Chemical Forums
Specialty Chemistry Forums => Chemical Education and Careers => Topic started by: Borek on April 13, 2006, 11:15:50 AM
-
I need an example of such situation - unfortunately whenever I try I am just balancing, can't locate a loop. I am just too good ;)
-
?? ???
-
When somebody not fluent in equation balancing starts to work on some 'complicated' case it sometimes end in a circle - "to balance A atoms I have to modify B coefficient, in turn there is not enough C atoms and I have to modify D coefficient - which makes my equation lacking A atoms which forces me to modify B coefficient" - and circle closes here.
I just have troubles constructing such example.
-
Oh, I think I've got it! You want us to propose you such examples, don't you?
-
Yes. I have approached several equations trying to make this mistake but I am too efficient in NOT making it :)
-
Try this one. When working with the adjustment method, as you mentioned above, this one is hard:
CH3ONa + NaClO2 + HCl --> COCl2 + NaCl + H2O
Balancing by means of linear algebra methods is easy, but balancing by hand may be quite hard for a beginner.
-
Good one, thanks!
-
I think I have found a better one - looks much simpler ot first sight:
P4O6 -> P4 + P2O4
but is surprisingly hard to balance by inspection.
-
I balanced it (I don't post the reaction, I don't want to spoil it ;) ... the other members should try to balance it as well...)
-
I find it very easy to attempt to balance the molecules with the most amount of elements, then work down to the molecules with the least amount. That way, when you get down to balancing the last molecule, you won't disrupt any other pre-determined coefficients.
I also balanced this, took me under 30 seconds. ;D
-
This one will probably take you longer:
P2I4 + P4 + H2O -> PH4I + H3PO4
-
Not so hard. I is only on the right in 1 place, so is O, then after that H is only on the right in one unknown place, then you just have to count the P's.
Kind of ugly coeffients though.
-
I tried for about 10 minutes, I can't figure that last one out. I can't get the ratio of H to O on the product side to be 2:1 like it is on the reactant side.
-
Check Jched, they love comparing methods for these types of things.
-
Many of these, although some not directly on the subject:
J.Chem.Educ. 22, 266, 461 (1945). 72, 716, 1125 (1995).
J.Chem.Educ. 36, 215 (1959). 74, 1359 (1997).
J.Chem.Educ. 38, 327 (1961). 65, 45 (1988). 74, 1270 (1997). 76, 362 (1999).
J.Chem.Educ. 69, 276 (1992)
J.Chem.Educ. 23, 550 (1946). 36, 77 (1959). 74, 69 (1987). 76, 362 (1999)
J.Chem.Educ. 38, 329 (1961). 74, 1365, 1365 (1997)
J.Chem.Educ. 62, 507 (1985). 72, 894 (1995).
J.Chem.Educ. 73, 507 (1996). 74, 1367, 1368 (1997).
J.Chem.Educ. 54, 704 (1977). 59, 728 (1982). 69, 279 (1992). 71, 295 (1994). 74, 1369 (1997).
-
I'll have to snoop around in those files. I'm new to the site so I haven't really explored.
-
http://www.chembuddy.com/?left=balancing-stoichiometry&right=toc
Comments welcomed.
-
Page error: http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-inspection
XML Parsing Error: mismatched tag. Expected: </p>. Location: http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-inspection Line Number 110, Column 5: </div> ----^
-
(2)P4O6 -> (1/2)P4 + (3)P2O4
That was a really easy one by inspection. :)
-
(5/16)P2I4 + (13/32)P4 + (4)H2O -> (5/4)PH4I + (1)H3PO4
The above is also a very easy one for me to do by inspection. Most people will not want to use fractions and will get caught up avoiding them. But if you just run with them and are comfortable enough with them its no big thing. And, then at the end you can multiply by 32 and get a more aesthetically pleasing form.
-
Thanks for that </p> tag. For some reason Opera behaves differently than before - instead of stopping of the first error it goes into quirks mode and renders everything, displaying error messages below, that's why I am missing these things now :( Have to check what's going on, perhaps something is wrong with the headers sent (XHTML/HTML)?
