2.00 g of a mixture of FeSO4 (molar mass 152) and Fe2(SO4)3 (molar mass 400) is dissolved in dilute H2SO4 and made up to 100 mL. A 20.0 mL portion needs 10.0 mL of 0.0200 M KMnO4. Another 20.0 mL portion is first treated with zinc to reduce all iron to Fe^2+, then needs 22.4 mL of the same KMnO4. Find the mass percentage of Fe2(SO4)3 in the mixture, to the nearest integer.

JEE Advanced levelNumerical answerDifficulty 3 of 3rx208

2.00 g of a mixture of $FeSO X_{4}$ (molar mass 152) and $Fe X_{2} (SO X_{4}) X_{3}$ (molar mass 400) is dissolved in dilute $H X_{2} SO X_{4}$ and made up to 100 mL. A 20.0 mL portion needs 10.0 mL of 0.0200 M $KMnO X_{4}$ . Another 20.0 mL portion is first treated with zinc to reduce all iron to $Fe X^{2 +}$ , then needs 22.4 mL of the same $KMnO X_{4}$ . Find the mass percentage of $Fe X_{2} (SO X_{4}) X_{3}$ in the mixture, to the nearest integer.

Numerical answer type, answer in %, tolerance 0.5.

First titration: mmol

Fe X^{2 +}

in 20 mL =

10.0 \times 0.0200 \times 5 = 1.00

, so 5.00 mmol

FeSO X_{4}

(760 mg) in 100 mL. Second: total Fe =

22.4 \times 0.0200 \times 5 = 2.24

mmol per 20 mL, so 11.2 mmol in 100 mL. Fe(III) originally =

11.2 - 5.00 = 6.20

mmol, so

Fe X_{2} (SO X_{4}) X_{3}

= 3.10 mmol = 1240 mg (and

760 + 1240 = 2000

mg checks). Percentage =

1240/2000 \times 100 = 62%