#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)Mn->stackrel(+2)Mn=>"5unit reduction"#
#stackrel(-1)O->stackrel(0)O=>"1unit oxidation"#
The interacting ratio of oxidant and rductant is
#stackrel(+7) Mn:stackrel(-1)O=1:5#
#=>stackrel(+7) Mn:stackrel(-2)O_2=1:5/2=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_2SO_4->1mol#
#MnSO_4->2mol#
#Na_2SO_4->5mol#
#" TOTAL "->8mols#
So salt former #H_2SO_4->8mols# 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
