A 3.33 g sample of iron ore is transformed to a solution of iron(II) sulfate, FeSO
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6FeSO 4 (aq) + K 2 Cr 2 O 7 (aq) +7H 2 SO 4 (aq)  PowerPoint PPT Presentation


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6FeSO 4 (aq) + K 2 Cr 2 O 7 (aq) +7H 2 SO 4 (aq) 

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6feso4aq k2cr2o7aq 7h2so4aq

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?


6feso4aq k2cr2o7aq 7h2so4aq

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 x 100 = %

Whole


6feso4aq k2cr2o7aq 7h2so4aq

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.

Part x 100 = %

Whole

the part = g of Fe

the whole = 3.33 g ore


6feso4aq k2cr2o7aq 7h2so4aq

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) 

41.7 mL

3Fe2(SO4)3(aq) + Cr2(SO4)3(aq) +7H2O(l) + K2SO4(aq)

grams of Fe???


6feso4aq k2cr2o7aq 7h2so4aq

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) 

41.7 mL

3Fe2(SO4)3(aq) + Cr2(SO4)3(aq) +7H2O(l) + K2SO4(aq)

K2Cr2O7(sol)

41.7 mL

= g Fe

2.096

%Fe =


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