# Question ec000

Nov 21, 2016

Kstackrel(+7)MnO_4 + Na_2stackrel(-1)O_2 + H_2SO_4 → K_2SO_4 + stackrel(+2)MnSO_4 + Na_2SO_4 + H_2O + stackrel(0)O_2

The changes in ON are as follows

$\stackrel{+ 7}{M} n \to \stackrel{+ 2}{M} n \implies \text{5unit reduction}$

$\stackrel{- 1}{O} \to \stackrel{0}{O} \implies \text{1unit oxidation}$

The interacting ratio of oxidant and rductant is

$\stackrel{+ 7}{M} n : \stackrel{- 1}{O} = 1 : 5$

$\implies \stackrel{+ 7}{M} n : {\stackrel{- 2}{O}}_{2} = 1 : \frac{5}{2} = 2 : 5$

The balanced equation becomes

2Kstackrel(+7)MnO_4 + 5Na_2stackrel(-1)O_2 + 8H_2SO_4 → K_2SO_4 + 2stackrel(+2)MnSO_4 + 5Na_2SO_4 + 8H_2O + 5stackrel(0)O_2#

Here salts produced by adjusting the coefficients at LHS imposing the required reacting ratio $2 : 5$ are

${K}_{2} S {O}_{4} \to 1 m o l$

$M n S {O}_{4} \to 2 m o l$

$N {a}_{2} S {O}_{4} \to 5 m o l$

$\text{ TOTAL } \to 8 m o l s$

So salt former ${H}_{2} S {O}_{4} \to 8 m o l s$ and the corresponding water produced will be 8mols.

The moles of ${O}_{2}$ produced will be same as moles of ${\stackrel{- 2}{O}}_{2}$ at LHS i.e.5