Chapter U4, Problem SV2RE

(a)

Interpretation Introduction

Interpretation:The given chemical reaction is to be balanced.

Concept introduction:The number of atoms on the left hand side and right hand side of the equation is equal in balanced chemical equation. The amount of product formed in the reaction depends upon the amount of limiting reactant.

Answer to Problem SV2RE

The given chemical reaction is balanced as shown below:

  $\underset{\_}{3}{\text{CaCl}}_{\text{2}}\left(aq\right)+\underset{\_}{2}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)\to \underset{\_}{1}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+\underset{\_}{6}\text{NaCl}\left(aq\right)$

Explanation of Solution

The given chemical reaction is shown below:

  $\underset{\_}{}\underset{\_}{}{\text{CaCl}}_{\text{2}}\left(aq\right)+\underset{\_}{}\underset{\_}{}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)\to \underset{\_}{}\underset{\_}{}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+\underset{\_}{}\underset{\_}{}\text{NaCl}\left(aq\right)$

In the above reaction, the number of calcium ion is three on the product side. Therefore, multiply ${\text{CaCl}}_2$ by 3 to balance calcium ion. Now phosphate ion is two on the product side. Therefore, multiply sodium phosphate $\left({\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\right)$ to 2. Now the number of sodium ion and chloride ion is six on the reactant side. Therefore, multiply $\text{NaCl}$ by 6 to balance.

Therefore, the given chemical reaction is balanced as shown below:

  $\underset{\_}{3}{\text{CaCl}}_{\text{2}}\left(aq\right)+\underset{\_}{2}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)\to \underset{\_}{1}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+\underset{\_}{6}\text{NaCl}\left(aq\right)$

(b)

Interpretation Introduction

Interpretation: The maximum amount of calcium phosphate, that can be produced from $13.5\text{g}$ of ${\text{CaCl}}_{\text{2}}$ and $9.80\text{g}$ of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ , is to be calculated.

Concept introduction:The amount of product formed in the reaction depends upon the amount of limiting reactant. After completion of chemical reaction, the limiting reagent is fully utilized.

Answer to Problem SV2RE

The maximum amount of calcium phosphate, that can be produced from $13.5\text{g}$ of ${\text{CaCl}}_{\text{2}}$ and $9.80\text{g}$ of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ , is $9.30\text{g}$ .

Explanation of Solution

The number of moles of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ is calculated as

  $n_{{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}}=\frac{m_{{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}}}{M_{{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}}}$

Where,

• $m_{{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}}$ is the given mass of sodium phosphate.
• $M_{{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}}$ is the molar mass of sodium phosphate.

The mass of sodium phosphate is $9.80\text{g}$ . The molar mass of sodium phosphate is $163.94\text{g/mol}$ .

Substitute the given mass and molar mass of sodium phosphate in the above formula.

  $\begin{array}{c}{n}_{{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}}=\frac{9.80\text{g}}{163.94\text{g/mol}}\\ =0.06\text{mol}\end{array}$

The number of moles of ${\text{CaCl}}_2$ is calculated as

  $n_{{\text{CaCl}}_2}=\frac{m_{{\text{CaCl}}_2}}{M_{{\text{CaCl}}_2}}$

Where,

• $m_{{\text{CaCl}}_2}$ is the given mass of calcium chloride.
• $M_{{\text{CaCl}}_2}$ is the molar mass of calcium chloride.

The mass of calcium chloride is $13.5\text{g}$ . The molar mass of calcium chloride is $110.98\text{g/mol}$ .

Substitute the given mass and molar mass ofcalcium chloride in the above formula.

  $\begin{array}{c}{n}_{{\text{CaCl}}_2}=\frac{13.5\text{g}}{110.98\text{g/mol}}\\ =0.12\text{mol}\end{array}$

The balanced chemical reaction is shown below:

  $\underset{\_}{3}{\text{CaCl}}_{\text{2}}\left(aq\right)+\underset{\_}{2}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)\to \underset{\_}{1}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+\underset{\_}{6}\text{NaCl}\left(aq\right)$

The molar ratio of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ and ${\text{CaCl}}_2$ is $2:3$ . Therefore, sodium phosphate is the limiting reactant.

