February 21, 2020, 05:11:29 AM
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### Topic: Balancing really difficult chemical equations  (Read 11763 times)

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#### Borek ##### Re: Balancing really difficult chemical equations
« Reply #15 on: January 11, 2013, 03:53:46 AM »
Gaussian elimination is just a step on the numerical approach to solving system of equations - that is, it is a tool that will help you find a solution like the one curiouscat posted: Now as everything is expressed using a, assume a=1 and see if all coefficients are small integers. If they are not - see what you can do to make them integers (that is, find the lowest number that will make them all integers once you multiply them by this number). Using rational numbers from the start helps enormously.
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#### curiouscat

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• Mole Snacks: +121/-35 ##### Re: Balancing really difficult chemical equations
« Reply #16 on: January 11, 2013, 04:07:04 AM »
Quite often we do miss one equation when balancing reactions, but we also have additional condition that the solution should consist of the smallest integers - and that's enough to find the unique solution.

Not "quite often" but always I think. Because we have a homogeneous system (Since we always have a+b=c etc. and never a+b=c+2) of linear equations and such a system would always admit all zeros as a trivial equation. So a unique solution would mean all coefficients zero.

So the fact that we get a non-unique solution (before applying the smallest integers constraint) is a feature and not a bug. If you ever got a full rank matrix, be worried! #### Borek ##### Re: Balancing really difficult chemical equations
« Reply #17 on: January 11, 2013, 04:38:01 AM »
Not "quite often" but always I think.

Sometimes we miss even more than one equation.

C7H6O3 + C4H6O3 C9H8O4 + HC2H3O2

(salicylic acid + acetic anhydride -> aspirin + acetic acid)

4 coefficients, but only two independent equations.

Quote
Because we have a homogeneous system (Since we always have a+b=c etc. and never a+b=c+2) of linear equations and such a system would always admit all zeros as a trivial equation. So a unique solution would mean all coefficients zero.

So the fact that we get a non-unique solution (before applying the smallest integers constraint) is a feature and not a bug. If you ever got a full rank matrix, be worried! Good point But I am not convinced - just because there exist trivial solution doesn't mean there is no other, non-trivial solution. I have no time to investigate now, but I have a feeling I have seen reactions with a full rank matrix.
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#### curiouscat

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• Mole Snacks: +121/-35 ##### Re: Balancing really difficult chemical equations
« Reply #18 on: January 11, 2013, 06:55:25 AM »

Sometimes we miss even more than one equation.

Ah! True. No oubt.

Quote
Good point But I am not convinced - just because there exist trivial solution doesn't mean there is no other, non-trivial solution. I have no time to investigate now, but I have a feeling I have seen reactions with a full rank matrix.

I'd love to see one if you come up with one.

#### kingkev101

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• Mole Snacks: +0/-0 ##### Re: Balancing really difficult chemical equations
« Reply #19 on: January 05, 2018, 04:27:51 PM »
aCu + bHNO3 -> cCu(NO3)2 + dNO + eNO2 + fH2O

Completing the solution for this:-

(Cu)  a = c....1)
(H)   b  =  2f......2)
(N)  b = 2c +d + e ..3)
(O) 3b = 6c +d + 2e + f....4)

Also Remember that in the initial post we are told that d:e must be 2:3 and so d/e = 2/3 which when removing denominators leaves 3d = 2e (basic maths).

3d = 2e...5)

We can derive one more simple equation. Notice that the LHS of eq 4) is 3 times that of eq 3)
so we can immediately say that 6c + 3d + 3e = 6c + d + 2e + f. Substituting 3d = 2e gives:
6c + 2e + 3e = 6c + d +2e +f. Next you can remove 6c and 2e from both sides giving 3e = d + f but we know d in relation e i.e. d = 2e/3 so 3e = 2e/3 +f ie

7e/3 = f.....6)

Now we have 2 simple equations with f in each so lets try f = 1 this => b =2 (eq 2) and e=3/7 (eq 6) and knowing e gives us d ie d = 2e/3 = 2/3 * 3/7 = 2/7 ie d = 2/7

knowing f,b,d & e then eq 4)...3 * 2 = 6c + 2/7 + 6/7 + 1 => 6 = 6c + 15/7 ie 6c = 27/7 => c = 9/14 and so eq 1 gives us a = 9/14

so a:b:c:d:e:f   =   9/14:2:9/14:2/7:3/7:1
Multiply throughout by 14 gives 9:28:9:4:6:14

9Cu + 28HNO3 -> 9Cu(NO3)2 + 4NO + 6NO2 + 14H2O
Double checking if it balances we see
Cu:   9 on LHS and 9 on RHS
H:   28 on LHS and 14*2 = 28 on RHS
N:   28 on LHS and 18+ 4 + 6 = 28 on RHS
O:   3 * 28 = 84 on LHS and 9 * 3* 2 + 4 + 12 + 14 = 54 + 4 + 12 + 14 = 84 on RHS
Ie balances …..!