Sciencemadness Discussion Board - Hard balancing

Hard balancing

Vosoryx - 17-11-2017 at 21:48

For context: I am in HS, not college or uni. (Yet)

Today, I had a chem test. At the bottom, there was a challenge on balancing for a bonus mark - Which I did complete. It was fairly challenging for me (Im sure some of you are laughing at was challenges me, but hey - i'm young) but it got me thinking - what is the hardest reasonable balancing problem you can think of?
The problem was:
NH4MnO4 + H2SO3 + (NH4)2SO3 ---> MnSO4 + (NH4)2SO4 + H2O
If you want to solve yourself, i'll leave the coefficient ratio under white text here: 2:3:2:2:3:3

Obviously, the ammonium could have been replaced with any positive Ion, say K.
So my thought is this: What's the most challenging balancing problem that is actually used in chem? They have to be reasonable to solve, so no:

"30448582 C2952H4664N812O832S8Fe4 + 10833308052 Na2C4H3O4SAu
+ 3899586588 Fe(SCN)2 + 1408848684 Fe(NH4)2(SO4)2∙6H2O + 5568665015C4H8Cl2S (3)
+ 1379870764 C8H12MgN2O8 → 1379870764 C55H72MgN4 + 5430229600 Na3.99Fe(CN)6
+ 10975996000 Au0.987SC6H11O5 + 11137330030 HClO4 + 16286436267 H2S."

That is literally only solvable by a computer. (Source)

Just maybe something else challenging/fun? What balancing problems have plagued you in the past?
Figured since this was whimsy this is an OK topic.

[Edited on 18-11-2017 by Vosoryx]

Geocachmaster - 18-11-2017 at 05:10

Edit: Spoiler warning if you are trying to solve it

Do you mean 2:3:2 2:3:3 ?

The answer you left doesn't balance, I'm guessing you mixed the order up.

[Edited on 11/18/2017 by Geocachmaster]

Vosoryx - 18-11-2017 at 10:06

Well, that's embarrasing.

Yes, you are right. I checked my notebook, i just copied it in wrong.

Changed it in the original post - thanks.

[Edited on 18-11-2017 by Vosoryx]

DraconicAcid - 18-11-2017 at 10:18

The reaction of dimethylhydrazine with nitric acid to give carbon dioxide, water, and nitrogen. I challenged some of my colleagues to balance it without resorting to the redox method, and they gave up.

aga - 18-11-2017 at 11:19

Wooo ! A Challenge !

aga - 18-11-2017 at 14:27

Now that was Hard.
Been at it for ages !

Chances are some cock will point out that you can just type the equation into an online calculator before i finish typing this tome, but here goes ...

Here's how i solved it :-

Step 1: Chemical reaction.
Wiki says dimethylhydrazine is C₂H₈N₂ so i got to go with that in the absence of any other info, AND assume it is pure nitric acid, so the reaction is :-

C₂H₈N₂ + HNO₃ => CO₂ + H₂O + N₂

Step 2: Assign algebraic values (a,b,x,y,z etc) to each item)

a C₂H₈N₂ + b HNO₃ => c CO₂ + d H₂O + e N₂
Step 3: Create the algebraic equations by counting atoms on each side

For Carbon: 2a = c
For Hydrogen: 8a + b = 2d
For Nitrogen: 2a + b = 2e
For Oxygen: 3b = 2c + d

So now we got 4 equations with 5 variables, which is a lot to mess with all at once.
If we make 'a' the Important One, then we can reduce the number of equations to 3. Better.
Now 2a = c, so if we substitute that into the Oxygen equation we get

3b = 4a + d

May as well forget variable 'c' for now as it doesn't appear anywhere else. Also Better.
Now we have just 3 equations to work on, with :-

8a + b = 2d
2a + b = 2e
3b = 4a + d

Step 4: The next commonest variable is 'b', so i (eventually) chose to reduce the remaining 3 equations so that 'a' and 'b' were on the same side.

