Balance This Equation (Answer solved Biyatches)

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Posted by Floorfiller on Oct-27-2003 23:59:

quote:
Originally posted by Evan Almae
I was just about to say that! Nice try though! Thanks everyone for helping by the way! Didn't want to leave that unsaid, your help is much appreciated!

i'll leave the chemistry to those that took notes...talk to you guys later hehehe...and enjoy slaming this answer in your teachers face tomorrow...give a big BOOOYA for me hehehe...

Posted by Turbonium on Oct-28-2003 00:00:

quote:
Originally posted by kewlness
hmm.. let me re-do that.. i think i screwed up somewhere... go linear algebra

Yea, sorry though, after that long explanation of yours too. But your method is still probably right, just did a mistake maybe like you mentioned.

Posted by Turbonium on Oct-28-2003 00:03:

kewlness: I still don't get what you mean by variables assigned as x and a number. Shouldn't it be like x, y, z, etc?

Posted by Steven Hays on Oct-28-2003 00:07:

okay, to make this any harder, is it legal to use decimal for the coefficients???

Posted by EriK_V on Oct-28-2003 00:09:

hmm, should have posted this last year and i would have solved it. all i can remember now is using oxidation numbers and adding water or oxygen to balance out the equation.

Posted by J.L. on Oct-28-2003 00:14:

The Real correct answer

3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O

How I did it:
assign the unknown coefficients as unknowns x1,x2,x3,x4,x5
x1Cu + x2HNO3 -> x3Cu(NO3)2 + x4NO + x5H2O

Now you know that
x1 - x3 = 0 for Copper
x2 - 2x5 = 0 for Hydrogen
3x2 - 6x3 - x4 -x5 = 0 for Oxygen
x2 - 2x3 - x4 = 0 for Nitrogen

Solving for the system of equations (i used linear algebra but you can use any method you want), I got the solutions:

x1= 3/4 t
x2 = 2t
x3= 3/4 t
x4= 1/2 t
x5= t

where t is a parameter
Now you have to choose a value t so that x1,x2,x3,x4,x5 are integers (1,2,3,4,5,6, etc...) while still being in the lowest denominator... The number that fits this category is where t=4

so therefore x1 = 3, x2 = 8 x3 = 3 x4 = 2 x5 = 4

I win for sure this time. You all lose

Posted by piggy on Oct-28-2003 00:19:

lol, I just took chem last year, (and did well too), and I've forgotten everything,

Posted by Steven Hays on Oct-28-2003 00:22:

lol, kewlness, i was just about there! I had everything but the 2 infront of the NO! Well done. Thanks for the help!

Posted by armandzadza on Oct-28-2003 00:26:

Crap, I'm late!

I'm an ox-redux whore. Give me another one!

Posted by Turbonium on Oct-28-2003 00:27:

quote:
Originally posted by kewlness
The Real correct answer

3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O

How I did it:
assign the unknown coefficients as unknowns x1,x2,x3,x4,x5
x1Cu + x2HNO3 -> x3Cu(NO3)2 + x4NO + x5H2O

Now you know that
x1 - x3 = 0 for Copper
x2 - 2x5 = 0 for Hydrogen
3x2 - 6x3 - x4 -x5 = 0 for Oxygen
x2 - 2x3 - x4 = 0 for Nitrogen

Solving for the system of equations (i used linear algebra but you can use any method you want), I got the solutions:

x1= 3/4 t
x2 = 2t
x3= 3/4 t
x4= 1/2 t
x5= t

where t is a parameter
Now you have to choose a value t so that x1,x2,x3,x4,x5 are integers (1,2,3,4,5,6, etc...) while still being in the lowest denominator... The number that fits this category is where t=4

so therefore x1 = 3, x2 = 8 x3 = 3 x4 = 2 x5 = 4

I win for sure this time. You all lose

Well I got the right answer too, just did the last step wrong. Oh well...

quote:
Originally posted by Turbonium
I get the initial part correct, but restoring it to normal eqn yield incorrect result, weird...

