Question #ec000

1 Answer
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)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