# How can I balance this reaction using the algebraic method?

How can I balance this reaction using the algebraic method?

$$\ce{K4Fe(CN)6 + KMnO4 + H2SO4 + H2SO4 -> KHSO4 + Fe2(SO4)3 + MnSO4 + HNO3 + CO2 + H2O}$$

My solution:

$$\ce{a K4Fe(CN)6 + b KMnO4 + c H2SO4 -> d KHSO4 + e Fe2(SO4)3 + f MnSO4 + g HNO3 + h CO2 + i H2O}$$

$\ce{K}{:~} 4a + b = d$
$\ce{Fe}{:~} a = 2e$
$\ce{C}{:~} 6a = h$
$\ce{N}{:~} 6a = g$
$\ce{Mn}{:~} b = f$
$\ce{O}{:~} 4b + 4c = 4d + 12e + 4f + 3g + 2h + i$
$\ce{H}{:~} 2c = d + g + 2i$
$\ce{S}{:~} c = d + 3e + f$

I couldn't solve further than that.

• I will just point it out - although you require the algebraic method - that the redox method instead is much easier than this. – Gaurang Tandon Jun 14 '18 at 11:18

## 2 Answers

You are lucky. From time to time, I take on a challenge like this, just to see if I can still do it. But normally, we expect that you show some effort yourself. See our homework policy.

We start off with some mathematics.

(1) $4a + b = d$
(2) $a = 2e$
(3) $6a = h$
(4) $6a = g$
(5) $b = f$
(6) $4b + 4c = 4d + 12e + 4f + 3g + 2h + i$
(7) $2c = d + g + 2i$
(8) $c = d + 3e + f$

and rewrite this in the "equals zero" form.

(1) $0 = 4a + b - d$
(2) $0 = a - 2e$
(3) $0 = 6a - h$
(4) $0 = 6a - g$
(5) $0 = b - f$
(6) $0 = 4b + 4c - 4d - 12e - 4f - 3g - 2h - i$
(7) $0 = 2c - d - g - 2i$
(8) $0 = c - d - 3e - f$

I fill this in a table. (I could have done this right from the original chemical equation.)

     a   b   c   d   e   f   g   h   i
(1)  4   1      -1
(2)  1              -2
(3)  6                          -1
(4)  6                      -1
(5)      1              -1
(6)      4   4  -4 -12  -4  -3  -2  -1
(7)          2  -1          -1      -2
(8)          1  -1  -3  -1


Now, (2) to (5) look simple enough to do some replacements. I start with (2) where $a = 2e$. In the first line, we replace $4a$ by $8e$ and so forth.

     a   b   c   d   e   f   g   h   i
(1)  0   1      -1  0+8
(2)  0             -2+2
(3)  0              0+12        -1
(4)  0              0+12    -1
(5)      1              -1
(6)      4   4  -4 -12  -4  -3  -2  -1
(7)          2  -1          -1      -2
(8)          1  -1  -3  -1


We got rid of $a$, (2) is cancelled out. (5) allows us to get rid of $b$ via $b=f$.

     a   b   c   d   e   f   g   h   i
(1)      0      -1   8  0+1
(3)                 12          -1
(4)                 12      -1
(5)      0             -1+1
(6)      0   4  -4 -12 -4+4 -3  -2  -1
(7)          2  -1          -1      -2
(8)          1  -1  -3  -1


Let's use (3) and (4) to remove $g=12e$ and $h=12e$.

     a   b   c   d   e   f   g   h   i
(1)             -1   8   1
(3)                12-12         0
(4)                12-12     0
(6)          4 -4 -12-36-24  0   0  -1
(7)          2  -1  0-12     0      -2
(8)          1  -1  -3  -1


i.e.

     a   b   c   d   e   f   g   h   i
(1)             -1   8   1
(6)          4  -4 -72              -1
(7)          2  -1 -12              -2
(8)          1  -1  -3  -1


Now we take (1) with $d = 8e+1f$

     a   b   c   d   e   f   g   h   i
(6)          4     -104 -4          -1
(7)          2      -20 -1          -2
(8)          1      -11 -2


Now we take (8) with $c = 11e+2f$

     a   b   c   d   e   f   g   h   i
(6)                -60   4          -1
(7)                  2   3          -2


One last step with (6) $i=-60e+4f$

     a   b   c   d   e   f   g   h   i
(7)                122  -5


That is $f=122e/5$

We cannot solve that further with mathematics. There is an infinite number of possible results. We need just one solution, so we guess one number. $e=5$ seems good so that we get a nice $f$. Now, from the many equations we had above, we search some that fit.

$e = 5$
$a = 2e = 10$
$f = 122e/5 = 122$
$g = 12e = 60$
$h = 12e = 60$
$b = f = 122$
$c = 11e+2f = 299$
$d = 4a+b = 162$
$i = -60e+4f = 188$

Now all you have to do is variable substitution, which is kinda tedious since it's such a large equation. Remember that solutions are defined up to multiplying everything by a constant factor, so we might as well pick

• $a = 2$ (this guess is inspired by the equation for $\ce{Fe}$)
• $\ce{Fe}{:~} e = 1$
• $\ce{C}{:~} h = 12$
• $\ce{N}{:~} g = 12$
• Because $\ce{Mn}{:~} b = f$, we can rewrite $\ce{O}$ as $4c = 4d + 12e + 3g + 2h + i$
• Subtract 4 times $\ce{S}$, and we get $0 = -4f + 3g + 2h + i$, so $4f = 60 + i$ (call this $\ce{O}'$)
• Now rewrite everything in terms of $b$:
• $\ce{K}{:~} 8 + b = d$, so $d = b + 8$
• $\ce{Mn}{:~} f = b$
• $\ce{O}'{:~} 4b = 60 + i$, so $i = 4b - 60$
• $\ce{S}{:~} c = (b + 8) + 3 + b = 2b + 11$
• $\ce{H}{:~} 2(2b + 11) = b + 8 + 12 + 2(4b - 60)$, so $4b + 22 = 9b - 100$, so $5b = 122$ and $b = 122/5$. This is a bit unfortunate since later on we have to multiply all variables by 5.
• We now have $a=2$, $b=122/5$, $c=299/5$, $d=162/5$, $e=1$, $f=122/5$, $g=12$, $h=12$ and $i=188/5$.
• By multiplying everything by 5 we get the final solution:
• $a = 10$
• $b = 122$
• $c = 299$
• $d = 162$
• $e = 5$
• $f = 122$
• $g = 60$
• $h = 60$
• $i = 188$

You can check that these figures will make the equation balanced.