Balancing Chemical Equations [LifeHack]

Its not often that we come across chemical equations with coefficients like 48 and 50.But when we do most of us are meant to screw up.Thats exactly why a friend of mine asked me to come up with something.(Its hard enough just to memorise all those equations.)

I'll show you how to balance complicated chemical equations with huge coefficients easily and quickly.All you need to know is how to solve linear equations in variables.

The method might seem long but the equations that we get are simple and the method is much quicker than manually trying to balance the atoms.

Steps: 

1.Assign coefficients to each term in the equation (a,b,c.....)

2.Now balance each element i.e. solve for the coefficients.

(You can skip some steps by assigning coefficients directly as shown below)

Lets start with..

S₈ + HNO₃ = H₂SO₄ + NO₂ + H₂O

Lets say there are x moles of S₈ then there will be 8x moles of H₂SO₄.(Balancing S)

xS₈ + HNO₃ = 8xH₂SO₄ + NO₂ + H₂O

Now say there are y moles of HNO₃ then there will be y moles of NO₂.(Balancing N)

xS₈ + yHNO₃ = 8xH₂SO₄ + yNO₂ + H₂O

Now put z for H₂O.

xS₈ + yHNO₃ = 8xH₂SO₄ + yNO₂ + zH₂O

Now equate the number of O on both sides.

3y = 32x + 2y + z

y = 32x + z               --- 1

Now equate the number of H on both sides.

y = 16x + 2z             --- 2

Solve 1 & 2 we get 

z = 16x

y = 48x

Now the equation become

xS₈ + 48xHNO₃ = 8xH₂SO₄ + 48xNO₂ + 16xH₂O

Putting least integer value for x

S₈ + 48HNO₃ = 8H₂SO₄ + 48NO₂ + 16H₂O

Now lets balance a humongous equation...

 K₄[Fe(SCN)₆] + K₂Cr₂O₇ +  H₂SO₄ =  Fe₂(SO₄)₃ +  Cr₂(SO₄)₃ +  CO₂ +  H₂O +  K₂SO₄ + KNO₃

Let the coefficient of  K₄[Fe(SCN)₆] be x then coefficient of Fe₂(SO₄)₃ will be x/2 (Balancing Fe)
The coefficient of CO₂ will be 6x (Balancing C)
The coefficient of KNO₃ will be 6x (Balancing N)
xK₄[Fe(SCN)₆] + K₂Cr₂O₇ +  H₂SO₄ =  x/2Fe₂(SO₄)₃ +  Cr₂(SO₄)₃ +  6xCO₂ +  H₂O +  K₂SO₄ + 6xKNO₃

Let the coefficient of K₂Cr₂O₇ be y then coefficient of Cr₂(SO₄)₃ will also be y (Balancing Cr)
xK₄[Fe(SCN)₆] + yK₂Cr₂O₇ +  H₂SO₄ =  x/2Fe₂(SO₄)₃ +  yCr₂(SO₄)₃ +  6xCO₂ +  H₂O +  K₂SO₄ + 6xKNO₃

Now put some random coefficient for others.
xK₄[Fe(SCN)₆] + yK₂Cr₂O₇ +  uH₂SO₄ =  x/2Fe₂(SO₄)₃ +  yCr₂(SO₄)₃ +  6xCO₂ +  vH₂O +  wK₂SO₄ + 6xKNO₃

Balance H.(This could have been done before assigning random coefficients)
2u = 2v
u = v   ---1

Balance O.
7y + 4u = 12x/2 + 12y + 12x + v + 4w +18x  
4u = 36x + 5y + v + 4w
3u = 36x + 5y +4w  (Since u = v )   ---2

Balance K.
4x + 2y = 2w + 6x
2y = 2w + 2x
y = w + x         ---3

Balance S.
6x + u = 3x/2 + 3y + w
9x + 2u = 6y + 2w      ---4

Solve 2 , 3 & 4.
We have 4 variables and 3 eqautions.So we can solve all other variables in terms of any of the variable.
Lets for instance solve all other variables in terms of y.
w = 91y/97    u=355y/97     x=6y/97

Now substituting in original chemical equation and put y =1 ..
6K₄[Fe(SCN)₆] + 97K₂Cr₂O₇ +  355H₂SO₄ =  3Fe₂(SO₄)₃ +  97Cr₂(SO₄)₃ +  36CO₂ +  355H₂O +  91K₂SO₄ + 36KNO₃


I've also developed a method for computers and will be publishing the algorithm as soon as my exams are over and i get hold of a computer.