Question Number 13260 by Tinkutara last updated on 17/May/17
$$\mathrm{Number}\:\mathrm{of}\:\mathrm{moles}\:\mathrm{of}\:\mathrm{KMnO}_{\mathrm{4}} \:\mathrm{required} \\ $$$$\mathrm{to}\:\mathrm{oxidise}\:\mathrm{one}\:\mathrm{mole}\:\mathrm{of}\:\mathrm{Fe}\left(\mathrm{C}_{\mathrm{2}} \mathrm{O}_{\mathrm{4}} \right)\:\mathrm{in} \\ $$$$\mathrm{acidic}\:\mathrm{medium}\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{0}.\mathrm{6} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{1}.\mathrm{67} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{0}.\mathrm{2} \\ $$$$\left(\mathrm{d}\right)\:\mathrm{0}.\mathrm{4} \\ $$
Answered by sandy_suhendra last updated on 17/May/17
![[MnO_4 ^− + 8H^+ + 5e → Mn^(2+) +4H2O]×3 [C_2 O_4 ^(2−) → 2CO_2 + 2e]×5 [Fe^(2+) → Fe^(3+) + e]×5 −−−−−−−−−−−−−−−−−−−−(+ 3MnO_4 ^− + 5C_2 O_4 ^(2−) + 5Fe^(2+) + 24H^+ → 3Mn^(2+) + 12H_2 O +10CO_2 + 5Fe^(3+) Fe(C_2 O_4 ) = 1 mole KMnO_4 = (3/5) × 1 mole = 0.6 mole (A)](https://www.tinkutara.com/question/Q13296.png)
$$\left[\mathrm{MnO}_{\mathrm{4}} ^{−} \:+\:\mathrm{8H}^{+} \:+\:\mathrm{5e}\:\rightarrow\:\mathrm{Mn}^{\mathrm{2}+} \:+\mathrm{4H2O}\right]×\mathrm{3} \\ $$$$\left[\mathrm{C}_{\mathrm{2}} \mathrm{O}_{\mathrm{4}} ^{\mathrm{2}−} \:\rightarrow\:\mathrm{2CO}_{\mathrm{2}} \:+\:\mathrm{2e}\right]×\mathrm{5} \\ $$$$\left[\mathrm{Fe}^{\mathrm{2}+} \:\rightarrow\:\mathrm{Fe}^{\mathrm{3}+} \:+\:\mathrm{e}\right]×\mathrm{5} \\ $$$$−−−−−−−−−−−−−−−−−−−−\left(+\right. \\ $$$$\mathrm{3MnO}_{\mathrm{4}} ^{−} \:+\:\mathrm{5C}_{\mathrm{2}} \mathrm{O}_{\mathrm{4}} ^{\mathrm{2}−} \:+\:\mathrm{5Fe}^{\mathrm{2}+} \:+\:\mathrm{24H}^{+} \:\rightarrow\:\mathrm{3Mn}^{\mathrm{2}+} \:+\:\mathrm{12H}_{\mathrm{2}} \mathrm{O}\:+\mathrm{10CO}_{\mathrm{2}} \:+\:\mathrm{5Fe}^{\mathrm{3}+} \:\:\:\:\: \\ $$$$\mathrm{Fe}\left(\mathrm{C}_{\mathrm{2}} \mathrm{O}_{\mathrm{4}} \right)\:=\:\mathrm{1}\:\mathrm{mole} \\ $$$$\mathrm{KMnO}_{\mathrm{4}} \:=\:\frac{\mathrm{3}}{\mathrm{5}}\:×\:\mathrm{1}\:\mathrm{mole}\:=\:\mathrm{0}.\mathrm{6}\:\mathrm{mole}\:\left(\mathrm{A}\right) \\ $$
Commented by Tinkutara last updated on 15/Jul/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Answered by Tinkutara last updated on 15/Jul/17
