Question

A sample weighing 3.109 g is a mixture of Fe2O3 (molar mass = 159.69 g/mol) and Al2O3 (molar mass = 101.96 g/mol). When heat and a stream of H2 gas is applied to the sample, the Fe2O3 reacts to form metallic Fe and H2O(g). The Al2O3 does not react. If the sample residue (the solid species remaining after the reaction) weighs 2.515 g, what is the mass fraction of Fe2O3 in the original sample?

Answer 1

Answer: The mass fraction of $Fe_2O_3$ in the original sample is 0.637

Explanation:

We are given:

Mass of sample = 3.109 g

Mass of solid residue (Aluminium oxide + iron) = 2.515 g

Amount of mass lost = (3.109 - 2.515) = 0.594 g

The amount of mass lost from the sample is equal to the mass of oxygen lost from iron (III) oxide.

The chemical equation for the reaction of iron (III) oxide with hydrogen follows:

$Fe_2O_3+3H_2\rightarrow 2Fe+3H_2O$

In 1 mole of iron (III) oxide, 2 moles of iron are present and 3 moles of oxygen are present.

Mass of oxygen lost from 1 mole of iron (III) oxide = (3 × 16) = 48 g

Calculating the mass of iron (III) oxide by using unitary method:

48 grams of oxygen is lost when 159.69 grams of iron (III) oxide is reacted with hydrogen gas

So, 0.594 grams of oxygen will be lost when = $\frac{159.69}{48}\times 0.594=1.98g$ of iron (III) oxide is reacted with hydrogen gas

• To calculate the mass fraction of iron (III) oxide, we use the equation:

$\text{Mass fraction of iron (III) oxide}=\frac{\text{Mass of iron (III) oxide}}{\text{Mass of sample}}$

Mass of iron (III) oxide = 1.98 grams

Mass of sample = 3.109 grams

Putting values in above equation, we get:

$\text{Mass fraction of iron (III) oxide}=\frac{1.98g}{3.109g}=0.637$

Hence, the mass fraction of $Fe_2O_3$ in the original sample is 0.637