Chapter 12, Problem 79A

Interpretation Introduction

Interpretation : The limiting reagent is to be calculated.

Concept Introduction : The reactant that limits how much product can be produced by a reaction is known as the limiting reagent. Only until the limiting reagent is consumed does the reaction take place.

Answer to Problem 79A

The limiting reagent is $\text{KOH}$ .

Explanation of Solution

Consider the given chemical reaction:

  ${\text{2MnO}}_{\text{2}}{\text{+4KOH+O}}_{\text{2}}{\text{+Cl}}_{\text{2}}\to {\text{2KMnO}}_{\text{4}}{\text{+2KCl+2H}}_{\text{2}}\text{O}$

The reactant available is $100\text{g}$ .

The molar mass of ${\text{MnO}}_{\text{2}}$ is $\text{86}\text{.936}\text{g/mol}$ .

The relation is given as

  $\begin{array}{l}\text{1}\text{}\text{mol}\to \text{86}\text{.936}\text{g/mol}\\ x\text{mol}\to \text{100}\text{g}\end{array}$

Then $x$ is given as

  $\begin{array}{c}x\times \text{86}\text{.936}\text{g/mol}=\text{100}\text{g}\\ x=\frac{\text{100}\text{g}}{\text{86}\text{.936}\text{g/mol}}\\ =1.15\text{mol}\end{array}$

The moles of water with respect to ${\text{MnO}}_{\text{2}}$ is given as

  $1.15\text{mol}\cancel{{\text{MnO}}_2}\times \frac{\cancel{2\text{mol}}{\text{H}}_2\text{O}}{\cancel{2\text{mol}}\cancel{{\text{MnO}}_2}}=1.15\text{mol}{\text{H}}_2\text{O}$

The molar mass of $\text{KOH}$ is $51.1\text{g/mol}$ .

The relation is given as

  $\begin{array}{l}\text{1}\text{}\text{mol}\to 51.1\text{g/mol}\\ x\text{mol}\to \text{100}\text{g}\end{array}$

Then $x$ is given as

  $\begin{array}{c}x\times 51.1\text{g/mol}=\text{100}\text{g}\\ x=\frac{\text{100}\text{g}}{51.1\text{g/mol}}\\ =1.78\text{mol}\end{array}$

The moles of water with respect to $\text{KOH}$ is given as

  $1.78\text{mol}\cancel{\text{KOH}}\times \frac{2\text{mol}{\text{H}}_2\text{O}}{4\text{mol}\cancel{\text{KOH}}}=0.89\text{mol}{\text{H}}_2\text{O}$

The molar mass of ${\text{O}}_2$ is $32\text{g/mol}$ .

The relation is given as

  $\begin{array}{l}\text{1}\text{}\text{mol}\to 32\text{g/mol}\\ x\text{mol}\to \text{100}\text{g}\end{array}$

Then $x$ is given as

  $\begin{array}{c}x\times 32\text{g/mol}=\text{100}\text{g}\\ x=\frac{\text{100}\text{g}}{32\text{g/mol}}\\ =3.125\text{mol}\end{array}$

The moles of water with respect to ${\text{O}}_2$ is given as

  $3.125\text{mol}\cancel{{\text{O}}_2}\times \frac{2\text{mol}{\text{H}}_2\text{O}}{1\text{mol}\cancel{{\text{O}}_2}}=6.25\text{mol}{\text{H}}_2\text{O}$

The molar mass of ${\text{Cl}}_2$ is $70\text{g/mol}$ .

The relation is given as

  $\begin{array}{l}\text{1}\text{}\text{mol}\to 70\text{g/mol}\\ x\text{mol}\to \text{100}\text{g}\end{array}$

Then $x$ is given as

  $\begin{array}{c}x\times 70\text{g/mol}=\text{100}\text{g}\\ x=\frac{\text{100}\text{g}}{70\text{g/mol}}\\ =1.42\text{mol}\end{array}$

The moles of water with respect to ${\text{Cl}}_2$ is given as

  $1.42\text{mol}\cancel{{\text{Cl}}_2}\times \frac{2\text{mol}{\text{H}}_2\text{O}}{1\text{mol}\cancel{{\text{Cl}}_2}}=2.85\text{mol}{\text{H}}_2\text{O}$

  $\text{KOH}$ is the limiting reagent as it readily dissolves in water.

  $\text{1}\text{.78}\text{mol}\text{KOH}$ dissolves in $0.89\text{mol}{\text{H}}_2\text{O}$ .