How do you balance the equation: #(A)FeSO_4+(B)KMnO+(C)H_2SO_4 -> (D)Fe_2(SO_4)_3+(E)K_2SO_4+(F)MnSO_4+(G)H_2O# ?

1 Answer
Feb 29, 2016

Answer:

#10FeSO_4+2KMnO+8H_2SO_4 -> 5Fe_2(SO_4)_3+K_2SO_4+2MnSO_4+8H_2O#

Explanation:

Let's eliminate unknowns backwards, starting with #G#.

From the equations for #H#, #Mn#, #K# and #Fe# we find:

#G=C#

#F=B#

#E=1/2 B#

#D=1/2 A#

Substituting these in the remaining equations we get:

#S: A+C=3(1/2 A)+(1/2 B)+B = 3/2A + 3/2B#

#O: 4A+4B+4C = 12 (1/2 A)+4 (1/2 B)+4B+C = 6A + 6B+C#

Subtracting #A# from both sides of the equation for #S# we get:

#C=1/2A+3/2B#

Substitute this in our equation for #O# to get:

#4A+4B+4(1/2A+3/2B) = 6A+6B+(1/2A+3/2B)#

That is:

#6A+10B = 13/2A+15/2B#

Multiplying both sides by #2# that becomes:

#12A+20B = 13A+15B#

Subtract #12A+15B# from both sides to get:

#5B=A#

So if we put #B=2# then we get:

#A=5B=10#

#B=2#

#C=1/2A+3/2B=5+3=8#

#D=1/2A = 5#

#E=1/2B = 1#

#F=B=2#

#G=C=8#

So:

#10FeSO_4+2KMnO+8H_2SO_4 -> 5Fe_2(SO_4)_3+K_2SO_4+2MnSO_4+8H_2O#