I know its not Chemistry group but hey... TO ALL OF THOSE WHO CANNOT OR ARE HAVING PROBLEMS WITH BALANCING CHEMICAL REACTIONS: Back in 2012 I had to find a way that would make it easier for me to balance the long and complex chemical reactions learnt about in Chemistry101 up to, and possibly beyond 302. So I tried to develop a one size fits all type of method to compute the coefficients of the compounds found in a chemical reaction without the trial and error tedious operation. Take a look at this and see if you think you might make use of it. To illustrate this method lets take, for example, a simple combustion reaction of butane. C₄H10 + O₂ → CO₂ + H₂O ① Here what Im going to do is to put variables in front of every compound so that we can easily mark the places we want to input the numbers. Ill use the variables x, y, z, and j. (Note: Ill also use x, y, z, and j to indicate the term IN THE 1st, 2nd, 3rd or 4th compound, respectively.) xC₄H10 + yO₂ → zCO₂ + jH₂O ② Now ask yourself, how many types of elements are there in the chemical reaction? The answer is three. You have Carbon, Hydrogen, Oxygen. The following step is to indicate how many of each elements are present in each compound. As I have said, instead of saying in the first, second, third or fourth compound Ill simply use x, y, z and j to denote that. ③ Now check the number of these elements in each compound. How many carbons are there in the compounds? C: 4x + 0y → 1z + 0j How many oxygens are there in the compounds? O: 0x + 2y → 2z + 1j How many hydrogen are there in the compounds? H: 10x + 0y → 0z + 2j ④ Now we rewrite the equations, eliminating all the parts with zeros. So, Carbon: 4x → 1z Oxygen: 2y → 2z + 1j Hydrogen 10x → 2j This is the same as 4x = 1z 2y = 2z + 1j 10x = 2j ⑤ Perform Algebra: Solve simulataneously. So if x = 2 (I chose 2 because 1 wont make a difference) Then 4x = 1z becomes 4(2) = 1z which is 8 = z If 10x = 2j and x is still = 2 Then 10x = 2j Becomes 10(2) = 2j 20 = 2j J = 10 If 2y = 2z + 1j Then 2y = 2(8) + 1(10) 2y = 16 + 10 2y = 26 y = 13 ∴ x = 2, y = 13, j = 10 and z = 8 ⑥ Going back to our equation and substitute these values: xC₄H10 + yO₂ → zCO₂ + jH₂O 2C₄H10 + 13O₂ → 8CO₂ + 10H₂O And there you go, its balanced.
Posted on: Sun, 12 Oct 2014 06:16:41 +0000