Chapter U4.25, Problem 4E

(a)

Interpretation Introduction

Interpretation: The type of given reaction is to be identified.

Concept introduction: The type of reaction, in which cations (positive ions) and anions (negative ions) of reacting species exchange their position to form two new compounds, is termed as double replacement reaction.

The general form of reaction is shown below:

Answer to Problem 4E

The given reaction is identified as double replacement reaction.

Explanation of Solution

The given is shown below:

  ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)+{\text{CaSO}}_{\text{4}}\left(aq\right)\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)+{\text{CaCO}}_{\text{3}}\left(s\right)$

In the given reaction, the sodium ion $\left({\text{Na}}^{+}\right)$ and calcium ion $\left({\text{Ca}}^{2+}\right)$ replace each other to form sodium sulfate and calcium carbonate. Therefore, the given reaction is a double replacement reaction.

(b)

Interpretation Introduction

Interpretation: The amount of sodium carbonate $\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ and calcium sulfate $\left({\text{CaSO}}_{\text{4}}\right)$ required to produce 4500 g of calcium carbonate $\left({\text{CaCO}}_{\text{3}}\right)$ is to be calculated.

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

Answer to Problem 4E

The amount of sodium carbonate $\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ and calcium sulfate $\left({\text{CaSO}}_{\text{4}}\right)$ required to produce $4500\text{g}$ of calcium carbonate $\left({\text{CaCO}}_{\text{3}}\right)$ is4765.3 g and 6120.85 g respectively.

Explanation of Solution

The given is shown below:

  ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)+{\text{CaSO}}_{\text{4}}\left(aq\right)\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)+{\text{CaCO}}_{\text{3}}\left(s\right)$

The mass of calcium carbonate $\left({\text{CaCO}}_{\text{3}}\right)$ is $4500\text{g}$ . The molar mass of calcium carbonate $\left({\text{CaCO}}_{\text{3}}\right)$ is $\text{100}\text{.0869}\text{g/mol}$ . The number of moles of calcium carbonate $\left({\text{CaCO}}_{\text{3}}\right)$ is calculated as shown below:

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\end{array}$

          $=\frac{4500\text{g}}{100.0869\text{g/mol}}$

          $=44.96\text{mol}$

In the given reaction, 1.0 mol of ${\text{CaCO}}_{\text{3}}$ is produced by 1.0 mol of sodium carbonate $\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ and calcium sulfate $\left({\text{CaSO}}_{\text{4}}\right)$ . Therefore, 44.96 mol of ${\text{CaCO}}_{\text{3}}$ will be produced by 44.96 mol of sodium carbonate $\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ and calcium sulfate $\left({\text{CaSO}}_{\text{4}}\right)$ .

Therefore, the required number of moles of sodium carbonate $\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ is 44.96 mol and the number of moles calcium sulfate $\left({\text{CaSO}}_{\text{4}}\right)$ is 44.96 mol.

The molar mass of sodium carbonate $\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ is 105.99 g/mol. The mass of sodium carbonate $\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ is calculated as shown below:

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\end{array}$

     $\begin{array}{c}44.96\text{mol}=\frac{\text{Mass}}{105.99\text{g/mol}}\end{array}$

       $\begin{array}{c}\text{Mass}=44.96\text{mol}\cdot 105.99\text{g/mol}\end{array}$

         $\begin{array}{c}=4765.3\text{g}\end{array}$

The molar mass of calcium sulfate $\left({\text{CaSO}}_{\text{4}}\right)$ is 136.14 g/mol. The mass of calcium sulfate $\left({\text{CaSO}}_{\text{4}}\right)$ is calculated as shown below:

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\end{array}$

     $\begin{array}{c}44.96\text{mol}=\frac{\text{Mass}}{136.14\text{g/mol}}\end{array}$

       $\begin{array}{c}\text{Mass}=44.96\text{mol}\cdot 136.14\text{g/mol}\end{array}$

         $\begin{array}{c}=6120.85\text{g}\end{array}$