Chapter 9, Problem 64A

Interpretation Introduction

Interpretation: The substance $\text{X}$ should be identified for the following reaction under given conditions:

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

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

The ratio of amount of any two compounds involved in a chemical reaction in terms of mole is said to be mole ratio, and used as a conversion factor between reactants and products.

Answer to Problem 64A

The substance $\text{X}$ is aluminum, $\text{Al}$ .

Explanation of Solution

Given:

Mass of $\text{X}$ that reacts with excess of iron(III) oxide is 77.7 g and produces 1.44 moles of ${\text{X}}_{\text{2}}{\text{O}}_3$ .

The given reaction is:

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

The above reaction is not balanced as the number of atoms of Fe and X on the reactant side is 2 and 1 and on the product side is 1 and 2 respectively. So, in order to balance the reaction, the coefficient 2 is written before X and Fe both on the reactant and product side respectively. Thus, the balanced reaction is:

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

From the balanced reaction, the mole ratio between $\text{X}$ and ${\text{X}}_{\text{2}}{\text{O}}_3$ is 2:1 that means 2 moles of $\text{X}$ produces 1 mole of ${\text{X}}_{\text{2}}{\text{O}}_3$ . The number of moles required for the production of 1.44 mol of ${\text{X}}_{\text{2}}{\text{O}}_3$ will be:

  $\text{1}{\text{.44 mol of X}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)×}\frac{\text{2 mol of X(s)}}{{\text{1 mol of X}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}}\text{ = 2}\text{.88 mol of X(s)}$

Now, calculating the molar mass of $\text{X}$ using relation:

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

Mass of $\text{X}$ that reacts with excess of iron(III) oxide is 77.7 g so,

  $\begin{array}{l}\text{2}\text{.88 mol}=\frac{\text{77}\text{.7 g}}{\text{Molar mass of X}}\\ \text{Molar mass of X = }\frac{\text{77}\text{.7 g}}{\text{2}\text{.88 mol}}\\ \text{Molar mass of X = 26}\text{.98 g/mol}\end{array}$

Since, the element having molar mass 26.98 g/mol is aluminum so, the substance $\text{X}$ is aluminum, $\text{Al}$ .