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

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

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  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) What are we looking for?

  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)  3Fe2(SO4)3(aq) + Cr2(SO4)3(aq) +7H2O(l) + K2SO4(aq) % Fe in the ore. Part x 100 = % Whole

  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)  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

  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)  41.7 mL 3Fe2(SO4)3(aq) + Cr2(SO4)3(aq) +7H2O(l) + K2SO4(aq) grams of Fe???

  5. 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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