a) Construct the balanced chemical equation for the reaction.
This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Here's a step-by-step solution to your chemistry problem:
a) Construct the balanced chemical equation for the reaction.
Iron (III) oxide ($\text{Fe}_2\text{O}_3$) reacts with sulfuric acid ($\text{H}_2\text{SO}_4$) to produce iron (III) sulfate ($\text{Fe}_2(\text{SO}_4)_3$) and water ($\text{H}_2\text{O}$).
The unbalanced equation is:
$\text{Fe}_2\text{O}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + \text{H}_2\text{O}$
To balance the equation:
1. Balance Fe: There are 2 Fe atoms on both sides.
2. Balance $\text{SO}_4$ groups: There is 1 $\text{SO}_4$ group on the left and 3 on the right. Place a coefficient of 3 in front of $\text{H}_2\text{SO}_4$.
$\text{Fe}_2\text{O}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + \text{H}_2\text{O}$
3. Balance H atoms: There are $3 \times 2 = 6$ H atoms on the left. Place a coefficient of 3 in front of $\text{H}_2\text{O}$.
$\text{Fe}_2\text{O}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 3\text{H}_2\text{O}$
4. Check O atoms (excluding those in $\text{SO}_4$): There are 3 O atoms in $\text{Fe}_2\text{O}_3$ on the left and 3 O atoms in $3\text{H}_2\text{O}$ on the right. The equation is balanced.
The balanced chemical equation is:
$$\boxed{\text{Fe}_2\text{O}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 3\text{H}_2\text{O}}$$
b)
i) Calculate the mass of the pure oxide that reacted with the acid.
Step 1: Calculate the moles of sulfuric acid reacted.
Given:
Volume of $\text{H}_2\text{SO}_4 = 600 \, \text{cm}^3 = 0.600 \, \text{dm}^3$
Concentration of $\text{H}_2\text{SO}_4 = 0.100 \, \text{mol/dm}^3$
$$\text{Moles of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume}$$
$$\text{Moles of H}_2\text{SO}_4 = 0.100 \, \text{mol/dm}^3 \times 0.600 \, \text{dm}^3 = 0.0600 \, \text{mol}$$
Step 2: Use stoichiometry to find the moles of pure iron (III) oxide reacted.
From the balanced equation: $1 \, \text{mol of Fe}_2\text{O}_3$ reacts with $3 \, \text{mol of H}_2\text{SO}_4$.
The mole ratio of $\text{Fe}_2\text{O}_3 : \text{H}_2\text{SO}_4$ is $1:3$.
$$\text{Moles of Fe}_2\text{O}_3 = \text{Moles of H}_2\text{SO}_4 \times \frac{1}{3}$$
$$\text{Moles of Fe}_2\text{O}_3 = 0.0600 \, \text{mol} \times \frac{1}{3} = 0.0200 \, \text{mol}$$
Step 3: Calculate the mass of pure iron (III) oxide.
Molar mass of $\text{Fe}_2\text{O}_3$:
$\text{M}(\text{Fe}_2\text{O}_3) = (2 \times \text{Ar of Fe}) + (3 \times \text{Ar of O})$
$\text{M}(\text{Fe}_2\text{O}_3) = (2 \times 55.8) + (3 \times 16.0) = 111.6 + 48.0 = 159.6 \, \text{g/mol}$
$$\text{Mass of Fe}_2\text{O}_3 = \text{Moles} \times \text{Molar Mass}$$
$$\text{Mass of Fe}_2\text{O}_3 = 0.0200 \, \text{mol} \times 159.6 \, \text{g/mol} = 3.192 \, \text{g}$$
Rounding to 3 significant figures:
$$\text{Mass of pure oxide} = \boxed{3.19 \, \text{g}}$$
ii) Calculate the percentage (%) purity of iron (III) oxide assuming that the impurities did not react with the acid.
Given:
Total mass of impure iron (III) oxide = $4.00 \, \text{g}$
Mass of pure iron (III) oxide reacted = $3.192 \, \text{g}$ (from part b(i))
$$\text{Percentage purity} = \frac{\text{Mass of pure substance}}{\text{Total mass of impure substance}} \times 100\%$$
$$\text{Percentage purity} = \frac{3.192 \, \text{g}}{4.00 \, \text{g}} \times 100\%$$
$$\text{Percentage purity} = 0.798 \times 100\%$$
$$\text{Percentage purity} = 79.8\%$$
Rounding to 3 significant figures:
$$\text{Percentage purity} = \boxed{79.8\%}$$
