Question
Number of moles of $$KMn{O_4}$$ required to oxidize one mole of $$Fe\left( {{C_2}{O_4}} \right)$$ in acidic medium is
A.
0.167
B.
0.6
C.
0.2
D.
0.4
Answer :
0.6
Solution :
The required equation is
$$\eqalign{
& 2KMn{O_4} + 3{H_2}S{O_4} \to \cr
& {K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O + \mathop {5\left[ O \right]}\limits_{{\text{nascent oxygen}}} \cr
& 2Fe\left( {{C_2}{O_4}} \right) + 3{H_2}S{O_4} + 3\left[ O \right] \to \cr
& \,\,\,\,\,\,\,\,\,\,\,\,F{e_2}{\left( {S{O_4}} \right)_3} + 2C{O_2} + 3{H_2}O \cr} $$
$$O$$ required for $$1\,mol.$$ of $$Fe\left( {{C_2}{O_4}} \right)$$ is 1.5, $$5$$ $$O$$ are obtained from $$2\,moles$$ of $$KMn{O_4}$$
$$\therefore \,\,1.5\,\left[ O \right]$$ will be obtained from
$$ = \frac{2}{5} \times 1.5 = 0.6\,moles$$ of $$KMn{O_4}.$$