Transcript No Slide Title

A 3.33 g sample of iron ore is transformed to a solution of iron(II) sulfate,
FeSO4, and this solution is titrated with 0.150 M K2Cr2O7 (potassium
dichromate). If it requires 41.7 mL of potassium dichromate solution to
titrate the iron(II) sulfate solution, what is the percentage of iron in the
ore?
6FeSO4(aq) + K2Cr2O7(aq) +7H2SO4(aq) 
3Fe2(SO4)3(aq) + Cr2(SO4)3(aq) +7H2O(l) + K2SO4(aq)
What are we looking for?
1
A 3.33 g sample of iron ore is transformed to a solution of iron(II) sulfate,
FeSO4, and this solution is titrated with 0.150 M K2Cr2O7 (potassium
dichromate). If it requires 41.7 mL of potassium dichromate solution to
titrate the iron(II) sulfate solution, what is the percentage of iron in the
ore?
6FeSO4(aq) + K2Cr2O7(aq) +7H2SO4(aq) 
3Fe2(SO4)3(aq) + Cr2(SO4)3(aq) +7H2O(l) + K2SO4(aq)
% Fe in the ore.
Part
Whole
x 100 = %
2
A 3.33 g sample of iron ore is transformed to a solution of iron(II) sulfate, FeSO4,
and this solution is titrated with 0.150 M K2Cr2O7 (potassium dichromate). If it
requires 41.7 mL of potassium dichromate solution to titrate the iron(II) sulfate
solution, what is the percentage of iron in the ore?
6FeSO4(aq) + K2Cr2O7(aq) +7H2SO4(aq) 
So we have to
find the
g of Fe.
3Fe2(SO4)3(aq) + Cr2(SO4)3(aq) +7H2O(l) + K2SO4(aq)
% Fe in the ore.
the part = g of Fe
Part
Whole
x 100 = %
the whole = 3.33 g ore
3
A 3.33 g sample of iron ore is transformed to a solution of iron(II) sulfate, FeSO4, and this
solution is titrated with 0.150 M K2Cr2O7 (potassium dichromate). If it requires 41.7 mL of
potassium dichromate solution to titrate the iron(II) sulfate solution, what is the percentage
of iron in the ore?
6FeSO4(aq) + K2Cr2O7(aq) +7H2SO4(aq) 
0.150 mol
L
41.7 mL
3Fe2(SO4)3(aq) + Cr2(SO4)3(aq) +7H2O(l) + K2SO4(aq)
grams of Fe???
4
A 3.33 g sample of iron ore is transformed to a solution of iron(II) sulfate, FeSO4, and this
solution is titrated with 0.150 M K2Cr2O7 (potassium dichromate). If it requires 41.7 mL of
potassium dichromate solution to titrate the iron(II) sulfate solution, what is the percentage
of iron in the ore?
6FeSO4(aq) + K2Cr2O7(aq) +7H2SO4(aq) 
0.150 mol
L
41.7 mL
3Fe2(SO4)3(aq) + Cr2(SO4)3(aq) +7H2O(l) + K2SO4(aq)
K2Cr2O7(sol)
41.7 mL
0.150mol K 2Cr2O 7 6 mol FeSO 4
1 mol Fe 55.85 g Fe
= 2.096 g Fe
1000mL
mol
Fe
1 mol K 2Cr2O 7 1 mol FeSO 4
K 2Cr2O 7 (sol)
%Fe =
2.096
x 100  62.9 % Fe
3.33
5