Chapter 9, Problem 40A

Interpretation Introduction

Interpretation: The mass of iron (III) oxide produced on heating 1.25 g of iron and 0.0204 mole of oxygen gas needs to be calculated.

Concept introduction: The ratio of the mass of a substance to its molar mass is said to be the number of moles of that substance.

Answer to Problem 40A

The mass of iron (III) oxide produced on heating 1.25 g of iron and 0.0204 mole of oxygen gas is $1.76\text{ g}$ .

Explanation of Solution

The chemical equation for the formation of iron(III) oxide from iron and oxygen gas is:

  ${\text{Fe(s) + O}}_{\text{2}}\text{(g)}\to {\text{Fe}}_{\text{2}}{\text{O}}_3\text{(s)}$

The above reaction is not balanced as the number of atoms of Fe and O is 1 and 2 on the reactant side and 2 and 3 on the product side.So in order to balance the reaction, coefficient 4 and 3 are written before Fe and O₂ respectively on the reactant side and coefficient 2 is written before Fe₂O₃ on the product side. Thus, the balanced equation is:

  ${\text{4Fe(s) + 3O}}_{\text{2}}\text{(g)}\to 2{\text{Fe}}_{\text{2}}{\text{O}}_3\text{(s)}$

The mass of iron (III) oxide produced will be calculated after determining the limiting reagent (The reagent in a reaction that controls the amount of product formed is termed as limiting reagent. After completion of chemical reaction, the limiting reagent is fully utilized.):

Calculating the moles of iron as:

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

The molar mass of iron is 55.845 g/mol. The number of moles of ironwill be:

  $\begin{array}{l}{\text{n}}_{\text{Fe}}=\frac{1.25\text{ g}}{\text{55}\text{.845 g/mol}}\\ {\text{n}}_{\text{Fe}}=0.022\text{ mol}\end{array}$

The given mole of oxygen gas is 0.0204 mol.

In the above-balanced reaction, 4.0 mol of $\text{Fe}$ reacts with 3.0 mol of ${\text{O}}_2$ . So, the number of moles of $\text{Fe}$ required to react with $\text{0}\text{.0204 mol}$ of ${\text{O}}_2$ is:

  $0.0204{\text{ mol of O}}_2\text{×}\frac{\text{4 mol of Fe}}{{\text{3 mol of O}}_2}\text{ = }0.0272\text{ mol of Fe}$

Since, the required moles of $\text{Fe}$ is $0.0272\text{ mol}$ and the available amount is $0.022\text{ mol}$ , so $\text{Fe}$ is the limiting reactant and ${\text{O}}_2$ is present in excess. So, the amount of iron(III) oxide formed will depend on the number of moles of $\text{Fe}$ :

From the balanced chemical reaction, the mole ratio of $\text{Fe}$ and iron(III) oxide is 4:2 so, the number of moles of water iron(III) oxide from $0.022\text{ mol}$ of $\text{Fe}$ is:

  $0.022\text{ mol of Fe×}\frac{{\text{2 mol of Fe}}_{\text{2}}{\text{O}}_3}{\text{4 mol of Fe}}\text{ = 0}{\text{.011 mol of Fe}}_{\text{2}}{\text{O}}_3$

Now, the mass of iron(III) oxide produced is calculated as:

The molar mass of iron(III) oxide is 159.69 g/mol. So,

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ 0.011\text{ mol}=\frac{\text{Mass}}{159.69\text{ g/mol}}\\ \text{Mass = }0.011\text{ mol}\times 159.69\text{ g/mol}\\ \text{Mass = }1.76\text{ g}\end{array}$

Hence, the mass of iron (III) oxide produced on heating 1.25 g of iron and 0.0204 mole of oxygen gas is $1.76\text{ g}$ .