The balanced chemical equation shows that the molar ratio between sodium phosphate and calcium phosphate is $2:1$ . Therefore, the number of moles of calcium phosphate produced by $0.06\text{mol}$ of sodium phosphate is $0.03\text{mol}$ .

The amount of calcium phosphate is calculated as

  $m_{{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_2}=n_{{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_2}\times M_{{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_2}$

Where,

• $n_{{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_2}$ is the number of moles of calcium phosphate.
• $M_{{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_2}$ is the molar mass of calcium phosphate $\left(310.18\text{g/mol}\right)$ .

Substitute the number of moles and molar mass of calcium phosphate in the above formula.

  $\begin{array}{c}{m}_{{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_2}=0.03\text{mol}\times 310.18\text{g/mol}\\ =9.30\text{g}\end{array}$

Therefore, the maximum amount of calcium phosphate, that can be produced from $13.5\text{g}$ of ${\text{CaCl}}_{\text{2}}$ and $9.80\text{g}$ of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ , is $9.30\text{g}$ .

(c)

Interpretation Introduction

Interpretation: The amount of excess reagent is to be calculated.

Concept introduction: The amount of product formed in the reaction depends upon the amount of limiting reactant.After completion of chemical reaction, the limiting reagent is fully utilized.

Answer to Problem SV2RE

The amount of excess reagent is $\text{3}\text{.3}\text{g}$ .

Explanation of Solution

The balanced chemical equation is shown below:

  $\underset{\_}{3}{\text{CaCl}}_{\text{2}}\left(aq\right)+\underset{\_}{2}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)\to \underset{\_}{1}{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+\underset{\_}{6}\text{NaCl}\left(aq\right)$

The number of moles of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $0.06\text{mol}$ . The number of moles of ${\text{CaCl}}_2$ is $0.12\text{mol}$ . It is already known that sodium phosphate is the limiting reactant. Now 2.0 mol of sodium phosphate reacts with 3.0 mol of calcium chloride. Therefore, $0.06\text{mol}$ of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ will reacts with-

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{CaCl}}_{\text{2}}=0.06\text{mol}\text{of}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(\frac{3.0\text{mol}\text{of}{\text{CaCl}}_{\text{2}}}{2.0\text{mol}\text{of}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}}\right)\\ =0.09\text{mol}\text{of}{\text{CaCl}}_{\text{2}}\end{array}$

Now, the remaining amount of calcium chloride is calculated as shown below:

  $\begin{array}{c}\text{Remaining}\text{amount}\text{of}{\text{CaCl}}_{\text{2}}=0.12\text{mol}-0.09\text{mol}\\ =\text{0}\text{.03}\text{mol}\end{array}$

The molar mass of calcium chloride is $110.98\text{g/mol}$ . The amount of excess reagent is calculated as shown below:

  $\begin{array}{c}{m}_{{\text{CaCl}}_{\text{2}}}=n_{{\text{CaCl}}_{\text{2}}}\cdot M_{{\text{CaCl}}_{\text{2}}}\\ m_{{\text{CaCl}}_{\text{2}}}=0.03\text{mol}\times 110.98\text{g/mol}\\ =3.3\text{g}\end{array}$

Therefore, the amount of excess reagent is $\text{3}\text{.3}\text{g}$ .

(d)

Interpretation Introduction

Interpretation: The percent yield of calcium phosphate is to be calculated.

Concept introduction: The ratio of actual yield to the theoretical yield is called percentage yield of the reaction.

The expression of the percent yield is shown below:

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

Answer to Problem SV2RE

The percent yield of calcium phosphate is $90.81\%$ .

Explanation of Solution

The expression of the percent yield is shown below:

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

The value of actual yield is $8.20\text{g}$ and the value of theoretical yield is $\text{9}\text{.30}\text{g}$ .

Substitute the given value in above equation.

  $\begin{array}{c}\text{Percent}\text{yield}=\frac{\text{Actual}\text{yield}}{\text{Theoretical}\text{yield}}\times 100\\ =\frac{8.20\text{g}}{9.30\text{g}}\times 100\\ =90.81\%\end{array}$

Therefore, the percent yield of calcium phosphate is $90.81\%$ .