# 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 ?

Feb 29, 2016

$10 F e S {O}_{4} + 2 K M n O + 8 {H}_{2} S {O}_{4} \to 5 F {e}_{2} {\left(S {O}_{4}\right)}_{3} + {K}_{2} S {O}_{4} + 2 M n S {O}_{4} + 8 {H}_{2} O$

#### Explanation:

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

From the equations for $H$, $M n$, $K$ and $F e$ we find:

$G = C$

$F = B$

$E = \frac{1}{2} B$

$D = \frac{1}{2} A$

Substituting these in the remaining equations we get:

$S : A + C = 3 \left(\frac{1}{2} A\right) + \left(\frac{1}{2} B\right) + B = \frac{3}{2} A + \frac{3}{2} B$

$O : 4 A + 4 B + 4 C = 12 \left(\frac{1}{2} A\right) + 4 \left(\frac{1}{2} B\right) + 4 B + C = 6 A + 6 B + C$

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

$C = \frac{1}{2} A + \frac{3}{2} B$

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

$4 A + 4 B + 4 \left(\frac{1}{2} A + \frac{3}{2} B\right) = 6 A + 6 B + \left(\frac{1}{2} A + \frac{3}{2} B\right)$

That is:

$6 A + 10 B = \frac{13}{2} A + \frac{15}{2} B$

Multiplying both sides by $2$ that becomes:

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

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

$5 B = A$

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

$A = 5 B = 10$

$B = 2$

$C = \frac{1}{2} A + \frac{3}{2} B = 5 + 3 = 8$

$D = \frac{1}{2} A = 5$

$E = \frac{1}{2} B = 1$

$F = B = 2$

$G = C = 8$

So:

$10 F e S {O}_{4} + 2 K M n O + 8 {H}_{2} S {O}_{4} \to 5 F {e}_{2} {\left(S {O}_{4}\right)}_{3} + {K}_{2} S {O}_{4} + 2 M n S {O}_{4} + 8 {H}_{2} O$