Answer:
Mass percent: 53.23%
Explanation:
Balancing the reaction in acidic medium:
Fe2+ → Fe3+ + 1e-
Cr₂O₇²⁻ → 2Cr³⁺
Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O (Balancing oxygen)
Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O (Balancing protons)
Cr₂O₇²⁻ + 14H⁺ + 6e-→ 2Cr³⁺ + 7H₂O (Balancing charges)
To cancel the 6e-:
6 Fe2+ → 6 Fe3+ + 6e-
The balanced redox equation in acidic medium is:
6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O
We need to find moles of potassium dichromate used to reach the final point in the titration. Then, using the chemical equation, we can find moles of Fe²⁺. With its molar mass (Molar mass Fe: 55.85g/mol)b we can find mass of Fe²⁺ in the iton ore and its mass percent:
Moles potassium dichromate:
28.72mL = 0.02872L * (0.05051mol / L) = 1.4506x10⁻³ moles K₂Cr₂O₇
Moles Fe²⁺:
As 6 moles of Fe²⁺ reacted per mole of K₂Cr₂O₇:
1.4506x10⁻³ moles K₂Cr₂O₇ * (6 moles Fe²⁺ / 1 mole K₂Cr₂O₇) = 8.7039x10⁻³ moles Fe²⁺
Mass Fe²⁺:
8.7039x10⁻³ moles Fe²⁺ * (55.845g / mol) = 0.4861g of Fe²⁺
Mass percent:
As the sample of iron ore weighs 0.9132g. Mass percent is:
0.4861g of Fe²⁺ / 0.9132g * 100
The mass percentage of the iron in the sample is 53.7%.
The balanced redox reaction equation is;
6Fe2+(aq) + Cr2O7^2-(aq) + 14H^+(aq)→ 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)
The number of moles of dicromate ion is; 28.72/1000 L × 0.05051M
= 0.00145 moles
Now;
6 moles of Fe2+ reacts with 1 mole of dichromate ion
x moles of Fe2+ reacts with 0.00145 moles of dichromate ion
x = 6 moles × 0.00145 moles/1 mole
x = 0.0087 moles
Mass of iron = 0.0087 moles × 56 = 0.49 g
Mass percent of iron = 0.49 g/0.9132 g × 100/1
= 53.7%
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