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Balance Chemical Equations, Help
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Chrysalis
post May 11 2011, 10:27 AM
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QUOTE(DoO_Quasar @ May 11 2011, 02:26 AM)
QUOTE(Chrysalis @ May 10 2011, 06:01 PM)
Figuring out the variables was the hard part of this one. (z-48x) is based on H, whereas (z-48x-y) is based on S.

x[Cr(N2H4CO)6]4[Cr(CN)6]3 + yKMnO4 + (z-48x)H2SO4 --> 7xK2CrO7 + yMnSO4 + 42xCO2 + 66xKNO3 + (z-48x-y)K2SO4 + zH2O

*


Can you explain how to do bolded for me please? I completely understood the first problem but I'm not quite sure how you got to your answers with what you described there. More help? xD
*



Assign x for [Cr(N2H4CO)6]4[Cr(CN)6]3 arbitrarily. K2CrO7 must have coefficient 7x because there's 7Cr's in [Cr(N2H4CO)6]4[Cr(CN)6]3. CO2 has coefficient 42x because there's 42 C's in [Cr(N2H4CO)6]4[Cr(CN)6]3. KNO3 has coefficient 66x because there's 66 N's in [Cr(N2H4CO)6]4[Cr(CN)6]3. You can do this because [Cr(N2H4CO)6]4[Cr(CN)6]3 has all of the Cr's, C's, and N's on its side. Then give KMnO4 and MnSO4 y because they're the only Mn's on their respective sides. Then give H2O a new variable z, and H2SO4 w. Well, 2w + 96x = 2z based on the H's in [Cr(N2H4CO)6]4[Cr(CN)6]3, H2SO4, and H2O. so w = z - 48x. Then give K2SO4 v. From this, y + z-48x = v based on the S's on both sides. so v = z - 48x - y. :flowers:

and yes, you can divide everything by 5, good call :thumbsup:

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AoL_Berserker
post May 11 2011, 11:25 AM
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fruit chemistry :flowers:

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MarquesDS
post May 11 2011, 11:28 AM
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QUOTE(AoL_Berserker @ May 11 2011, 04:25 PM)
fruit chemistry  :flowers:
*


But if you are a desginer its obvious that you need to know how To balance chemical equations.

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Oedy
post May 11 2011, 12:19 PM
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The real question is: are you sure this is the only valuable solution?
If you have a redox process you can get more than 1 reaction quite often, and then you need to know more about the environment the process is considered in to be sure...

I'm too lazy to check if there is another solution here though, this is hardly any chemistry anyways, just basic algebra

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[DoD]Quasar
post May 11 2011, 03:18 PM
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Ok, I did make that mistake, my bad, good catch >.<

QUOTE(Chrysalis @ May 11 2011, 09:27 AM)
QUOTE(DoO_Quasar @ May 11 2011, 02:26 AM)
QUOTE(Chrysalis @ May 10 2011, 06:01 PM)
Figuring out the variables was the hard part of this one. (z-48x) is based on H, whereas (z-48x-y) is based on S.

x[Cr(N2H4CO)6]4[Cr(CN)6]3 + yKMnO4 + (z-48x)H2SO4 --> 7xK2CrO7 + yMnSO4 + 42xCO2 + 66xKNO3 + (z-48x-y)K2SO4 + zH2O

*


Can you explain how to do bolded for me please? I completely understood the first problem but I'm not quite sure how you got to your answers with what you described there. More help? xD
*



Assign x for [Cr(N2H4CO)6]4[Cr(CN)6]3 arbitrarily. K2CrO7 must have coefficient 7x because there's 7Cr's in [Cr(N2H4CO)6]4[Cr(CN)6]3. CO2 has coefficient 42x because there's 42 C's in [Cr(N2H4CO)6]4[Cr(CN)6]3. KNO3 has coefficient 66x because there's 66 N's in [Cr(N2H4CO)6]4[Cr(CN)6]3. You can do this because [Cr(N2H4CO)6]4[Cr(CN)6]3 has all of the Cr's, C's, and N's on its side. Then give KMnO4 and MnSO4 y because they're the only Mn's on their respective sides. Then give H2O a new variable z, and H2SO4 w. Well, 2w + 96x = 2z based on the H's in [Cr(N2H4CO)6]4[Cr(CN)6]3, H2SO4, and H2O. so w = z - 48x. Then give K2SO4 v. From this, y + z-48x = v based on the S's on both sides. so v = z - 48x - y. :flowers:

and yes, you can divide everything by 5, good call :thumbsup:
*


Oh, ok that makes perfect sense now. Excellent explanation, muchas gracias :thumbsup:

@Oedy: Yeah, this is just a worksheet with equation balancing, it's all that needs to be done :)

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Chrysalis
post May 11 2011, 04:43 PM
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QUOTE(Oedy @ May 11 2011, 12:19 PM)
The real question is: are you sure this is the only valuable solution?
If you have a redox process you can get more than 1 reaction quite often, and then you need to know more about the environment the process is considered in to be sure...

