Chapter U4.93, Problem 5E

(a)

Interpretation Introduction

Interpretation: The balanced chemical equation for the given reaction is to be written.

Concept introduction: The reactions having equal number of atoms of each element present on reactants and products side is said to be balanced chemical reaction.

Answer to Problem 5E

The balanced chemical equation for the given reaction is written below.

  ${\text{AgNO}}_{\text{3}}\left(aq\right)+\text{NaCl}\left(aq\right)\to \text{AgCl}\left(s\right)+{\text{NaNO}}_{\text{3}}\left(aq\right)$

Explanation of Solution

Silver nitrate $\left({\text{AgNO}}_{\text{3}}\right)$ reacts with sodium chloride $\left(\text{NaCl}\right)$ in aqueous solution to form solid silver chloride $\left(\text{AgCl}\right)$ and aqueous sodium nitrate $\left({\text{NaNO}}_{\text{3}}\right)$ . The chemical equation for this reaction is shown below.

  ${\text{AgNO}}_{\text{3}}\left(aq\right)+\text{NaCl}\left(aq\right)\to \text{AgCl}\left(s\right)+{\text{NaNO}}_{\text{3}}\left(aq\right)$

(b)

Interpretation Introduction

Interpretation: The number of moles of each reactant is to be calculated.

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

Answer to Problem 5E

The number of moles of silver nitrate $\left({\text{AgNO}}_{\text{3}}\right)$ is 0.0371 mol.

The number of moles of sodium chloride $\left(\text{NaCl}\right)$ is 0.077 mol.

Explanation of Solution

The chemical equation for this reaction is shown below.

  ${\text{AgNO}}_{\text{3}}\left(aq\right)+\text{NaCl}\left(aq\right)\to \text{AgCl}\left(s\right)+{\text{NaNO}}_{\text{3}}\left(aq\right)$

In the given reaction, the reactant is silver nitrate $\left({\text{AgNO}}_{\text{3}}\right)$ and sodium chloride $\left(\text{NaCl}\right)$ .

The mass of silver nitrate is 6.3 g. The molar mass of silver nitrate is 169.87 g/mol. The number of moles of silver nitrate is calculated as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{6.3\text{g}}{169.87\text{g/mol}}\\ =0.0371\text{mol}\end{array}$

The mass of sodium chloride (NaCl) is 4.5 g. The molar mass of sodium chloride (NaCl) is 58.44 g/mol. The number of moles of NaCl is calculated as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{4.5\text{g}}{58.44\text{g/mol}}\\ =0.077\text{mol}\end{array}$

Therefore, the number of moles of silver nitrate $\left({\text{AgNO}}_{\text{3}}\right)$ is 0.0371 mol. The number of moles of sodium chloride $\left(\text{NaCl}\right)$ is 0.077 mol.

(c)

Interpretation Introduction

Interpretation: The limiting reagent in the given reaction is to be identified.

Concept introduction: The reagent in reaction that controls the amount of product formed is termed as limiting reagent. After completion of chemical reaction, the limiting reagent is fully utilized.

Answer to Problem 5E

The limiting reagent in the given reaction is silver nitrate.

Explanation of Solution

The chemical equation for this reaction is shown below.

  ${\text{AgNO}}_{\text{3}}\left(aq\right)+\text{NaCl}\left(aq\right)\to \text{AgCl}\left(s\right)+{\text{NaNO}}_{\text{3}}\left(aq\right)$

The mass of silver nitrate is 6.3 g. The molar mass of silver nitrate is 169.87 g/mol. The number of moles of silver nitrate is calculated as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{6.3\text{g}}{169.87\text{g/mol}}\\ =0.0371\text{mol}\end{array}$

The mass of sodium chloride (NaCl) is 4.5 g. The molar mass of sodium chloride (NaCl) is 58.44 g/mol. The number of moles of NaCl is calculated as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{4.5\text{g}}{58.44\text{g/mol}}\\ =0.077\text{mol}\end{array}$

In the given reaction, 1.0 mol of silver nitrate reacts with 1.0 mol of sodium chloride. Therefore, 0.077 mol of sodium chloride will react with 0.077 mol of silver nitrate but the available amount is 0.0371 mol of silver nitrate. Therefore, silver nitrate is the limiting reagent.

(d)

Interpretation Introduction

Interpretation: The amount of each product formed is to be calculated.

Concept introduction: The reagent in reaction that controls the amount of product formed is termed as limiting reagent. After completion of chemical reaction, the limiting reagent is fully utilized.

