Chapter 11, Problem 6STP

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 6STP

The correct option is (d), $86.8\text{ \%}$.

Explanation of Solution

Given:

The reaction:

$\text{2HgO(s) }\to {\text{ 2Hg(l) + O}}_{\text{2}}\text{(g)}$

The number of moles of $\text{HgO}$ decomposed = 3.55 mol

The number of moles of ${\text{O}}_2$ produced = 1.54 mol

The mass of $\text{Hg}$ produced is $\text{618 g}$ (actual yield).

From balanced reaction, the mole ratio of $\text{HgO}$ and $\text{Hg}$ is 2:2 that is 1:1 that means 1 mole of $\text{HgO}$ produces 1 mole of $\text{Hg}$. So, the number of moles of $\text{Hg}$ produced from 3.55 of $\text{HgO}$ is 3.55 mol.

Now, the mass of produced $\text{Hg}$ is calculated as:

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

Molar mass of $\text{Hg}$ is 200.59 g/mol. So,

$\begin{array}{l}\text{3}\text{.55 mol}=\frac{\text{Mass}}{\text{200}\text{.59 g/mol}}\\ \text{Mass = 3}\text{.55 mol}\times \text{200}\text{.59 g/mol}\\ \text{Mass = }712.09\text{ g}\end{array}$

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:

$\begin{array}{l}\text{Percent yield}=\frac{\text{618 g }}{\text{712}\text{.09 g}}\times 100\text{ \%}\\ \text{Percent yield}=86.8\text{ \%}\end{array}$

Hence, the percent yield for the given reaction is $86.8\text{ \%}$.