I'm too lazy to check if there is another solution here though, this is hardly any chemistry anyways, just basic algebra
*



well obviously, all high school "science" is complete bs 11

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[DoD]Hellsravage
post May 11 2011, 04:47 PM
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i found physics 10x more interesting than chem

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[DoD]Quasar
post May 11 2011, 05:02 PM
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Yeah, the class is really boring.

@Chrysalis: Ok, I inserted that 4 that I missed into the equation on this site and it came up with this answer:

10 [Cr(N2H4CO)6]4[Cr(CN)6]3 + 1176 KMnO4 + 1399 H2SO4 = 35 K2Cr2O7 + 1176 MnSO4 + 420 CO2 + 660 KNO3 + 223 K2SO4 + 1879 H2O

:unsure:

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DoD_W0WzAP0WzA
post May 11 2011, 05:05 PM
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QUOTE(DoO_Quasar @ May 10 2011, 10:33 PM)
Is that supposed to mean 'CUZ I GET 5s on all the rest'? :huh:

I just took the AP English test (we can take it for American Lit.  Honours and for the 12th grade AP English class) last Thursday and I feel I at least got a 3, most likely a 4, unlikely a 5, which I find EXTREMELY good for not having read a single book of the 14 we did in class. All my essays I had good topics to write about and I felt I was on my game with a lot of adrenaline. The multiple choice I feel I at least got like 70% on and you only need like 50% to pass so... :teehee:

Last year, I had the AP History test and I ended up with a 4 on that, which I was also really grateful for considering I didn't get to the conclusion paragraph on any of my essays and only got to the third paragraph of my 5-paragraph essay on one of the 3 essays (not good at writing/reading fast :().

Next year, I'll take the AP Bio test and I'm confident I'll do well at that considering theoretical stuff and memorizing facts are my strong-points.

Those are the only AP tests I'll get to take though...kind of sad those 3 + AP Stats are the only AP tests my school offers :(
*


I was more sad about it cause I was getting 5's on the practice tests fairly consistently, but the problems that were on the free response were just a lot harder than usual

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Chrysalis
post May 11 2011, 05:14 PM
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QUOTE(DoO_Quasar @ May 11 2011, 05:02 PM)
Yeah, the class is really boring.

@Chrysalis: Ok, I inserted that 4 that I missed into the equation on this site and it came up with this answer:

10 [Cr(N2H4CO)6]4[Cr(CN)6]3 + 1176 KMnO4 + 1399 H2SO4 = 35 K2Cr2O7 + 1176 MnSO4 + 420 CO2 + 660 KNO3 + 223 K2SO4 + 1879 H2O

:unsure:
*



Try again, first molecule on the right should be K2CrO7, not K2Cr2O7 :P

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Chrysalis
post May 11 2011, 05:17 PM
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Site gave me this:

[Cr(N2H4CO)6]4[Cr(CN)6]3 + 126 KMnO4 + 149 H2SO4 = 7 K2CrO7 + 126 MnSO4 + 42 CO2 + 66 KNO3 + 23 K2SO4 + 197 H2O

which is incidentally the same answer I originally got (dividing everything by 5) :chinese:

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[DoD]Quasar
post May 11 2011, 05:19 PM
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Oh, damn then I guess I copied it wrong to you in the first place because it shows a 2 there on my worksheet :(

Any chance you wouldn't mind doing it again? :ph34r:

:bowdown: :bowdown: :bowdown:

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Chrysalis
post May 11 2011, 05:22 PM
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1111111111111111111111 sigh

luckily it's a pretty minor change, just change the very first coefficient to 2x, then accordingly change -48x to -96x, 42x to 84x, and 66x to 132x:

2x[Cr(N2H4CO)6]4[Cr(CN)6]3 + yKMnO4 + (z-96x)H2SO4 --> 7xK2Cr2O7 + yMnSO4 + 84xCO2 + 132xKNO3 + (z-96x-y)K2SO4 + zH2O

Cr N H C O K Mn S

From O coefficients: 48x + 4y + 4z - 384x = 49x + 4y + 168x + 396x + 4z - 384x - 4y + z
48x + 4y + 4z = 613x + 5z
4y = 565x + z
8y = 1130x + 2z

From K coefficients: y = 14x + 132x + 2z - 192x - 2y
3y = -46x + 2z

12y = 1695x + 3z
12y = -184x + 8z

5z = 1879x
5y = 1176x

x = 5, y = 1176, z = 1879

10[Cr(N2H4CO)6]4[Cr(CN)6]3 + 1176KMnO4 + 1399H2SO4 --> 35K2CrO7 + 1176MnSO4 + 420CO2 + 660KNO3 + 223K2SO4 + 1879H2O

;)

This post has been edited by Chrysalis: May 11 2011, 08:38 PM

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[DoD]Quasar
post May 11 2011, 05:29 PM
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:bowdown::bowdown::bowdown::bowdown::bowdown::bowdown::bowdown::bowdown::bowdown
::bowdown::bowdown::bowdown::bowdown:

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LoSt ReDEyE
post May 11 2011, 08:05 PM
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Chrysalis just gg'ed the thread. No doubt oedy would have done the same however :D




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