Question

The metal content of iron in ores can be determined by a redox procedure in which the sample is first oxidized with Br2 to convert all the iron to Fe3+ and then titrated with Sn2+ to reduce the Fe3+ to Fe2+.
The balanced equation is: 2Fe3+(aq) + Sn2+(aq) -> 2Fe2+(aq) + Sn4+(aq)

What is the mass percentage Fe in a 0.1875g sample if 13.28 mL of 0.1015 M Sn2+ solution is needed to titrate the Fe3+?

Answer 1

Answer:

80.27%

Explanation:

Let's consider the following balanced equation.

2 Fe³⁺(aq) + Sn²⁺(aq) ⇒ 2Fe²⁺(aq) + Sn⁴⁺(aq)

First, we have to calculate the moles of Sn²⁺ that react.

$\frac{0.1015molSn^{2+} }{1L} .13.28 \times 10^{-3} L=1.348\times 10^{-3}molSn^{2+}$

We also know the following relations:

• According to the balanced equation, 1 mole of Sn²⁺ reacts with 2 moles of Fe³⁺.
• 1 mole of Fe³⁺ is oxidized from 1 mole of Fe.
• The molar mass of Fe is 55.84 g/mol.

Then, for 1.348 × 10⁻3 moles of Sn²⁺:

$1.348\times 10^{-3}molSn^{2+}.\frac{2molFe^{3+} }{1molSn^{2+} } .\frac{1molFe}{1molFe^{3+} } .\frac{55.84gFe}{1molFe} =0.1505gFe$

If there are 0.1505 g of Fe in a 0.1875 g sample, the mass percentage of Fe is:

$\frac{0.1505g}{0.1875g} \times 100 \% = 80.27\%$