Please don\'t rush, I need the right answers with steps. The metal content of ir

Question

Please don't rush, I need the right answers with steps. 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+?
Again please don't rush, I need the right answers with all of the steps.

Explanation / Answer

Solution :-

Lets first calculate the moles of the Sn2+ reacted

Moles = molarity * volume in liter

           = 0.1015 mol per L * 0.01328 L

           = 0.001348 mol Sn2+

Now using the mole ratio calculate the moles of Fe3+ reacted

0.001348 mol Sn^2+ * 2 mol Fe^3+ / 1 mol Sn^2+ = 0.002696 mol Fe^3+

Now lets calculate the mass of the Fe3+

Mass of fe^3+ = moles * molar mass

                          = 0.002696 mol * 55.845 g per mol

                          = 0.1506 g Fe^3+

Now lets calculate the percent of the Fe

% Fe = (mass of Fe/ sample mass )*100%

          = (0.1506 g / 0.1875 g)*100 %

           = 80.3 % Fe

So the percent mass of Fe = 80.3 %

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