Hydrogen equation:
$$8a + b = 2d$$
$$\therefore d = \frac {8a +b}2$$

Nitrogen equation:
$$2a + b = 2e$$
$$\therefore e = \frac{2a + b}2$$

Substituted Oxygen equation:
$$3b = 4a + d$$
$$\therefore d = 3b - 4a$$

Step 5: Noticing things. The 'd' variable has two equations now, containing only 'a' and 'b' so we can solve 'b' in terms of 'a'

$$d = \frac {8a +b}2 = 3b - 4a$$
$$8a +b = 6b - 8a$$
$$16a = 5b$$
$$b = \frac{16a}5$$

Step 6: Rearrange, then Plug the 'b' equation from Step 5 into the Hydrogen equation

$$8a +b = 2d$$
$$\therefore b = 2d - 8a$$
$$\therefore \frac{16a}5 = 2d - 8a$$
$$\therefore \frac{16a}5 + 8a = 2d$$
$$\therefore \frac{16a +40a}5 =2d$$
$$\frac{56a}5 = 2d$$
$$d= \frac{56a}{10}$$

Step 7: Now do the same to the Nitrogen equation
$$2a + b = 2e$$
$$2a + \frac{16a}{5} = 2e$$
$$\frac{10a}{5} + \frac{16a}{5} = 2e$$
$$\frac{26a}{5} = 2e$$
$$e=\frac{26a}{10}$$

Phew. That's the algebra over with and the Solution calculated ... Almost.
All it gives is ratios, not Numbers.

Step 8: Assuming a = 1 we get :-

$$a = 1$$
$$b = \frac{16}5$$
$$c = 2$$
$$d = \frac{56}{10}$$
$$e = \frac{26}{10}$$

Step 9: Clearly we need to multiply by 5 to get rid of those god awful fractions, so we get

a = 5
b = 16
c = 10
d = 28
e = 13

Hence the Balanced reaction of dimethylhydrazine and 100% nitric acid is :-

5 C₂H₈N₂ + 16 HNO₃ => 10 CO₂ + 28 H₂O + 13 N₂

It is MUCH quicker to just type it into an online calculator, although somebody has to know how to do it, or who could build an online calculator ?

(any further typos are just typos. it was all done on paper)

[Edited on 18-11-2017 by aga]

DraconicAcid - 18-11-2017 at 15:24

Well done. I did it this way:

C2H8N2 + HNO3 => CO2 + H2O + N2 is tough to balance. So I cheat, and balance..

C2H8N2 + N2O5 => CO2 + H2O + N2 instead. One element at a time; carbon first, then hydrogen, then oxygen, then nitrogen (fewest compounds first, then elements last).

C2H8N2 + N2O5 => 2 CO2 + 4 H2O + N2

C2H8N2 + 8/5 N2O5 => 2 CO2 + 4 H2O + N2

(Now multiply by five to get rid of the fractions)

5 C2H8N2 + 8 N2O5 => 10 CO2 + 20 H2O + N2

Balance nitrogen....

5 C2H8N2 + 8 N2O5 => 10 CO2 + 20 H2O + 13 N2

Adding 8 waters to each side, and reacting the water with N2O5 to give nitric acids...

5 C2H8N2 + 16 HNO3 => 10 CO2 + 28 H2O + 13 N2

aga - 18-11-2017 at 15:32

Bugger.

Therein lies the difference between a Chemist and an Amateur i guess.

Took me the best part of 3 hours of ending up with 0 = 0 or 3a = 3a,

Admittedly some of that time was re-learning LaTex syntax for showing the maths,

Edit:

At least an hour was wasted by not tracking which derived formula came from which original formula - no point substituting back into the same formula.

[Edited on 18-11-2017 by aga]

DraconicAcid - 18-11-2017 at 15:57

It's actually easy if you do it by the redox half-reaction method.

aga - 18-11-2017 at 16:01

Redox with an OC ?

