Question

1. 1.456 g sample of unknown alloy was dissolved in acid and diluted to 100 mL. What was the percentage (by weight) of iron in the sample, if titration of 20.00 mL aliquot of the diluted solution requires on average 33.45 mL of 0.01005 M solution of potassium permanganate?
​​5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O

Answer 1

The balanced reaction equation is:

​​5Fe²⁺ + MnO₄^- + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O.

As one can see from the stoichiometry of the reaction, 5 ions of iron react with 1 ion of permanganate. Therefore, the number of the moles of iron ions and permanganate ions relate as:

$\frac{n(Fe^{2+})}{5} = n(MnO_4^-)$ .

The number of the moles of permanganate is:

$n(MnO_4^-) = cV = 0.01005 \text{ M}·33.45·10^{-3}\text{ L}$

$n(MnO_4^-) = 3.36·10^{-4}$ mol.

Thus, the number of the moles of iron ions in the 20.00 mL aliquot is:

$n(Fe^{2+}) = n(MnO_4^-)·5 = 1.68·10^{-3}$ mol.

The aliquot represents 1/5 of the sample content (100/20 = 5), so the total iron quantity is:

$n_{tot} = 5·n(Fe^{2+}) = 8.40·10^{-3}$ mol.

The mass of iron in the sample can be calculated as a product of its number of the moles and its molar mass $M$ :

$m = nM = 8.40·10^{-3}\text{ mol}·55.85\text{ g/mol} = 0.469$ g.

The mass percentage of iron in the sample is:

$w = \frac{0.469\text{ g}}{1.456 \text{ g}} = 32\%$ .

Answer: 32%