Can anyone help me in this problem?

The iron in a 6.675 g sample containing some Fe2O3 is reduced to Fe2+. The Fe2+ is titrated with 14.53 mL of 0.1467 M K2Cr2O7 in an acid solution.

Question

Can anyone help me in this problem?

The iron in a 6.675 g sample containing some Fe2O3 is reduced to Fe2+. The Fe2+ is titrated with 14.53 mL of 0.1467 M K2Cr2O7 in an acid solution.

anonymous · Answer

Find a. mass of Fe
       b. the percentage of Fe in the sample

anonymous · Answer

$$6Fe ^{2+}+Cr _{2}0_{7}^{2-}+14H ^{+}\rightarrow6Fe ^{3+}2Cr ^{3+}+7H _{2}O(l)  $$

aaronq · Answer

Use the same approach as with any stoichiometry problem.

Convert whats given to moles, use the mole ratio to find the moles of Iron. 

Convert moles to mass

anonymous · Answer

(0.01453 L) x (0.1467 mol/L K2Cr2O7) x (6 mol Fe / 1 mol K2Cr2O7) x 
(55.8452 g Fe/mol) =  0.7142 g Fe

(0.7142 g Fe) / (6.675 g) = 0.1070 = 10.70% Fe

anonymous · Answer

thank you @aaronq and @Tech1