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### Topic: How to balance this equation?  (Read 22588 times)

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#### Borek

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##### Re: How to balance this equation?
« Reply #15 on: March 10, 2010, 05:29:37 PM »
You don't have to use 6 & 3, to cancel electrons it is enough to multiply first reaction by 2.

However, let's assume I have not told you multiplying by 2 will do. Write reaction as you think you should - that is, multpiplying by 6 & 3.
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#### Evaldas

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##### Re: How to balance this equation?
« Reply #16 on: March 11, 2010, 04:32:22 AM »
KClO3 + CrCl3 + KOH  6K2CrO4 + 3KCl + H2O Doesn't make any sense

#### Borek

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##### Re: How to balance this equation?
« Reply #17 on: March 11, 2010, 05:09:46 AM »
But you have not followed the rule. You should multiply equations on both sides. That means 6 CrCl3 on the left and 6 K2CrO4 on the right. Do the same to chlorate and chlorides.

Note: you are trying to balance full equation, that means you will need to spend substantial amount of time trying to balance spectators.
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#### Evaldas

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##### Re: How to balance this equation?
« Reply #18 on: March 11, 2010, 05:30:21 AM »
3KClO3 + 6CrCl3 + KOH  6K2CrO4 + 3KCl + H2O

#### Borek

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##### Re: How to balance this equation?
« Reply #19 on: March 11, 2010, 06:43:13 AM »
Now, you need to balance everything else - that is, oxygen, hydrogen and potassium. Do it with KOH and water, don't change coefficients in compounds that are getting reduced or oxidized, as they are already in correct ratio.

Unfortunately after that - as you are trying to balance full reaction equation - you also have to balance chlorine. However, chlorides from CrCl3 are just spectators, so they don't take part in the redox process. You may treat them as separate from chlorides that are being produced in chlorate reduction (imagine you have started not with chromium chloride, but with chromium nitrate - then you will have to balance NO3- as spectator and it will not confuse you by being identical with chlorides produced in the reduction process).
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#### Evaldas

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##### Re: How to balance this equation?
« Reply #20 on: March 11, 2010, 01:48:23 PM »
Now, you need to balance everything else - that is, oxygen, hydrogen and potassium. Do it with KOH and water, don't change coefficients in compounds that are getting reduced or oxidized, as they are already in correct ratio.

Unfortunately after that - as you are trying to balance full reaction equation - you also have to balance chlorine. However, chlorides from CrCl3 are just spectators, so they don't take part in the redox process. You may treat them as separate from chlorides that are being produced in chlorate reduction (imagine you have started not with chromium chloride, but with chromium nitrate - then you will have to balance NO3- as spectator and it will not confuse you by being identical with chlorides produced in the reduction process).
I can't manage to do this... I'm lost.

#### Borek

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##### Re: How to balance this equation?
« Reply #21 on: March 11, 2010, 02:16:42 PM »
I suppose chlorides confuse you. So let's get rid of those that are not involved in the redox part of the reaction.

Try now:

3KClO3 + 6CrBr3 + KOH  =  6K2CrO4 + 3KCl + KBr + H2O

Just remember - don't touch 3KClO3, 6CrBr3, 6K2CrO4 and 3KCl - these are already balanced.
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#### Evaldas

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##### Re: How to balance this equation?
« Reply #22 on: March 11, 2010, 02:25:33 PM »
Ok:
3KClO3 + 6CrBr3 + 30KOH = 6K2CrO4 + 3KCl + 18KBr + 15H2O ?

#### Borek

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##### Re: How to balance this equation?
« Reply #23 on: March 11, 2010, 03:09:40 PM »
1. Replace now all Br by Cl - just a moment ago we did the reverse to not confuse you, time to get back to the original equation.

2. There is something wrong with this equation - coefficients can be smaller. Do you see it?
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#### Evaldas

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##### Re: How to balance this equation?
« Reply #24 on: March 11, 2010, 03:18:57 PM »
1. 3KClO3 + 6CrCl3 + 30KOH = 6K2CrO4 + 3KCl + 18KCl + 15H2O
2. Yes, I see that it can be divided by 3:
3KClO3 + 6CrCl3 + 30KOH = 6K2CrO4 + 3KCl + 18KCl + 15H2O /:3 =>
KClO3 + 2CrCl3 + 10KOH = 2K2CrO4 + KCl + 6KCl + 5H2O

#### Evaldas

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##### Re: How to balance this equation?
« Reply #25 on: March 11, 2010, 03:34:27 PM »
But see now, I don't get one thing: we have to write that number of electrons as a coefficient on both sides, but how do I act in this situation:
P0(s) + O02(g)  P+52O-25(g)
P0  P+5 + 5e       | 5 | 4 |
O02 + 2x2e  2O-2 | 4 | 5 |
So I write 4 next to solid phosphorus and a 5 next to oxygen, but obviously I won't write a 4 neither a 5 next to phosphorus(V) oxide, right?
4P0(s) + 5O02(g)  2P+52O-25(g)
How can this be explained, when does this rule apply?

#### Borek

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##### Re: How to balance this equation?
« Reply #26 on: March 11, 2010, 04:15:07 PM »
There are two P atoms in a product molecule, so it will be easier to write phosphorus part as

2P0 -> 2P+5 + 10e-

(you did the same for oxygen, just the other way around).
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#### Evaldas

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##### Re: How to balance this equation?
« Reply #27 on: March 11, 2010, 04:25:37 PM »
But wouldn't it then be:
2P0 -> 2P+5 + 10e-  | 10 | 5 | 2 |
O02 + 2x2e- -> 2O-2 |  4 | 2 | 5 |
So accordingly I'd have to write:
2P0(s) + 5O02(g)  ?P+52O-25(g)
Or not? Or for some reason I'm supposed to know that the five will go to oxygen, the two to phosphorus(V) oxide, and then accordingly I'd have to count that I have to write 4 next to phosphorus?

#### Borek

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##### Re: How to balance this equation?
« Reply #28 on: March 11, 2010, 06:01:19 PM »
2P+5 is equivalent to P+52 which you have in the oxide molecule.
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#### Evaldas

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##### Re: How to balance this equation?
« Reply #29 on: March 12, 2010, 02:56:22 PM »
Ok. Let's go through another reaction
Cr2+3(SO4)3 + Br20 + NaOH  Na2Cr+6O4 + 3NaBr- + Na2SO4 + H2O
Br20 + 2x1e-  2Br-     | 2 | 1 | 3
Cr2+3 - 2x3e-  2Cr+6 | 6 | 3 | 1
So you said that I should follow the rules and write on both sides the numbers, so I'd have to write on both sides 3s, and I don't need to write 1s:
Cr2(SO4)3 + 3Br2 + NaOH  Na2CrO4 + 3NaBr + Na2SO4 + H2O
But now I don't have a choice but to write a six next to NaBr instead of 3. IS IT OK TO DO SO?