Use Fe = 56, S = 32, O = 16. A $5.00 g$ sample contains $FeSO X_{4}$ , $Fe X_{2} (SO X_{4}) X_{3}$ and inert matter. Titration of one portion (the whole sample, acidic) needs $40.0 mL$ of $0.0200 M$ $KMnO X_{4}$ . A second identical sample is shaken with zinc to reduce all $Fe X^{3 +}$ to $Fe X^{2 +}$ , the excess zinc is removed, and the solution needs $92.0 mL$ of the same $KMnO X_{4}$ . Find the mass percentage of $Fe X_{2} (SO X_{4}) X_{3}$ in the sample (two decimals).

Numerical answer type, answer in %, tolerance 0.05.

Answer: 20.8 %

First titration:

Fe X^{2 +}

= 40.0 \times 0.0200 \times 5 = 4.0

mmol. Second: total Fe

= 92.0 \times 0.0200 \times 5 = 9.2

mmol.

Fe X^{3 +} = 5.2

mmol, so

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

= 2.6

mmol (

M = 400

)

= 1.040 g

. Percentage

= 1.040/5.00 \times 100 = 20.80%