Chapter 9, Problem 69AP

Interpretation Introduction

Interpretation:

To determine the theoretical yield of iron (II) sulphide.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

The limiting reactant in a particular reaction has due to following properties:

1. Limiting reactant completely reacted in a particular reaction.
2. Limiting reactant determines the amount of the product in mole.

The theoretical yield of iron (II) sulphide is $8.26\text{g}\text{F}\text{e}\text{S}$.

Theoretical yield is also known as calculated yield. Theoretical yield is the amount of product which is theatrically calculates by the amount of limiting agent. It is the maximum amount of product which is formed in any reaction.

Other name of actual yield is observed yield.

The percentage yield is calculated by the actual or observed experimental yield divided by the theoretical yield and multiplied by 100.

The percentage yield can be calculates by the use of following expression:

$\frac{\text{Actual yield or given yield }}{\text{Theoretical yield or calculated yield }}\times \text{100\% }$

Theoretical yield is the greatest amount of a product possible from a reaction, when assume that the reaction goes to 100% competition.

The balance equation for the calcium and oxalate ions is as follows:

$\text{F}\text{e}(s)+\text{S}(s)\to \text{F}\text{e}\text{S}(s)$

Given:

Amount of $\text{F}\text{e}$ -=5.25 g

Amount of $\text{S}$ = 12.7 g

Calculation:

Number of moles of $\text{F}\text{e}$ and $\text{S}$ are calculated as follows:

$\begin{array}{l}\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}=\frac{\text{m}\text{a}\text{s}\text{s}\text{i}\text{n}\text{g}}{\text{m}\text{o}\text{l}\text{a}\text{r}\text{m}\text{a}\text{s}\text{s}}=\frac{5.25\text{g}}{\text{55}\text{.845 }\text{g}/\text{m}\text{o}\text{l}}=0.094\text{m}\text{o}\text{l}\text{e}\text{s}\text{F}\text{e}\\ \text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}=\frac{12.7\text{g}}{\text{32}\text{.065 }\text{g}/\text{m}\text{o}\text{l}}=0.3966\text{m}\text{o}\text{l}\text{e}\text{s}\text{S}\end{array}$

In the above reaction Fe is a limiting reagent because it reacted completely. Now calculated the theoretical amount of $\text{F}\text{e}\text{S}$ as follows:

$0.094\text{m}\text{o}\text{l}\text{e}\text{s}\text{F}\text{e}\times \frac{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{F}\text{e}\text{S}}}{1.00\text{m}\text{o}\text{l}\text{e}\text{s}\text{F}\text{e}}\times \frac{\text{87}\text{.91 }\text{g}\text{F}\text{e}\text{S}}{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{F}\text{e}\text{S}}}=8.26\text{g}\text{F}\text{e}\text{S}$.

The theoretical yield of iron (II) sulphide is $8.26\text{g}\text{F}\text{e}\text{S}$.