1 6FeSO 4 (aq) + K 2 Cr 2 O 7 (aq) +7H 2 SO 4 (aq)  3Fe 2 (SO 4 ) 3 (aq) + Cr 2 (SO 4 ) 3 (aq) +7H 2 O(l) + K 2 SO 4 (aq) What are we looking for? A.

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1 6FeSO 4 (aq) + K 2 Cr 2 O 7 (aq) +7H 2 SO 4 (aq)  3Fe 2 (SO 4 ) 3 (aq) + Cr 2 (SO 4 ) 3 (aq) +7H 2 O(l) + K 2 SO 4 (aq) What are we looking for? A 3.33 g sample of iron ore is transformed to a solution of iron(II) sulfate, FeSO 4, and this solution is titrated with M K 2 Cr 2 O 7 (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?

2 6FeSO 4 (aq) + K 2 Cr 2 O 7 (aq) +7H 2 SO 4 (aq)  3Fe 2 (SO 4 ) 3 (aq) + Cr 2 (SO 4 ) 3 (aq) +7H 2 O(l) + K 2 SO 4 (aq) % Fe in the ore. Part x 100 = % Whole A 3.33 g sample of iron ore is transformed to a solution of iron(II) sulfate, FeSO 4, and this solution is titrated with M K 2 Cr 2 O 7 (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?

3 6FeSO 4 (aq) + K 2 Cr 2 O 7 (aq) +7H 2 SO 4 (aq)  3Fe 2 (SO 4 ) 3 (aq) + Cr 2 (SO 4 ) 3 (aq) +7H 2 O(l) + K 2 SO 4 (aq) % Fe in the ore. Part x 100 = % Whole the whole = 3.33 g ore the part = g of Fe So we have to find the g of Fe. A 3.33 g sample of iron ore is transformed to a solution of iron(II) sulfate, FeSO 4, and this solution is titrated with M K 2 Cr 2 O 7 (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?

4 6FeSO 4 (aq) + K 2 Cr 2 O 7 (aq) +7H 2 SO 4 (aq)  3Fe 2 (SO 4 ) 3 (aq) + Cr 2 (SO 4 ) 3 (aq) +7H 2 O(l) + K 2 SO 4 (aq) 41.7 mL grams of Fe??? A 3.33 g sample of iron ore is transformed to a solution of iron(II) sulfate, FeSO 4, and this solution is titrated with M K 2 Cr 2 O 7 (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?

5 6FeSO 4 (aq) + K 2 Cr 2 O 7 (aq) +7H 2 SO 4 (aq)  3Fe 2 (SO 4 ) 3 (aq) + Cr 2 (SO 4 ) 3 (aq) +7H 2 O(l) + K 2 SO 4 (aq) 41.7 mL = g Fe K 2 Cr 2 O 7 (sol) 41.7 mL %Fe = A 3.33 g sample of iron ore is transformed to a solution of iron(II) sulfate, FeSO 4, and this solution is titrated with M K 2 Cr 2 O 7 (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?