Chapter 11, Problem 4STP

Interpretation Introduction

Interpretation:

The percent yield for the given reaction under given conditions needs to be calculated.

Concept introduction:

In a chemical reaction, the amount of product formed in the relation to the amount of reactant consumed is term as yield. Yield is generally expressed in percentage.

Answer to Problem 4STP

The percent yield for the given reaction is (b) $32.7\text{ \%}$.

Explanation of Solution

Given:

${\text{2KClO}}_{\text{3}}\text{(s) }\to {\text{ 2KCl(s) + 3O}}_{\text{2}}\text{(g)}$

The mass of sample decomposed is, ${\text{KClO}}_{\text{3}}=\frac{\text{200 g}}{2}\text{ = 100 g}$.

The mass of ${\text{O}}_{\text{2}}$ produced is $\text{12}\text{.8 g}$.

The number of moles of ${\text{KClO}}_{\text{3}}$ is calculated as:

$\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

The molar mass of ${\text{KClO}}_{\text{3}}$ is 122.55 g/mol. So, the number of moles of ${\text{KClO}}_{\text{3}}$ is:

$\begin{array}{l}{\text{n}}_{{\text{KClO}}_{\text{3}}}=\frac{100\text{ g}}{\text{122}\text{.55 g/mol}}\\ {\text{n}}_{{\text{KClO}}_{\text{3}}}=0.816\text{ mol}\end{array}$

From balanced reaction, the mole ratio of ${\text{KClO}}_{\text{3}}$ and ${\text{O}}_{\text{2}}$ is 2:3 that means 2 moles of ${\text{KClO}}_{\text{3}}$ produces 3 moles of ${\text{O}}_{\text{2}}$. So, the number of moles of ${\text{O}}_{\text{2}}$ produced from 0.816 mol of ${\text{KClO}}_{\text{3}}$ is:

$\text{0}{\text{.816 mol of KClO}}_{\text{3}}\text{ ×}\frac{{\text{3 mol of O}}_2}{{\text{2 mol of KClO}}_{\text{3}}}\text{ = 1}{\text{.224 mol of O}}_2$

Now, the mass of produced ${\text{O}}_{\text{2}}$ is calculated as:

$\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

Molar mass of ${\text{O}}_{\text{2}}$ is 31.998 g/mol. So,

$\begin{array}{l}\text{1}\text{.224 mol}=\frac{\text{Mass}}{\text{31}\text{.998 g/mol}}\\ \text{Mass = 1}\text{.224 mol}\times \text{31}\text{.998 g/mol}\\ \text{Mass = }39.17\text{ g}\end{array}$

Now, the percent yield is calculated using formula:

$\text{Percent}\text{yield}=\frac{\text{Actual}\text{yield}}{\text{Theoretical}\text{yield}}\times 100$

Substituting the values:

$\begin{array}{l}\text{Percent yield}=\frac{\text{12}\text{.8 g}}{\text{39}\text{.17 g}}\times 100\text{ \%}\\ \text{Percent yield}=32.7\text{ \%}\end{array}$

Hence, the percent yield for the given reaction is $32.7\text{ \%}$.