Black Magic ! Save Yourselves! Flee !

Maybe i can risk One eye,

Please show how that method works with a hell-spawned OC.

DraconicAcid - 18-11-2017 at 16:31

C2H8N2 is getting oxidized, so

C2H8N2 --> 2 CO2 + N2

Balance O with water

4 H2O + C2H8N2 --> 2 CO2 + N2

Balance H with H+, then charge with electrons.

4 H2O + C2H8N2 --> 2 CO2 + N2 + 16 H+ + 16 e-

Nitric acid is being reduced; balance O with H2O, H with H+, charge with electrons.

2 HNO3 --> N2
2 HNO3 --> N2 + 6 H2O
10 H+ + 2 HNO3 --> N2 + 6 H2O
10 e- + 10 H+ + 2 HNO3 --> N2 + 6 H2O

5 x (4 H2O + C2H8N2 --> 2 CO2 + N2 + 16 H+ + 16 e-)
8 x (10 e- + 10 H+ + 2 HNO3 --> N2 + 6 H2O)

20 H2O + 5 C2H8N2 + 80 e- + 80 H+ + 16 HNO3 --> 10 CO2 + 13 N2 + 80 H+ + 48 H2O + 80 e-

5 C2H8N2 + 16 HNO3 --> 10 CO2 + 13 N2 + 28 H2O

gdflp - 18-11-2017 at 17:10

An alternative way to solve the problem without "cheating" is to use linear algebra.

Starting from the equations aga figured out, a matrix can be constructed by pushing all of the variables to one side:
$$ \left[
\begin{array}{ccccc|c}
2 & 0 & -1 & 0 & 0 & 0 \\
8 & 1 & 0 & -2 & 0 & 0 \\
2 & 1 & 0 & 0 & -2 & 0 \\
0 & 3 & -2 & -1 & 0 & 0 \\
\end{array}
\right] $$

Then comes the Gaussian elimination(apologies, I know there's probably a much faster way to do this, but my linear algebra is rusty):
$$
r_3 = r_3 - r_1 \\
r_4 = r_4 - 3r_3 \\
r_2 = r_2 - 4r_1 \\
r_3 = r_3 - r_2 \\
r_4 = r_4 - \frac{5r_3}{3} \\
r_4 = 3r_4 \\
r_3 = 13r_3 + 2r_4 \\
r_3 = \frac{r_3}{3} \\
r_1 = -r_1 + r_3 \\
r_1 = \frac{r_1}{2} \\
r_2 = 13r_2 + 4r_3 \\
r_2 = r_2 - 2r_4 \\
r_x = \frac{r_x}{13} \\
r_1 = -r_1 \\
r_3 = -r_3 \\
r_4 = -r_4 \\
$$

Yielding the following matrix :
$$
\left[
\begin{array}{ccccc|c}
1 & 0 & 0 & 0 & \frac{-5}{13} & 0 \\
0 & 1 & 0 & 0 & \frac{-16}{13} & 0 \\
0 & 0 & 1 & 0 & \frac{-10}{13} & 0 \\
0 & 0 & 0 & 1 & \frac{-28}{13} & 0 \\
\end{array}
\right]
$$

This can be turned back into a series of equations :
$$
a = \frac{5e}{13} \\
b = \frac{16e}{13} \\
c = \frac{10e}{13} \\
d = \frac{28e}{13} \\
$$
The common denominator leads to a reasonable assumption of e = 13. Plugging this into each of the equations yields the answer.

This is a more systematic way of doing it, and it could be implemented algorithmically if desired. Online calculators probably don't solve it this way, but I see no reason that they couldn't(other than efficiency).

clearly_not_atara - 18-11-2017 at 20:21

^indeed, all balancing problems are reducible to Gaussian elimination. But I think the bigger issue is not the length of your solution but the fact that it's very hard to see how it works if you're not already familiar with GE.
Online calculators probably use matrix inversion which is like GE but faster on a computer due to some optimizations. But it's the same basic idea.

