally expressed in percentage.

Answer to Problem 30PP

The percent yield for the given reaction is $88.29\text{ \%}$ of $\text{Ag}$.

Explanation of Solution

The complete balanced reaction is:

${\text{Cu(s) + 2AgNO}}_3\text{(aq) }\to {\text{ 2Ag(s) + Cu(NO}}_3{\text{)}}_2\text{(aq)}$

The percent yield is calculated using formula:

$\text{Percent}\text{yield}=\frac{\text{Actual}\text{yield}}{\text{Theoretical}\text{yield}}\times 100$

Since, the theoretical yield of silver is $\text{67}\text{.96 g}$ and actual yield is 60.0 g so,

Substituting the values:

$\begin{array}{l}\text{Percent yield}=\frac{\text{60}\text{.0 g}}{\text{67}\text{.96 g}}\times 100\text{ \%}\\ \text{Percent yield}=88.29\text{ \%}\end{array}$

Hence, the percent yield for the given reaction is $88.29\text{ \%}$ of $\text{Ag}$.