Answer:
[tex]\% Fe^{+2}=70%[/tex]
Explanation:
Hello,
In this case, we could considering this as a redox titration:
[tex]Fe^{+2}+(Cr_2O_7)^{-2}\rightarrow Fe^{+3}+Cr^{+3}[/tex]
Thus, the balance turns out (by adding both hydrogen ions and water):
[tex]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^-[/tex]
Thus, by stoichiometry, the grams of Fe+2 ions result:
[tex]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}[/tex]
Finally, the mass percent is:
[tex]\% Fe^{+2}=\frac{0.388g}{0.5562g}*100\%\\ \% Fe^{+2}=70%[/tex]
Best regards.
