A 0.9132-gram sample of iron ore is dissolved in aqueous hydrochloric acid. It dissolves completely and in the process, all of the iron goes into solution as Fe2+ ion. The following chemical reaction is a redox reaction and it is the focus of the question: The acid solution containing all of the Fe2+ is placed in a beaker. A solution of aqueous potassium dichromate is used to fill a burette. A redox titration is performed. A volume of exactly 28.72 mL of 0.05051M K2Cr2O7 (aq) is required to complete the redox titration. The unbalanced redox equation is:

Fe2+(aq) + K2Cr2O7(aq) (in acid) → Fe3+(aq) + Cr3+(aq)

Required:
What is the mass percent iron in the original sample of iron ore? Please show all your work.

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

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