can someone help me with this chemistry problem (solution)

Question

A mixture consists of FeSO4 and Fe2(SO4)3. A 0.1223 g sample of this mixture was dissolved in water and titrated with a 0.00420 M KMnO4 solution. The titration required 18.85 mL of the KMnO4 solution. What percent, by mass, of the mixture was FeSO4?

Answer 1

To solve this problem, we need to understand the reaction that occurs between FeSO₄ and KMnO₄ in acidic solution. In this reaction, Fe²⁺ ions are oxidized to Fe³⁺ ions by permanganate ions (MnO₄⁻) in acidic medium. The balanced chemical equation for this reaction is:

5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

From the balanced equation, we can see that 5 moles of Fe²⁺ ions react with 1 mole of MnO₄⁻ ions.

Given that 0.1223 g of the mixture was used and it took 18.85 mL of 0.00420 M KMnO₄ solution to titrate it, we can first find the number of moles of KMnO₄ used:

Moles of KMnO₄ = (Volume of KMnO₄ solution used) × (Molarity of KMnO₄)

= 18.85 mL × (0.00420 mol/mL)

= 0.07917 mol

According to the stoichiometry of the reaction, 1 mole of KMnO₄ reacts with 5 moles of FeSO₄. So, the number of moles of FeSO₄ in the mixture is:

Moles of FeSO₄ = (Moles of KMnO₄) / 5

= 0.07917 mol / 5

= 0.01583 mol

Now, we need to find the mass of FeSO₄ in the mixture. The molar mass of FeSO₄ is approximately 151.91 g/mol. Thus:

Mass of FeSO₄ = (Moles of FeSO₄) × (Molar mass of FeSO₄)

= 0.01583 mol × 151.91 g/mol

= 2.403 g

To find the percent by mass of FeSO₄ in the mixture, we use the formula:

Percent by mass = (Mass of FeSO₄ / Total mass of mixture) × 100%

Total mass of mixture = 0.1223 g (given)

So,

Percent by mass = (2.403 g / 0.1223 g) × 100%

≈ 19.63%

Therefore, approximately 19.63% by mass of the mixture is FeSO₄.