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Crank
2 years ago
13

An iron ore sample weighing 0.5562 g is dissolved HCl (aq), and the iron is obtained as Fe2 in solution. This solution is then t

itrated with 28.72 mL of 0.04021 M K2Cr2O7 (aq). What is the percent by mass iron in the ore sample
Chemistry
1 answer:
erik [133]2 years ago
8 0

Answer:

\% Fe^{+2}=70%

Explanation:

Hello,

In this case, we could considering this as a redox titration:

Fe^{+2}+(Cr_2O_7)^{-2}\rightarrow Fe^{+3}+Cr^{+3}

Thus, the balance turns out (by adding both hydrogen ions and water):

Fe^{+2}\rightarrow Fe^{+3}+1e^-\\(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\6Fe^{+2}+(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow Cr^{+3}+7H_2O+6Fe^{+3}+6e^-

Thus, by stoichiometry, the grams of Fe+2 ions result:

m_{Fe^{+2}}=0.04021\frac{molK_2Cr_2O_7}{L}*0.02872L*\frac{6molFe^{+2}}{1molK_2Cr_2O_7}*\frac{56gFe^{+2}}{1molFe^{+2}}=0.388gFe^{+2}

Finally, the mass percent is:

\% Fe^{+2}=\frac{0.388g}{0.5562g}*100\%\\  \% Fe^{+2}=70%

Best regards.

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