Chapter 12, Problem 49A

Interpretation Introduction

Interpretation: To determine the amount of iron oxide which are produced when 12 grams of iron rust.

Concept Introduction: When iron is treated with oxygen then it will form iron oxide. It is a kind of combination reaction. Iron oxide is also known as rust.

Answer to Problem 49A

The amount of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is 10.g .

Explanation of Solution

It is given that,

Mass of iron = 12 g.

The balanced chemical reaction can be written as:

  ${\text{4Fe + 3O}}_{\text{2}}\to {\text{ 2Fe}}_{\text{2}}{\text{O}}_{\text{3}}$

Number of moles can be determined by using the formula:

  $\text{Number}\text{moles}\text{=}\frac{\text{Mass}}{\text{molar}\text{mass}}$

Put the given data in the above formula:

  $\begin{array}{l}\text{Number of moles = }\frac{\text{mass}}{\text{molar mass}}\hfill \\ \text{Number of moles = }\frac{\text{12}\text{.0 g }}{\text{55}\text{.85 g/mol}}\hfill \\ \text{Number of moles = 0}\text{.215 mol}\hfill \end{array}$

It is found that,

Mole of iron = 4

Mole of iron oxide = 2

So,

  $\begin{array}{c}\text{The mole of}{\text{Fe}}_2{\text{O}}_3\text{=2/4×0}\text{.125 }\\ \text{= 0}\text{.063 mol}\end{array}$

Now, the mass of iron oxide can be calculated by using the formula:

  $\begin{array}{l}\text{Mass = number of moles × molar mass}\hfill \\ \text{Mass = 0}\text{.063 mol × 159}\text{.7 g/mol}\hfill \\ \text{Mass = 10}\text{.1 g}\hfill \end{array}$

Therefore, the mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ will be 10.1 g.