A total of 32.15 ml of K_2Cr_2O_7 is required for the titration of a 0.3502 gram

Question

A total of 32.15 ml of K_2Cr_2O_7 is required for the titration of a 0.3502 gram sample of FeSO_4, that is 84.12% pure. If the products are Cr^+3 and Fe^+3, what is the Molarity of the K_2Cr_2O_7? You take a sample of iron ore, dry it, and then weigh it at 0.6556 grams. After dissolving it and reducing the natural Fe^+3 to Fe^+2, this sample was titrated with KMnO_4. This sample required 19.22 ml of 0.08322 M KMnO_4 to reach the equivalence point, where the permanganate was reduced to Mn^+2. Calculate the % Fe in the original ore. How else might you analyze this solution for %Fe??? Which is easier?

Explanation / Answer

7.a) Volume of K2Cr2O7 (V1) = 32.15mL

Mass of FeSO4 =0.3502g

Purity of Fe sample= 84.12%

molecular weight of FeSO4= 151.9g/mol

Number of moles of FeSO4 in the sample= (0.3502/151.9) *82/100= 0.0019 moles

and the total reaction is:

Cr2O72- + 6 Fe2+ + 14H+ 2Cr3+ + 6 Fe3+ + 7H2O

Number of moles of Cr2O7(2-) required= 0.0019/6= 0.00032moles

Molarity of chromate solution= Number of moles*1000/32.15= 0.01M

b) Weight of iron ore (Fe3+)= 0.6556g

Volume of KMnO4 (V1)= 19.22mL

Molarity of KMnO4 (M1)=0.08322M

Nmber of moles of KMnO4= 0.08322*19.22/1000=0.0016moles

reaction taking place during titration is

5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ + 4H2O

Number of moles of Fe2+ reacted dring titration=0.0016*5=0.008 moles

We suppose the number of moles of Fe3+is also the same

Molar mass of Fe3+= 0.6556/0.008 = 81.95g/mol

Since the iron ore composition is not given exactly moleclar weight is unknown.

c) Gravimetric analysis could be used as an alternative Other instrumental techniques like Atomic absorption spectra could also be used

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