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Question Number 13260 by Tinkutara last updated on 17/May/17

$$\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{KMnO}}_4\mathrm{required}$$$$\mathrm{to}\mathrm{oxidise}\mathrm{one}\mathrm{mole}\mathrm{of}\mathrm{Fe}\left({\mathrm{C}}_2{\mathrm{O}}_4\right)\mathrm{in}$$$$\mathrm{acidic}\mathrm{medium}\mathrm{is}$$$$\left(\mathrm{a}\right)0.6$$$$\left(\mathrm{b}\right)1.67$$$$\left(\mathrm{c}\right)0.2$$$$\left(\mathrm{d}\right)0.4$$

Answered by sandy_suhendra last updated on 17/May/17

$$[{\mathrm{MnO}}_4^{-}+{8\mathrm{H}}^{+}+5\mathrm{e}\to {\mathrm{Mn}}^{2+}+4\mathrm{H}2\mathrm{O}]\times 3$$$$[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\to {2\mathrm{CO}}_2+2\mathrm{e}]\times 5$$$$[{\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}+\mathrm{e}]\times 5$$$$--------------------(+$$$${3\mathrm{MnO}}_4^{-}+{5\mathrm{C}}_2{\mathrm{O}}_4^{2-}+{5\mathrm{Fe}}^{2+}+{24\mathrm{H}}^{+}\to {3\mathrm{Mn}}^{2+}+{12\mathrm{H}}_2\mathrm{O}+{10\mathrm{CO}}_2+{5\mathrm{Fe}}^{3+}$$$$\mathrm{Fe}\left({\mathrm{C}}_2{\mathrm{O}}_4\right)=1\mathrm{mole}$$$${\mathrm{KMnO}}_4=\frac{3}{5}\times 1\mathrm{mole}=0.6\mathrm{mole}\left(\mathrm{A}\right)$$

Commented by Tinkutara last updated on 15/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

Answered by Tinkutara last updated on 15/Jul/17

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