Chapter 11, Problem 70A

Interpretation Introduction

Interpretation:

The balanced chemical equation for the reaction between lead, lead(IV) oxide, and sulfuric acid to produce lead(II) sulfate and water should be determined.

Concept introduction:

Chemical equation is written in such a way that the symbolic representation of reaction represents the reaction taking place in the system. The reactants are written on the left-hand side and the products are written on the right-hand side of the equation and are separated by an arrow, two or more reactants and products are separated by “+”. The reactions for those the number of atoms of each element in the reactant and in the product, side are equal, such reactions are said to be a balanced chemical equation.

Answer to Problem 70A

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

Explanation of Solution

The elemental formula for lead is $\text{Pb}$, molecular formula of lead(IV) oxide is ${\text{PbO}}_{\text{2}}$, for sulfuric acid is ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, for lead(II) sulfate is ${\text{PbSO}}_{\text{4}}$ and for water is ${\text{H}}_{\text{2}}\text{O}$.

The physical state of the reactants and products are shown by writing the symbols after each reactant and product in the reaction. The symbol “l” represents liquid, “aq” represents aqueous, “s” represents solid, and “g” represents gas.

The reaction between lead, lead(IV) oxide, and sulfuric acid to produce lead(II) sulfate and water is written as:

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

This reaction is not balanced as the number of Pb and O atoms on the reactant side is 2 and 6 whereas in the product side is 1 and 5 respectively. So, in order to balance the reaction, coefficient 2 is written before ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}$ in the reactant side and 2 before both ${\text{PbSO}}_{\text{4}}\text{(aq)}$ and ${\text{H}}_{\text{2}}\text{O(l)}$ on the product side. Hence, the balanced reaction is:

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

Interpretation Introduction

Interpretation:

The mass of lead(II) sulfate produced should be calculated on reacting 25.0 g of lead with excess of lead(IV) oxide and sulfuric acid.

Concept introduction:

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

The ratio of amount of any two compounds involved in a chemical reaction in terms of mole is said to be mole ratio, and used as a conversion factor between reactants and products.

Answer to Problem 70A

The mass of lead(II) sulfate produced from 25.0 g lead is $73.39\text{ g}$.

Explanation of Solution

The complete balanced reaction is:

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

The molar mass of lead is 207.2 g/mol. The number of moles of lead is calculated as shown:

$\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Number of moles}=\frac{25.0\text{ g}}{\text{207}\text{.2 g/mol}}\\ \text{Number of moles}=0.121\text{ mol}\end{array}$

From the balanced chemical reaction, the mole ratio of lead and lead(II) sulfate is 1:2 that means 1 mole of lead produces 2 moles of lead(II) sulfate. So, the number of moles of lead(II) sulfate produced from 0.121 mol of lead is:

$0.121\text{ mol of Pb×}\frac{{\text{2 mol of PbSO}}_{\text{4}}}{\text{1 mol of Pb}}\text{ = }0.242{\text{ mol of PbSO}}_{\text{4}}$

Now, the mass of lead(II) sulfate produced from 25.0 g of lead is calculated as:

The molar mass of lead(II) sulfate is 303.26 g/mol. So,

$\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ 0.242\text{ mol}=\frac{\text{Mass}}{\text{303}\text{.26 g/mol}}\\ \text{Mass = 0}\text{.242 mol}\times 303.26\text{ g/mol}\\ \text{Mass = 73}\text{.39 g}\end{array}$

Hence, the mass of lead(II) sulfate produced from 25.0 g lead is $73.39\text{ g}$.