3Cu + 8H(+) + 2NO3(-) ---> 3Cu(2+) + 2NO + 4H2O

full eqn (if I did it right, which according to this, I didn't):

3Cu + 2HNO3 + 6H(+) ---> 3Cu(NO3)2 + 2NO + 4H2O

Obviously it was 8HNO3, stupid mistake, meh, I aced it on the exams, that's all that matters

Posted by Steven Hays on Oct-28-2003 00:27:

quote:
Originally posted by armandzadza
Crap, I'm late!

I'm an ox-redux whore. Give me another one!

Gimme a few minutes. I'm doing a paper, the ones I can't figure out I'll post. Or you can just do my homework for me???lol!

Posted by armandzadza on Oct-28-2003 00:33:

quote:
Originally posted by Evan Almae
Gimme a few minutes. I'm doing a paper, the ones I can't figure out I'll post. Or you can just do my homework for me???lol!

*cracking knuckles to get ready*

Posted by Steven Hays on Oct-28-2003 00:36:

NO2 +H20 ---> HNO3 + NO

Seemed easy, but I didn't get it fast at all. *scratches head*

Posted by armandzadza on Oct-28-2003 00:43:

quote:
Originally posted by Evan Almae
NO2 +H20 ---> HNO3 + NO

Seemed easy, but I didn't get it fast at all. *scratches head*

3NO2 +H20 ---> 2HNO3 + NO

Posted by Steven Hays on Oct-28-2003 00:46:

quote:
Originally posted by armandzadza
3NO2 +H20 ---> 2HNO3 + NO

hell yeah!

Posted by armandzadza on Oct-28-2003 00:46:

quote:
Originally posted by armandzadza
3NO2 +H20 ---> 2HNO3 + NO

Method:
NO2 -> N(+4)
HNO3 -> N(+5)
NO -> N(+2)

N(+4) - 1e = N(+5)
N(+4) - 2e = N(+2)

Switching the equations, N(+5) requires "2" up-front, and N(+2) requires "1". Then,

XNO2 + H20 ---> 2HNO3 + NO

and X=3

Posted by Steven Hays on Oct-28-2003 00:48:

Next one:

Bi2S3 + 02 --> Bi2O3 + SO2

see if I can beat you on this one!

Posted by armandzadza on Oct-28-2003 00:55:

quote:
Originally posted by Evan Almae
Next one:

Bi2S3 + 02 --> Bi2O3 + SO2

see if I can beat you on this one!

2Bi2S3 + 902 --> 2Bi2O3 + 6SO2

Posted by Steven Hays on Oct-28-2003 00:57:

quote:
Originally posted by armandzadza
2Bi2S3 + 902 --> 2Bi2O3 + 6SO2

bastard!!! LOL, good one. That was a hard one!

Posted by armandzadza on Oct-28-2003 00:59:

LOL, things I do to procrastinate studying French!

Posted by Steven Hays on Oct-28-2003 00:59:

(NH4)2Cr2O7 ---> N2 + H20 + Cr2O3

Looks tricky!

Posted by Steven Hays on Oct-28-2003 01:02:

(NH4)2Cr2O7 ---> N2 + 4H20 + Cr2O3

I think, that was fast!

Posted by armandzadza on Oct-28-2003 01:03:

quote:
Originally posted by Evan Almae
(NH4)2Cr2O7 ---> N2 + H20 + Cr2O3

Looks tricky!

Actually, this was the most straight-forward:

(NH4)2Cr2O7 ---> N2 + 4H20 + Cr2O3

EDIT: Damn it, you beat me!

Posted by Steven Hays on Oct-28-2003 01:05:

Do you find these disturbingly fun??? Cause I do!

Posted by armandzadza on Oct-28-2003 01:08:

A lot more fun than French, that's for sure

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