Chapter 12, Problem 66A

(a)

Interpretation Introduction

Interpretation: The chemical equation is to be balanced.

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 66A

The balanced chemical equation is as follows:

  ${\text{2PbSO}}_{\text{4}}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right) \to {\text{PbO}}_{\text{2}}\left(s\right)+\text{Pb}\left(s\right)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)$

Explanation of Solution

The given chemical equation is as follows:

  ${\text{PbSO}}_{\text{4}}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right) \to {\text{PbO}}_{\text{2}}\left(s\right)+\text{Pb}\left(s\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)$

The balanced chemical equation is given below:

  ${\text{2PbSO}}_{\text{4}}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right) \to {\text{PbO}}_{\text{2}}\left(s\right)+\text{Pb}\left(s\right)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)$

(b)

Interpretation Introduction

Interpretation: The grams of sulfuric acid 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 66A

The grams of sulfuric acid are $22.026\text{g}$ .

Explanation of Solution

Mass of lead (II) sulfate reacted is $68.1\text{g}$ .

The balanced chemical equation is as follows:

  ${\text{2PbSO}}_{\text{4}}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right) \to {\text{PbO}}_{\text{2}}\left(s\right)+\text{Pb}\left(s\right)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)$

Molar mass of $\text{Pb}$ is $207.2\text{g/mol}$ .

Molar mass of $\text{S}$ is $32.07\text{g/mol}$ .

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

Mass of ${\text{PbSO}}_4$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{PbSO}}_4=\text{M}\left(\text{Pb}\right)\text{+M}\left(\text{S}\right)+4\text{M}\left(\text{O}\right)\\ =\left(207.2+32.07+\left(4\times 16.00\right)\right)\text{g/mol}\\ \text{= 303}\text{.27}\text{g/mol}\end{array}$

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

  $\begin{array}{c}\text{Mass of}{\text{H}}_2{\text{SO}}_4=2\text{M}\left(\text{H}\right)\text{+M}\left(\text{S}\right)+4\text{M}\left(\text{O}\right)\\ =\left(\left(2\times 1.008\right)+32.07+\left(4\times 16.00\right)\right)\text{g/mol}\\ \text{= 98}\text{.07}\text{g/mol}\end{array}$

To calculate the number of moles of ${\text{PbSO}}_4$, substitute the values in equation (1):

  $\begin{array}{c}{\text{n}}_{{\text{PbSO}}_4}\text{=}\frac{{\text{m}}_{{\text{PbSO}}_4}}{\text{M}\left({\text{PbSO}}_4\right)}\\ =\frac{68.1\cancel{\text{g}}}{\text{303}\text{.27}\cancel{\text{g}}\text{/mol}}\\ =0.2246\text{mol}\end{array}$

From the chemical equation, moles of ${\text{PbSO}}_4$ 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 $0.2246\text{mol}$ .

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

  $\begin{array}{c}{\text{m}}_{{\text{H}}_2{\text{SO}}_4}={\text{n}}_{{\text{H}}_2{\text{SO}}_4}\times \text{M}\left({\text{H}}_2{\text{SO}}_4\right)\\ \text{=}0.2246\cancel{\text{mol}}\times 98.07\text{g/}\cancel{\text{mol}}\\ =22.026\text{g}\end{array}$