Question

A sample of iron ore weighing 0.6428g is dissolved in acid, the iron is reduced to Fe2+, and the solution is titrated with 36.30 mL of 0.01753 M K2Cr2O7 solution.

The reaction is

6Fe2+ + Cr2O72- +14H+ à 6Fe3+ + 2Cr3+ + 7H2O

What is the Percentage of iron (55.85g/mol) in the sample?

Answer 1

Balanced chemical equation is:

6Fe2+ + Cr2O72- +14H+ à 6Fe3+ + 2Cr3+ + 7H2O

lets calculate the mol of Cr2O72-

volume , V = 36.3 mL

= 3.63*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 1.753*10^-2*3.63*10^-2

= 6.363*10^-4 mol

According to balanced equation

mol of Fe2+ reacted = (6/1)* moles of Cr2O72-

= (6/1)*6.363*10^-4

= 3.818*10^-3 mol

This is number of moles of Fe2+

Molar mass of Fe = 55.85 g/mol

use:

mass of Fe,

m = number of mol * molar mass

= 3.818*10^-3 mol * 55.85 g/mol

= 0.2132 g

Now use:

Mass % of Fe = mass of Fe * 100 / mass of sample

= 0.2132 * 100 / 0.6428

= 33.17 %

Answer: 33.17 %