How do you balance tricky chemical equations

can someone tell me their process to solve this

can someone tell me their process to solve this

can someone tell me their process to solve this

Prehaps not the simplest, but:

Start by assuming 1 molecule of glucose, becomes a,b,c molecules of propanone, carbon dioxide, and water respectively.

Then consider the no. of C atoms. On the left we have 6, and on the right 3a+b. Giving us one equation:

6=3a+b.

Repeat for H and O.

Solve the 3 simultaneous equations, to get an initial chemical equation.

Your values for a,b,c may be, and in this case are, fractions, so multiply through by the lowest common multiple of the denominators of any fractions to get your desired result.

I've gotten
6= 3a +b
12 = 6a +2c
6 = a +2b +c

but i dont know how to solve three simultaneous equations

Lol are these supposed to be integers?
Well must've made a mistake but you get the overall method right?

First you want a
From first equation we have
b = 6 - 3a. (i)
From the second equation we have
c = 6 - 3a. (ii)
Sub this into the third equation to get
6 = a + 2(6-3a) + (6 - 3a)
6 = a + 12 - 6a + 6 - 3a
-12 = - 8 a
12/8 = a
Lol are these supposed to be integers?
Well must've made a mistake but you get the overall method right?
Then you sub the value of a into (i) then (ii) to find the values of b and c

Ngl what I got taught to do is to make columns with the number of each element present in that molecule. And then trial by fire.

So here goes:
(1) is C6H12O6

(2) is CH3COCH3
(3) is CO2
(4) is H2O
So the total number of each element on each side should be equal therefore: (1) = (2) + (3) + (4)

(1) - C=6 ; H=12 ; O=6

(2) - C=3 ; H=6 ; O=1
(3) - C=1 ; O=2
(4) - H=2 ; O=1
I'm going to multiply (1) by 2.
So (1) - C=12 ; H=24 ; O=12
(2)+(3) [carbon] is 4 carbons. And 12/4 is 3. So let's say that (2) and (3) are potentially each multiplied by 3.
That would mean we have 18 Hydrogens in (2) now but we need 24. 24 - 18 is 6. There's 2 Hydrogens in (4) so (4) must also be multiplied by 3?
So a potential balances equation could be:
2(1) -> 3(2) + 3(3) + 3(4)

Wtf ur a mad brainiac too Can I fall harder?

it'll be more difficult to catch you 🥺

Ur right about that
Tbf, I weigh a feather so might not be that hard

😂😂 I'll go to all lengths to catch you dw

Hahahaha you better get rdy cus I'm comingggg

Slxxxxxx 🙈 - let's leave this thread railed shall we

Why would you guys derail this thread? It’s just a guy asking help on a chem question.

Also, I condone everything that’s in post #3.

Why would you guys derail this thread? It’s just a guy asking help on a chem question.

Also, I condone everything that’s in post #3.

Read my previous post - in full.

@val7322 If you're not familiar with solving 3 simulaneous equations, then you can't use this method, though the poster above has given one way to do it. Can't see another easy method, given the oddity of this particular chemical decomposition. Perhaps you can make use of the other posters' suggestions.