Are there any easier ways to balance an equation?

[beachbum]

our teacher tells us to just keep on adding coefficients and dividing or multiplying it by 2 until its balanced and it takes FOREVER to balance an equation.

[Primarygun]

Solving several linear equation by assigning the amount of each element when it is really too difficult.

The amount from left side is exactly equal to the amount from right side.

[beachbum]

yea but is there an easier calculation? like finding LCM for example (although thats not the way to do it) but something like that? or else it takes a while to figure out the coefficients especially for more complicated molecules.

[jdurg]

What I tend to do is split the entire reaction into tiny little reactions. I'll look at each side and try and figure out what the 'easiest' 'small reaction' would be to balance. This is a reaction where the balancing of one element is accomplished by addition of a simply binary compound or pure element. Generally, anything with a hydrogen in it can be balanced fairly easy as you just add an H+ to one side, or H2O if the overall equation is neutral. Knowing the oxidation states of the elements involved in the reaction will also help out tremendously as you'll be able to easily balance the oxidation numbers. Once you have the 'main' reaction balanced, you slowly add back in the other reactants/products and balance those out.

[jdurg]

Here's a good example:

 

We're going to try and balance the reaction between potassium chlorate and hydrochloric acid which gives potassium chloride, water, chlorine dioxide and chlorine gas.

 

KClO3 + HCl --> KCl + H2O + ClO2 + Cl2

 

Remember, you can always start out by balancing it with fractions as coefficients. Then you can just use some simple mathematics to eliminate the fractions and get a whole number. It's also important to try and see which is the 'key' compound in the reaction. Every reaction has one compound which really controls the progress of the balancing. In this case, it would be KClO3 since it contains K, Cl, and O in its formula. It's only missing the H which is easy to balance since H is only present in the water on the right side. So you'd start out by eliminating pretty much everything but the KClO3. So I would start like this:

 

KClO3 -> ClO2 + KCl

 

With this step the potassium is balanced, but the chlorine is one too many on the right and the oxygen is one two few. So to correct this we will bring in the HCl portion and the water. This gives:

 

KClO3 + HCl -> KCl + H2O + ClO2

 

Now the potassium is balanced, the oxygen is balanced, and the chlorine is balanced. But the hydrogen is short one on the left side. So we'll add another molecule of HCl to give us:

 

KClO3 + 2HCl -> KCl + H2O + ClO2

 

Now the potassium is balanced, the hydrogen is balanced, and the oxygen is balanced. Sadly, the chlorine is off with 3 on the left side and only two on the right. So now we'll bring in our final part of this equation which is the chlorine gas. This can be balanced by using a 1/2 coefficient which gives us:

 

KClO3 + 2HCl -> KCl + H2O + ClO2 + (1/2)Cl2

 

Now everything is balanced. So we just need to multiply every coefficient by two to get a balanced, whole number coefficient for each substance.

 

2KClO3 + 4HCl --> 2KCl + 2H2O + 2ClO2 + Cl2

 

It is now balanced.

[woelen]

Here's a good example:

 

We're going to try and balance the reaction between potassium chlorate and hydrochloric acid which gives potassium chloride' date=' water, chlorine dioxide and chlorine gas.

 

KClO3 + HCl --> KCl + H2O + ClO2 + Cl2

 

<text snipped for brevity>

 

Now everything is balanced. So we just need to multiply every coefficient by two to get a balanced, whole number coefficient for each substance.

 

2KClO3 + 4HCl --> 2KCl + 2H2O + 2ClO2 + Cl2

 

It is now balanced.[/quote']

Jdurg, the method you present is nice, but it is not always correct. In this example, the answer you get does not necessarily reflect what is really happening. The solution you find is not the only solution, it is just a single element from an infinetely large set of solutions. The one you have found has no special meaning and might even be wrong for the experimental conditions. Why not use the solution

 

7KClO3 + 18HCl --> 7KCl + 9H2O + 6ClO2 + 6Cl2 ?

 

 

In general, balancing equations, either at molecule-level or at ionic level, can be reduced to the problem of finding the null-space of a matrix. For many chemical reactions, the null-space is one-dimensional and in that case, the reaction is fully determined, except for a multiplicative factor. An example of this is the reaction between H2 and O2, forming H2O.

 

The reaction equation can be

 

2H2 + O2 ---> 2H2O,

 

but also

 

4H2 + 2O2 ---> 4H2O

 

But, the second one is just the first one, multiplied with 2. When a constraint is added, that the gcd of all coeffcients in the equation must be equal to 1, then the equation is determined uniquely, but only if the null-space of the equation-matrix has dimension 1.

 

Now we take your example: The null-space of this has dimension equal to 2. I'll leave out all the math, but in terms of your equation, there are two linearly independent equations, which satisfy:

 

5KClO3 + 6HCl ---> 5KCl + 3H2O + 6ClO2 (1a)

KClO3 + 6HCl ---> KCl + 3H2O + 3Cl2 (1b)

 

Any linear combination of these makes the equation balanced, some examples:

 

6KClO3 + 12HCl ----> 6KCl + 6H2O + 6ClO2 + 3Cl2, which can be simplified to your equation:

2KClO3 + 4HCl --> 2KCl + 2H2O + 2ClO2 + Cl2

 

Others are equally suitable:

11KClO3 + 18HCl --> 11KCl + 9H2O + 12ClO2 + 3Cl2

7KClO3 + 18HCl --> 7KCl + 9H2O + 6ClO2 + 6Cl2

When you look carefully to these, then you'll see that they cannot be reduced to your solution.

 

The chemical/physical meaning of this type of solution structure is that the real outcome of a chemical reaction cannot be predicted precisely. The reaction between KClO3 and HCl can be anywhere between the two extremes, as described by equations (1a) and (1b). The molar ratio between formed chlorine and chlorine dioxide depends on temperature and concentration.

 

 

In fact, the occurrence of this kind of chemical reactions is quite common. Another well known example is the reaction between copper metal and nitric acid:

 

Cu + HNO3 ----> Cu(NO3)2 + H2O + NO2 + NO

 

When you balance this equation, then you also see that the solution space is 2-dimensional.

 

Even higher dimensional solutions spaces can occur. When thiocyanate is oxidized by acidified permanganate, then the reaction products are sulfate, cyanogen, hydrogen cyanide and thiocyanogen.

 

MnO4(-) + SCN(-) + H(+) ---> H2O + Mn(2+) + SO4(2-) + HCN + (SCN)2 + (CN)2

 

The solution space of this reaction is 3-dimensional.

 

 

Just for fun, as an exercise, I have written a C-program, which balances any chemical equation, and detects the dimension of the null-space. This program gives all possible linearly independent solutions and you can generate any solution, by simply adding the different solutions. The program can be found on my website:

 

http://81.207.88.128/science/chem/chemeq/index.html