Balancing Equations: Tips and Tricks for Mastering the Process

[synkk]

Homework Statement

Cl2 + NaOH --> NaCl + NaClO3 + H2O

I'm having troubles in balancing questions, is there a certain method to doing it or is it all just about practice?

I managed to get to 3Cl2 + 6NaOH --> 4NaCl + 2NaClO3 + 3H20

But now I am completely stuck and have no idea how to balance it.

 

[CompuChip]

Generally, the system is to start with something that only occurs once on each side, because that will fix at least one number. However, in this case there is no such compound.

As a mathematically inclinded person, I would just go and make it a mathematical problem. You can write variables for the coefficients:
a Cl2 + b NaOH = c NaCl + d NaClO3 + e H2O

Then the balancing gives you a set of equations:
Cl: 2a = c + d
Na: b = c + d
O: b = 3d + e
H: b = 2e

You have one degree of freedom (4 equations, 5 unknowns), so let's just set a = 1. Then you get:
a = 1
b = 2a = 2 (from the Cl equation)
2 = 2e, so e = 1 (from the H equation)
2 = 3d + 1, so d = 1/3 (from the O equation)
2 = c + 1/3, so c = 5/3 (from the Na equation)

Multiply it all by a suitable constant (3) to get rid of the fractions:
a = 3, b = 6, c = 5, d = 1, e = 3

I don't know what you level of math is, but using matrices there is a general way to solve these problems.

 

[synkk]

Generally, the system is to start with something that only occurs once on each side, because that will fix at least one number. However, in this case there is no such compound.

As a mathematically inclinded person, I would just go and make it a mathematical problem. You can write variables for the coefficients:
a Cl2 + b NaOH = c NaCl + d NaClO3 + e H2O

Then the balancing gives you a set of equations:
Cl: 2a = c + d
Na: b = c + d
O: b = 3d + e
H: b = 2e

You have one degree of freedom (4 equations, 5 unknowns), so let's just set a = 1. Then you get:
a = 1
b = 2a = 2 (from the Cl equation)
2 = 2e, so e = 1 (from the H equation)
2 = 3d + 1, so d = 1/3 (from the O equation)
2 = c + 1/3, so c = 5/3 (from the Na equation)

Multiply it all by a suitable constant (3) to get rid of the fractions:
a = 3, b = 6, c = 5, d = 1, e = 3

I don't know what you level of math is, but using matrices there is a general way to solve these problems.

Wow that was amazing, and the correct answer, how did you get to that level of mathematics to solve it just like that. Thanks, it's the correct answer.

 

[CompuChip]

As I said, I don't know what your current level of maths is.

Have you learned to solve multiple equations (systems of equations)? Do you know about matrices?

 

[synkk]

As I said, I don't know what your current level of maths is.

Have you learned to solve multiple equations (systems of equations)? Do you know about matrices?

No the only type of "multiple equations" would be simultaneous equations. I haven't learned matrices yet, but I'm pretty sure they're on the syllabus.

 

[CompuChip]

Sorry, that's what I meant.

Note that
2a = c + d
b = c + d
b = 3d + e
b = 2e
is just a system of simultaneous equations. It is straightforward to write them down (just stick a variable in front of all the components of the chemical formula), solving them is the hardest part (but that, too, is quite straightforward - it doesn't require a lot of insight, just good computational skills) :-)

 

[CompuChip]

And I just noticed that my answer was different from yours... so I basically gave the solution away. You're welcome :)

 

[synkk]

And I just noticed that my answer was different from yours... so I basically gave the solution away. You're welcome :)

Thank you, i got the answer before you posted the solution but i done it by just trial and error.

 

[synkk]

Sorry, that's what I meant.

Note that
2a = c + d
b = c + d
b = 3d + e
b = 2e
is just a system of simultaneous equations. It is straightforward to write them down (just stick a variable in front of all the components of the chemical formula), solving them is the hardest part (but that, too, is quite straightforward - it doesn't require a lot of insight, just good computational skills) :-)

Sorry to bump it again, but why did you let a = 1?

 

[Borek]

There are more variables than equations, so this system can't be solved. Assuming one of the variables has known value (be it 1 or anything else) allows calculation of all other values. That still means that the resulting solution is not unique, as we could put a=2 or a=7 as initial value, and each set of the coefficients found this way will give balanced equation. However, by convention we want our equation to have all coefficients to be the smallest possible integers - this is additional information that allows us to select the correct set of coefficients from the all possible solutions. Putting a=1 is just the simplest way to do. Sometimes it will mean your coefficients will be not integers, but fractions - that's not a problem, you just have to multiply them later by the least common multiple of denominators.

See balancing chemical equations and balancing chemical equations using algebraic method for some more information about equation balancing in general, and this method in particular.

 

[synkk]

Thank a lot guys, i used the method to solve all other equations i need to do and works perfectly.