Answer:
a. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O
b. 18.17% of Fe in the sample
Explanation:
a. In the reaction, Fe²⁺ is oxidized to Fe³⁺ and permanganate, MnO₄⁺ reduced to Mn²⁺, thus:
Fe²⁺ → Fe³⁺ + 1e⁻
MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O
5 times the iron and suming the manganese reaction:
MnO₄⁻ + 5e⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O
<h3>MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O</h3>
b. Moles of permanganate in the titration are:
0.03942L × (0.0281 moles / L) = <em>1.108x10⁻³ moles of MnO₄⁻</em>
Based on the reaction, 1 mole of permanganate reacts with 5 moles of iron, if 1.108x10⁻³ moles of MnO₄⁻ reacts, moles of iron are:
1.108x10⁻³ moles of MnO₄⁻ × (5 moles Fe²⁺ / 1 mole MnO₄⁻) =
<em>4.431x10⁻³ moles of Fe²⁺</em>. Molar mass of Fe is 55.845g/mol. 4.431x10⁻³ moles of Fe²⁺ are:
4.431x10⁻³ moles of Fe²⁺ ₓ (55.845g / mol) =
<h3>0.2474g of Fe you have in your sample.</h3>
Percent mass is:
0.2474g Fe / 1.362g sample ₓ 100 =
<h3>18.17% of Fe in the sample</h3>
