Chapter 7, Problem 10STP

Interpretation Introduction

Interpretation:

A balanced chemical equation corresponding to the reaction between aqueous solutions of lead nitrate and sodium iodide needs to be presented.

Concept Introduction:

• A chemical reaction is represented in terms of a chemical equation with the reactants on the left and the products on the right

  $\text{Reactants }\to \text{ Products}$

• A balanced chemical equation is one in which the number of atoms of various elements on the reactants side is equal to that on the product side. It depicts the stoichiometry of a given reaction.

• The equations are balanced in two steps:

1) Balance the atoms by matching the number of atoms of each type on the reactants and product side.

2) Adjust the coefficients to the smallest whole number. A coefficient of '1' is implied and therefore not explicitly written.

Answer to Problem 10STP

  ${\text{Pb(NO}}_{\text{3}}{\text{)}}_2\text{(aq) + 2NaI (aq)}\to {\text{PbI}}_{\text{2}}({\text{s) + 2NaNO}}_{\text{3}}\text{(aq)}$

Explanation of Solution

The given equation involves the reaction between aqueous solutions of lead nitrate and sodium iodide to form solid lead iodide and an aqueous solution of sodium nitrate. The reactants and products are:

Reactants:

  $\begin{array}{l}{\text{Lead nitrate, aqueous - Pb(NO}}_{\text{3}}{\text{)}}_2\text{ (aq)}\\ \text{Sodium iodide, aqueous - NaI (aq)}\end{array}$

Products:

  $\begin{array}{l}{\text{Lead iodide , solid - PbI}}_{\text{2}}(\text{s)}\\ {\text{Sodium nitrate, aqueous - NaNO}}_{\text{3}}\text{ (aq)}\end{array}$

Unbalanced equation:

  ${\text{Pb(NO}}_{\text{3}}{\text{)}}_2\text{(aq) + NaI (aq)}\to {\text{PbI}}_{\text{2}}({\text{s) + NaNO}}_{\text{3}}\text{(aq)}$

The above equation is not balanced since the number of N, O and I atoms on the reactants side is 2, 6 and 1 respectively while that on the products side is 1, 3 and 2. In order to balance the above equation, multiply both NaI and NaNO₃ by 2 to get:

  ${\text{Pb(NO}}_{\text{3}}{\text{)}}_2\text{(aq) + 2NaI (aq)}\to {\text{PbI}}_{\text{2}}({\text{s) + 2NaNO}}_{\text{3}}\text{(aq)}$

Conclusion

Therefore, 1 mole of lead nitrate reacts with 2 moles of sodium iodide to form 1 mole of lead iodide and 2 moles of sodium nitrate.