Techarena
Unregistered
Posts: N/A
Registered: N/A
Member Is Offline


How to Balance Chemical Equation
Hi, Need help in balancing chemical equations,
CH3OH + O2 = CO2 + H2O
I dont want simple resources to find the answer as i already got it from
https://www.easycalculation.com/chemistry/balancingequation...
I want to know how to find it


Milan
Harmless
Posts: 30
Registered: 1432015
Location: Europe
Member Is Offline
Mood: No Mood


Ok, first here's the equation:
2 CH3OH + 3 O2 = 2 CO2 + 4 H2O
I'll use it as an example.
So what you need to remember when balancing equations is the law of conservation of mass, which says "You can not create, nor destroy mass".
So if we apply it to the equation then the number of atoms of the same kind on the left side should be the same as the one on the right.
In this formula we have 2 Carbon atoms, 8 Hydrogen atoms and 8 Oxygen atoms on the left side, it is the same on the right side.
[Edited on 242015 by Milan]


woelen
Super Administrator
Posts: 8011
Registered: 2082005
Location: Netherlands
Member Is Offline
Mood: interested


You also need to have the same charge on the left and right side.
For many simple equations, it is a matter of guessing, but for more complicated equations where guessing is not easy, you can always find the right
coefficients by writing down equations for the number of atoms and the amount of charge on the left side and right side.
Example:
a Cr2O7(2) + b Fe(2+) + c H(+) > x Fe(3+) + y Cr(3+) + z H2O
This leads to a set of equations:
For Cr: 2a = y
For O: 7a = z
For Fe: b = x
For H: c = 2z
For charge: 2a + 2b + c = 3x + 3y
You now have 5 equations for 6 unknowns. Just fix one of them, e.g. to 1. Let's say a = 1. Now you can solve the system.
For Cr: a = 1 ==> y = 2
For O: a = 1 ==? z = 7
For H: z = 7 ==> c = 14
The equations for Fe and H give two simultaneous equations for unknowns b and x. I leave it as an exercise for you to solve these.
Normally, for all chemical equations, you will have N unknowns and N+1 equations. Fixing one of them usually gives a solution. If the solution has
fractional answers, then simply multiply all of them with the same factor to get rid of the fractions. If there is no solution, or there are
infinitely many solutions, then problem is either illposed, or there are multiple possible reactions, which can occur simultaneously.
I have written some tools, which solve chemical equations:
A small windows program: http://woelen.homescience.net/science/chem/chemeq/index.html
An online calculator: http://woelen.homescience.net:18080/chemeq/
These tools recognize situations of illposed problems or the possibility of multiple reactions.


morganbw
National Hazard
Posts: 561
Registered: 23112014
Member Is Offline
Mood: No Mood


@woelen
very informative post.


DraconicAcid
International Hazard
Posts: 4320
Registered: 122013
Location: The tiniest college campus ever....
Member Is Offline
Mood: Semivictorious.


When you are balancing equations, always work on one element at a time. Start with the elements that appear in the fewest compounds first, and leave
anything that appears as the element until last.
Consider this one:
C2H5OH + I2 > CHI3 + HI + H2O
Hydrogen appears in four compounds, iodine appears as the element, so do carbon and oxygen, then hydrogen, then iodine.
? C2H5OH + ? I2 > ? CHI3 + ? HI + ? H2O
Balance the carbon: C2H5OH + ? I2 > 2 CHI3 + ? HI + ? H2O
Balance the oxygen: C2H5OH + ? I2 > 2 CHI3 + ? HI + H2O
Now, when we balance the hydrogen, we've only got one coefficient to change. We've got six hydrogens on the reactants side, and four accounted for on
the products, so we need two more, so we need 2 HI.
C2H5OH + ? I2 > 2 CHI3 + 2 HI + H2O
Now we've got eight iodines on the products side, so we need 4 I2.
C2H5OH + 4 I2 > 2 CHI3 + 2 HI + H2O
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your exinstructor wants them.


Nicodem

Thread Moved 242015 at 08:00 
woelen
Super Administrator
Posts: 8011
Registered: 2082005
Location: Netherlands
Member Is Offline
Mood: interested


DraconicAcid, this is a good strategy, which I also frequenty use to get to a solution very quickly, but it unfortunately does not always work.
Sometimes there are cyclic dependencies. In many cases these can be resolved by guessing, but this is not always the case, especially if you have
equations with high coefficients and low GCD's between coefficients.
So my advice is to first try guessing, then apply DraconicAcid's strategy (optionally augmented with guessing on the cyclic dependencies) and if all
of these fail, then use my general method of equation solving. The latter always works, but it is not the easiest way. For software purposes, however,
it is the only surefire way.
I implemented it in a slightly different way in my software, I determine the nullspace of the full set of equations. If it has dimension 1, then we
have a welldefined solution. If it has dimension 0, then the set cannot be balanced (e.g. ? PCl3 + ? H2O > ? H3PO4 + ? HCl) and if it has
dimension larger than 1, then there are multiple independent solutions (e.g. ? Cu + ? HNO3 > ? Cu(NO3)2 + ? NO + ? NO2 + ? H2O has nullspace
dimension equal to 2).


brubei
Hazard to Others
Posts: 188
Registered: 832015
Location: France
Member Is Offline
Mood: No Mood


please can you explain what a cyclic dependency is ?


woelen
Super Administrator
Posts: 8011
Registered: 2082005
Location: Netherlands
Member Is Offline
Mood: interested


I'll give a certain example. NO2 gives nitric acid with water and as a sideproduct, NO is produced.
? NO2 + ? H2O > ? HNO3 + ? NO
We can balance for hydrogen:
? NO2 + 1 H2O > 2 HNO3 + ? NO
Now, if we want to balance the oxygen, then we have a problem. It depends on the number of nitrogens. So, we could think that we first try to balance
the nitrogen. However, that is not possible, because it depends on the number of oxygens. Ergo, we have a cyclic dependency, in this situation just
two mutually dependent ones. In more complicated situations we might have that for balancing element A we need to have balanced B, for balancing B we
need to have balanced C and for balancing C we need to have balanced A.
Solving this requires either guessing, or solving a set of equations. Guessing is fairly easy for this one. Start with 3 NO2 and 1 NO in order to make
the nitrogens balanced. Just check whether the oxygens are balanced as well in that case. That indeed is the case, so the end result is
3NO2 + H2O > 2HNO3 + NO
This is a somewhat trivial example, but it demonstrates the idea.
If you cannot guess, then write the following for the unknowns:
x NO2 + 1 H2O > 2 HNO3 + y NO
For nitrogen: x = 2 + y
For oxygen: 2x + 1 = 6 + y
Solving this simultaneous set of equations is easy. It yields x = 3, y = 1. So, if you cannot guess a suitable set of values, then solve the set. It
is more work, but it is a surefire technique.


aga
Forum Drunkard
Posts: 7030
Registered: 2532014
Member Is Offline


Here's a thing i typed up for someone once :
Attachment: Balancing Equations.pdf (44kB) This file has been downloaded 555 times
Sometimes just seeing the answer in different words helps.


brubei
Hazard to Others
Posts: 188
Registered: 832015
Location: France
Member Is Offline
Mood: No Mood


Ok thx
I usually balance the equation element by element and i'm pretty good at it. by the way, the mathematic way is nice too.


diddi
National Hazard
Posts: 723
Registered: 2392014
Location: Victoria, Australia
Member Is Offline
Mood: Fluorescent


@woelen
in you program, how have you implemented the solution of the matrix? I am thinking for a complex set of values you would need Gauss Seidel
elimination...
Beginning construction of periodic table display

