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I need help
Topic Started: May 30 2009, 08:31 PM (87 Views)
Brian May 30 2009, 08:31 PM Post #1
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I'm trying to study for an acceptance exam, and I've come across a couple of science problems that stump the hell out of me. Well here goes.

Balance These Equations:

(a) NaClO = NaCl + O
----------3--------------2

(b) MnCl + Al = Mn + AlCl
----------2------------------3

(c) C H + O = H O + CO
-----4-10--2-----2-------2


In case that was too confusing, I'll try again.

(a) NaClO (divided by 3) = NaCl + O (divided by 2)

(b) MnCl (Divided by 2) + Al = Mn + AlCl (divided by 3)
(c) C (divided by 4) multiplied by H (divided by 10) + O (divided by 2) = H (divided by 2) multiplied by O + CO (divided by 2)

Alright, I'm already confused just trying to explain this, and I doubt I even explained it right. Help me?
Edited by Brian, May 30 2009, 08:40 PM.
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Rebel X May 31 2009, 01:03 AM Post #2
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Alright, from what I understand your trying to balance the atoms of the Chemical
For the First One,
"(a) NaClO = NaCl + O
----------3--------------2
" means 1 atom of NaCl, 3 atoms of Oxygen equals 1 atom of NaCl and 1 atom of Oxygen. Which should look like :
NaClO3 = NaCl + O2
in order to balance the atoms on both sides of the equation first you balance the Oxygens, so on the both sides you multiply by 2. This gives you NaCl2O6 = 2NaCL+ O2. The leaves you NaCl balanced and you Oxygen unbalanced, so then you multiply by 3 to balance the Oxygen. Your answer being,

2NaClO3 = 2NaCl + 3O2


The Second,
(b) MnCl + Al = Mn + AlCl
----------2------------------3
" Looks like:
MnCl2 + Al = Mn + AlCl3
You answer being,
3MnCl2 + 2Al = 3Mn + 2AlCl3

As for the last one, it's a bit complicated. I don't think I can do it, but I can try and give you an indefinite answer later
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Trickster May 31 2009, 01:23 AM Post #3
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Wow, i remember doing these in school, and they were somewhat easy, but it has been many years, i remember a little bit, but not a lot.
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Kotetsu May 31 2009, 01:46 AM Post #4
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(c) C H + O = H O + CO
-----4-10--2-----2-------2

--------------------------------

2C(4)H(10) + 13O(2) = 10H(2)O + 8CO(2)

The easiest way to solve this problem is by starting at a multiple of 1 at the CH compound. Work your way up to about 4. You'll notice that both 2 and 4 work when you multiply everything out by balancing it. 4 is not correct because it can be reduced.

When doing these problems, starting at the compound with the largest amount of molecules is the best method.
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Brian May 31 2009, 01:54 AM Post #5
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Alright, these seem pretty accurate from what I can see. I could barely remember them either, Josh. Thanks a lot Dan, Kotetsu.
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Kotetsu May 31 2009, 02:12 AM Post #6
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No problem. I'm terrible at chemistry unless it's the math portions.

PV=nRT / Ideal Gas laws are my favorite questions.
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