Gaussian elimination can be made systematic:

First clear the first column:
r2 = r2 - 4r1
r3 = r3 - r1

[r1]
[0 1 4 -2 0]
[0 1 1 0 -2]
[r4]

Now clear the second:
r3 = r3 - r2
r4 = r4 - 3r2

[r1]
[r2]
[0 0 -3 2 -2]
[0 0 -14 5 0]

Now clear the third column:
r1 = 3r1 - r3
r2 = 3r2 + 4r3
r4 = 3r4 - 14r3

[6 0 0 -2 2]
[0 3 0 2 -8]
[r3]
[0 0 0 -13 28]

Eliminate common factors (common step, not that important here but useful in general):
r1 = r1/2

[3 0 0 -1 1]
[r2]
[r3]
[r4]

Finally clear the fourth column and eliminate common factors:
r1 = 13r1 - r4
r2 = 13r2 + 2r4
r3 = 13r3 + 2r4

[39 0 0 0 -15]
[0 39 0 0 -48]
[0 0 -39 0 30]
[0 0 0 -13 28]

r1 = r1/3
r2 = r2/3
r3 = -r3/3
r4 = -r4

[13 0 0 0 -5]
[0 13 0 0 -16]
[0 0 13 0 -10]
[0 0 0 13 -28]

Which gives [5 16 10 28 13] as expected.

EDIT: typos and formatting

[Edited on 19-11-2017 by clearly_not_atara]

aga - 19-11-2017 at 01:22

Stunning !

Thanks to each for their explanations, also to the OP for starting this thread.

Vosoryx - 19-11-2017 at 09:18

Well, that all went over my head in an amount of time equal to 0, but it forced me to spend a few hours learning the topics talked about here. Thanks, that was really cool.

aga - 19-11-2017 at 10:59

When there's a good post, such as yours, we all learn something.

Until gdflp mentioned it, i had never heard of linear algebra, nor attempted to understand how a matrix works.

This should be in Beginnings really - it's not Whimsical !

Vosoryx - 19-11-2017 at 14:33

Yeah I had to watch a few hours of explaining matrices to not understand it at all.

I figured because I wasn't actually asking a question and was more entering a discussion it belonged here, but you could be right.

aga - 19-11-2017 at 14:51

I have a lot of research to do about linear algebra to get an idea how that works, not all of which can be videos.

Until i saw the post, Linear Algebra didn't exist, now it's a new item on a never-ending list of things-to-do (which is Good - always have stuff to do).

'Beginnings' would have been 'proper', even 'Chemistry in General' maybe.

You had a Real Chemistry question that led onto some interesting stuff, which is cool.

Take a look at some of the utter Crap posted elsewhere on the forum.

Yes, er, OK. Admittedly some were my crap - a high post count does not imply accuracy, nor knowledge, nor expertise

[Edited on 19-11-2017 by aga]

clearly_not_atara - 19-11-2017 at 19:24

The empirical chemical composition of charcoal is approximately C7H4O. So one equation for the complete combustion (although real combustion is usually incomplete) of gunpowder containing sulfur is:

a KNO3 + b S + c C7H4O >> d K2SO4 + e CO2 + f N2 + g H2O

Some solid-fuel hobby rockets use ammonium perchlorate to overcome the oxygen deficit of nitrocellulose propellants. The reaction is:

a NH4ClO4 + b C6H8(NO2)2O5 >> c CO2 + d H2O + e N2 + f HCl

ninhydric1 - 19-11-2017 at 19:31

You could probably apply Cramer's Rule to these matrices but finding the determinants for 5x5 matrices aren't bound to be enjoyable.

EDIT: Actually, it shouldn't be too hard. If we use gdflp's matrix, we find the determinant of the 4x4 matrix formed by the first four columns and use this value when Cramer's rule is used for each four rows. The fifth row just makes it easier due to the three zeroes.

[Edited on 11-20-2017 by ninhydric1]