Answer to Problem 5E

The mass of silver chloride is 5.317 g.

The mass of sodium nitrate is 3.15 g.

Explanation of Solution

The chemical reaction is shown below.

  ${\text{AgNO}}_{\text{3}}\left(aq\right)+\text{NaCl}\left(aq\right)\to \text{AgCl}\left(s\right)+{\text{NaNO}}_{\text{3}}\left(aq\right)$

The product of the given reaction is silver chloride $\left(\text{AgCl}\right)$ and sodium nitrate $\left({\text{NaNO}}_{\text{3}}\right)$ .

The mass of silver nitrate is 6.3 g. The molar mass of silver nitrate is 169.87 g/mol. The number of moles of silver nitrate is calculated as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{6.3\text{g}}{169.87\text{g/mol}}\\ =0.0371\text{mol}\end{array}$

The mass of sodium chloride (NaCl) is 4.5 g. The molar mass of sodium chloride (NaCl) is 58.44 g/mol. The number of moles of NaCl is calculated as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{4.5\text{g}}{58.44\text{g/mol}}\\ =0.077\text{mol}\end{array}$

In the given reaction, 1.0 mol of silver nitrate reacts with 1.0 mol of sodium chloride. Therefore, 0.077 mol of sodium chloride will react with 0.077 mol of silver nitrate but the available amount is 0.0371 mol of silver nitrate. Therefore, silver nitrate is the limiting reagent.

From the given balanced given reaction, 1.0 mol of silver nitrate produces 1.0 mol of silver chloride. Therefore, 0.0371 mol of silver nitrate will produce 0.0371 mol of silver chloride.

The molar mass of silver chloride is 143.32 g/mol. The mass of silver chloride is calculated as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ 0.037\text{1}\text{mol}=\frac{\text{Mass}}{143.32\text{g/mol}}\\ \text{Mass}=0.037\text{1}\text{mol}\cdot 143.32\text{g/mol}\\ =\text{5}\text{.317}\text{g}\end{array}$

In the given reaction, 1.0 mol of silver nitrate produces 1.0 mol of silver chloride. Therefore, 0.0371 mol of silver nitrate will produce 0.0371 mol of sodium nitrate.

The molar mass of sodium nitrate is 85.0 g/mol. The mass of sodium nitrate is calculated as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ 0.037\text{1}\text{mol}=\frac{\text{Mass}}{85.0\text{g/mol}}\\ \text{Mass}=0.037\text{1}\text{mol}\cdot 85.0\text{g/mol}\\ =\text{3}\text{.15}\text{g}\end{array}$

(e)

Interpretation Introduction

Interpretation: The amount of excess reactant after completion of reaction is to be calculated.

Concept introduction: The reagent in reaction that controls the amount of product formed is termed as limiting reagent. After completion of chemical reaction, the limiting reagent is fully utilized.

Answer to Problem 5E

The amount of excess reactant after completion of reaction is 2.33 g.

Explanation of Solution

The chemical equation for this reaction is shown below.

  ${\text{AgNO}}_{\text{3}}\left(aq\right)+\text{NaCl}\left(aq\right)\to \text{AgCl}\left(s\right)+{\text{NaNO}}_{\text{3}}\left(aq\right)$

The number of moles of silver nitrate $\left({\text{AgNO}}_{\text{3}}\right)$ and sodium chloride $\left(\text{NaCl}\right)$ is 0.0371 mol and 0.077 mol respectively.

Here, the limiting reagent is silver nitrate $\left({\text{AgNO}}_{\text{3}}\right)$ . In the given reaction, 1.0 mol of silver nitrate reacts with 1.0 mol of sodium chloride. Therefore, 0.077 mol of sodium chloride will react with 0.077 mol of silver nitrate. Theremaining number of moles of sodium chloride is calculated as shown below.

  $\begin{array}{c}\text{Remaining}\text{mole}\text{of}\text{NaCl}=0.077\text{mol}-0.0371\text{mol}\\ =0.0399\text{mol}\end{array}$

The molar mass of NaCl is 58.44 g/mol. Therefore, the mass of NaCl is calculated as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ 0.0399\text{mol}=\frac{\text{Mass}}{58.4\text{4}\text{g/mol}}\\ \text{Mass}=0.0399\text{mol}\times 58.4\text{4}\text{g/mol}\\ =\text{2}\text{.33}\text{g}\end{array}$

Therefore, the remaining mass of NaCl is 2.33 g.