Chapter 12, Problem 73A

(a)

Interpretation Introduction

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

Concept Introduction: Reactants, products, and an arrow indicating the reaction's direction make up a chemical equation. A balanced chemical equation is one in which the number of atoms in each of the molecules is the same on both sides of the equation.

Answer to Problem 73A

The balanced equation for the reaction is as follows:

  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{+Ca}{\left(\text{OH}\right)}_{\text{2}}\to {\text{CaSO}}_{\text{4}}{\text{+H}}_{\text{2}}\text{O}$

Explanation of Solution

Sulfuric acid reacts with calcium hydroxide to form calcium sulfate and water.

  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{+Ca}{\left(\text{OH}\right)}_{\text{2}}\to {\text{CaSO}}_{\text{4}}{\text{+H}}_{\text{2}}\text{O}$

The balanced equation for the reaction is as follows:

  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{+Ca}{\left(\text{OH}\right)}_{\text{2}}\to {\text{CaSO}}_{\text{4}}{\text{+2H}}_{\text{2}}\text{O}$

(b)

Interpretation Introduction

Interpretation: The mass of unreacted starting material is to be calculated.

Concept Introduction: Reactants, products, and an arrow indicating the reaction's direction make up a chemical equation. A balanced chemical equation is one in which the number of atoms in each of the molecules is the same on both sides of the equation.

The formula to calculate the number of moles is given as

  $\text{n}\text{=}\frac{\text{given}\text{mass}}{\text{molar mass}}$   ...... (1)

Answer to Problem 73A

The mass of unreacted starting material is $\text{2}\text{.17}\text{g}$ .

Explanation of Solution

Mass of sulfuric acid is $75\text{g}$ .

Mass of calcium hydroxide is $55\text{g}$ .

The limiting reagent is calcium hydroxide. The excess reagent is sulfuric acid whose unreacted mass has to be found.

Molar mass of $\text{H}$ is $1.00\text{g/mol}$ .

Molar mass of $\text{Ca}$ is $40\text{g/mol}$ .

Molar mass of $\text{S}$ is $32.00\text{g/mol}$ .

Molar mass of $\text{O}$ is $\text{16}\text{.00}\text{g/mol}$ .

Mass of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is given as follows:

  $\begin{array}{c}\text{Mass of}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}=\text{M}\left(\text{Ca}\right)\text{+2M}\left(\text{OH}\right)\\ =\left(40+\left(2\times 16.00\times 1.00\right)\right)\text{g/mol}\\ \text{= 74}\text{g/mol}\end{array}$

Mass of ${\text{H}}_2{\text{SO}}_4$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{H}}_2{\text{SO}}_4=2\text{M}\left(\text{H}\right)\text{+M}\left(\text{S}\right)+4\text{M}\left(\text{O}\right)\\ =\left(\left(2\times 1.00\right)+32.00+\left(4\times 16.00\right)\right)\text{g/mol}\\ \text{= 98}\text{.00}\text{g/mol}\end{array}$

To calculate the number of moles of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$, substitute the values in equation (1):

  $\begin{array}{c}{\text{n}}_{\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}\text{=}\frac{{\text{m}}_{\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}}{\text{M}\left(\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\right)}\\ =\frac{55\cancel{\text{g}}}{74\cancel{\text{g}}\text{/mol}}\\ =0.743\text{mol}\end{array}$

From the chemical equation, moles of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is equal to the moles of ${\text{H}}_2{\text{SO}}_4$ .

So, the number of moles of ${\text{H}}_2{\text{SO}}_4$ is also $0.743\text{mol}$ .

To calculate the mass of ${\text{H}}_2{\text{SO}}_4$, substitute the values in equation (1):

  $\begin{array}{c}{\text{m}}_{{\text{H}}_2{\text{SO}}_4}={\text{n}}_{{\text{H}}_2{\text{SO}}_4}\times \text{M}\left({\text{H}}_2{\text{SO}}_4\right)\\ \text{=}0.743\text{mol}\cancel{\text{mol}}\times 98\text{g/}\cancel{\text{mol}}\\ =72.83\text{g}\end{array}$

To calculate the unreacted mass of reactant (sulfuric acid), find the difference between the taken mass of sulfuric acid and calculate mass of sulfuric acid.

  $\begin{array}{c}{\text{m}}_{\text{unreacted}}\text{=75}\text{g}-\text{72}\text{.83}\text{g}\\ \text{= 2}\text{.17}\text{g}\end{array}$