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

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

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:

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

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