Question

A 1.7316 gram sample of a steel is analyzed for iron content. The sample is dissolved in hydrochloric acid, giving

Fe2+ ion, which is titrated with 0.168 M K2Cr2O7 according to the following reaction.

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

If the titration required 23.77 mL of potassium dichromate solution, what was the mass percentage of iron in the

sample?

Answer 1

Balanced chemical equation is:

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

lets calculate the mol of Cr2O72-

volume , V = 23.77 mL

= 2.377*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.168*2.377*10^-2

= 3.993*10^-3 mol

According to balanced equation

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

= (6/1)*3.993*10^-3

= 2.396*10^-2 mol

This is number of moles of Fe

Molar mass of Fe = 55.85 g/mol

use:

mass of Fe,

m = number of mol * molar mass

= 2.396*10^-2 mol * 55.85 g/mol

= 1.338 g

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

= 1.338 * 100 / 1.7316

= 77.27 %

Answer: 77.27 %