Chapter 9, Problem 39A

Interpretation Introduction

Interpretation: The mass (g) of elemental iodine produced in each of the given reaction needs to be determined, if 25.0 g of NaI reacts with 5.00 g of ${\mathrm{Cl}}_2$ gas and liquid bromine.

  $\begin{array}{l}{\text{Cl}}_{\text{2}}{}_{\text{(g) }}{\text{+ NaI}}_{\text{(aq)}}\stackrel{}{\to }{\text{NaCl}}_{\text{(aq)}}{\text{ + I}}_{\text{2(s)}}\\ {\text{Br}}_{\text{2(l) }}{\text{+ NaI}}_{\text{(aq)}}\stackrel{}{\to }{\text{NaBr}}_{\text{(aq}}{}_{\text{) }}{\text{+ I}}_{\text{2(s)}}\end{array}$

Concept Introduction: A chemical reaction involves the conversion of reactant to product by breaking and making of chemical bonds.

There can be more than one reactant molecules involve in a chemical reaction to form the products. The reactant which is present in limited amount in a chemical reaction is called as limiting reactant.

Answer to Problem 39A

  ${\text{Br}}_{\text{2(l) }}{\text{+ NaI}}_{\text{(aq)}}\stackrel{}{\to }{\text{NaBr}}_{\text{(aq}}{}_{\text{) }}{\text{+ I}}_{\text{2(s)}}\text{ = 7}{\text{.92 g I}}_{\text{2}}$

  ${\text{Cl}}_{\text{2}}{}_{\text{(g) }}{\text{+ NaI}}_{\text{(aq)}}\stackrel{}{\to }{\text{NaCl}}_{\text{(aq)}}{\text{ + I}}_{\text{2(s)}}\text{ = 18}{\text{.0 g I}}_{\text{2}}$

Explanation of Solution

Given information:

  $\begin{array}{l}{\text{Cl}}_{\text{2}}{}_{\text{(g) }}{\text{+ NaI}}_{\text{(aq)}}\stackrel{}{\to }{\text{NaCl}}_{\text{(aq)}}{\text{ + I}}_{\text{2(s)}}\\ {\text{Br}}_{\text{2(l) }}{\text{+ NaI}}_{\text{(aq)}}\stackrel{}{\to }{\text{NaBr}}_{\text{(aq}}{}_{\text{) }}{\text{+ I}}_{\text{2(s)}}\end{array}$

Mass of NaI = 25.0 g

Mass of ${\mathrm{Cl}}_2$ = 5.00 g

Mass of ${\text{Br}}_{\text{2}}$ = 5.00 g

Molar mass of NaI = 149.89 g/mol

Molar mass of ${\text{Br}}_{\text{2}}$ = 159.8 g/mol

Molar mass of ${\mathrm{Cl}}_2$ = 70.9 g/mol

Calculate moles of each:

  $\begin{array}{l}\text{moles of NaI = }\frac{\text{25}\text{.0g}}{\text{149}\text{.89g/mol}}\text{=0}\text{.167moles}\\ {\text{moles of Br}}_{\text{2 }}\text{ = }\frac{\text{5}\text{.00g}}{\text{159}\text{.8 g/mol}}\text{=0}\text{.0312moles}\\ {\text{moles of Cl}}_{\text{2 }}\text{ = }\frac{\text{5}\text{.00g}}{\text{70}\text{.9 g/mol}}\text{=0}\text{.071moles}\end{array}$

From the balance chemical equation:

1 mole of ${\mathrm{Cl}}_2$ = 1 mole of NaI = 1 mole of ${\text{I}}_{\text{2}}$

1 mole of ${\text{Br}}_{\text{2}}$ = 1 mole of NaI = 1 mole of ${\text{I}}_{\text{2}}$

  $\begin{array}{l}{\text{Cl}}_{\text{2}}{}_{\text{(g) }}{\text{+ NaI}}_{\text{(aq)}}\stackrel{}{\to }{\text{NaCl}}_{\text{(aq)}}{\text{ + I}}_{\text{2(s)}}\\ \text{0}{\text{.071moles Cl}}_{\text{2 }}\text{×}\frac{\text{1 mole NaI}}{{\text{1 mole Cl}}_{\text{2 }}}\text{×}\frac{{\text{1 mole I}}_{\text{2}}}{\text{1 mole NaI}}\text{ ×}\frac{\text{253}{\text{.8 g I}}_{\text{2}}}{{\text{1 mole I}}_{\text{2}}}\text{= 18}{\text{.0 g I}}_{\text{2}}\end{array}$

  $\begin{array}{l}{\text{Br}}_{\text{2(l) }}{\text{+ NaI}}_{\text{(aq)}}\stackrel{}{\to }{\text{NaBr}}_{\text{(aq}}{}_{\text{) }}{\text{+ I}}_{\text{2(s)}}\\ \text{0}{\text{.0312 moles Br}}_{\text{2 }}\text{×}\frac{\text{1 mole NaI}}{{\text{1 mole Br}}_{\text{2 }}}\text{×}\frac{{\text{1 mole I}}_{\text{2}}}{\text{1 mole NaI}}\text{ ×}\frac{\text{253}{\text{.8 g I}}_{\text{2}}}{{\text{1 mole I}}_{\text{2}}}\text{= 7}{\text{.92 g I}}_{\text{2}}\end{array}$

Conclusion

  ${\text{Br}}_{\text{2(l) }}{\text{+ NaI}}_{\text{(aq)}}\stackrel{}{\to }{\text{NaBr}}_{\text{(aq}}{}_{\text{) }}{\text{+ I}}_{\text{2(s)}}\text{ = 7}{\text{.92 g I}}_{\text{2}}$

  ${\text{Cl}}_{\text{2}}{}_{\text{(g) }}{\text{+ NaI}}_{\text{(aq)}}\stackrel{}{\to }{\text{NaCl}}_{\text{(aq)}}{\text{ + I}}_{\text{2(s)}}\text{ = 18}{\text{.0 g I}}_{\text{2}}$