As for the equations - I am not stating they are impossible, but they ARE harder than most other examples. But I am open to suggestions - just bear in mind that the equation must fit the screen (that's why I am not using Stout:
10(Cr(N2H4CO)6)4(Cr(CN)6)3 + 1176KMnO4 + 1399H2SO4 -> 35K2Cr2O7 + 1176MnSO4 + 420CO2 + 660KNO3 + 223K2SO4 + 1879H2O
nor Blakley:
88H2 + 15Ca(CN)2 + 6NaAlF4 + 10FeSO4 + 3MgSiO3 + 6KI + 2H3PO4 + 6PbCrO4 + 12BrCl + 3CF2Cl2 + 20SO2 -> 6PbBr2 + 6CrCl3 + 3MgCO3 + 6KAl(OH)4 + 10Fe(SCN)3 + 2PI3 + 3Na2SiO3 + 15CaF2 + 79H2O
nor this nice oxidation example:
10K4Fe(CN)6 + 122KMnO4 + 299H2SO4 -> 162KHSO4 + 5Fe2(SO4)3 + 122MnSO4 + 60HNO3 + 60CO2 + 188H2O
-
Those look like great extra credit questions for chem studesnts. ::Evil grin::
-
K3[Fe(SCN)6] + 16Na2Cr2O7 + 58H2SO4 -> Fe(NO3)3 + 16Cr2(SO4)3 + 6CO2 + 58H2O + 16Na2SO4 + 3KNO3
EBAS rulez ;)
-
I found all the equations above (not Boreks' directly above!) relatively easy. Here are a couple which I had to balance in this year's UK Chemistry Olympiad competition, which I got a Ag in :'( (missed the Au by a couple of marks :'(), but I was happy to get these right :):
NH3 + NO --> N2 + H2O
NH3 + NO2 --> N2 + H2O
I found them easy but harder than a couple on this page so I thought I might as well mention them!
I came across this equation while messing around with a chemistry experiment simulation program (called CrocChem I think):
Zn + HNO3 --> H2O + N2O + NO + Zn(NO3)2
I checked this reaction out online and different reactions happen with Zn and HNO3 in different conditions, so this equation may not be very accurate. I haven't tried to balance it myself because I've already seen the coefficients so its hard to try to balance it without 'cheating'! I'm sure most of you will find it easy to balance ;).
-
Zn + HNO3 --> H2O + N2O + NO + Zn(NO3)2
4Zn + 10HNO3 -> 5H2O + N2O + 4Zn(NO3)2
3Zn + 8HNO3 -> 4H2O + 2NO + 3Zn(NO3)2
http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-failure
(read the section about KClO3 and HCl).
-
88H2 + 15Ca(CN)2 + 6NaAlF4 + 10FeSO4 + 3MgSiO3 + 6KI + 2H3PO4 + 6PbCrO4 + 12BrCl + 3CF2Cl2 + 20SO2 -> 6PbBr2 + 6CrCl3 + 3MgCO3 + 6KAl(OH)4 + 10Fe(SCN)3 + 2PI3 + 3Na2SiO3 + 15CaF2 + 79H2O
I can't imagine a situation in which you'd need to balance this monstrosity. In real life, you only balance single reaction steps and then add them all together to get a big overall equation like this. Or am I wrong?
-
I can't imagine a situation in which you'd need to balance this monstrosity. In real life, you only balance single reaction steps and then add them all together to get a big overall equation like this. Or am I wrong?
Exam with mad teacher? ;)
I believe it was an example of how efficient algebraic method of equation balancing (http://www.chembuddy.com/?left=balancing-stoichiometry&right=algebraic-method) is, in the paper published by GR Blakley in JCE in 1982 (never seen original paper though so I can be wrong).
-
4Zn + 10HNO3 -> 5H2O + N2O + 4Zn(NO3)2
3Zn + 8HNO3 -> 4H2O + 2NO + 3Zn(NO3)2
The simulator went for the 2A + B equation! (11Zn + 28HNO3 --> 14H2O + 2N2O + 2NO + 11Zn(NO3)2)
I wonder if you asked a group of students to balance: HNO2 + HN3 --> N2 + N2O + H2O, how many would come up with: 2HNO2 + 4HN3 --> 6N2 + N2O + 3H2O!
-
I wonder if you asked a group of students to balance: HNO2 + HN3 --> N2 + N2O + H2O, how many would come up with: 2HNO2 + 4HN3 --> 6N2 + N2O + 3H2O!
Personally, I would not have come up with that answer. I would have stopped right at the beginning, because it's already balanced...
HNO2 + HN3 --> N2 + N2O + H2O
2 H , 4 N , 2 O on left side
2 H , 4 N , 2 O on right side
-
Personally, I would not have come up with that answer. I would have stopped right at the beginning, because it's already balanced.
Yup!
Is the 'other' reaction thats happening/could happen: HNO2 + 3HN3 --> 5N2 + 2H2O?
-
HNO2 + 3HN3 -> 5N2 + 2H2O
4HNO2 + 2HN3 -> 5N2O + 3H2O
so, for example:
4169HNO2 + 4637HN3 -> 5105N2 + 3935N2O + 4403H2O
is correctly balanced ;)
-
4169HNO2 + 4637HN3 -> 5105N2 + 3935N2O + 4403H2O
:o
As lemonoman said, I wonder how many times in in my life (which is most likely going to be a chemist-life ;) ) I'll have to balance such equations... :P
-
That's a trick that have nothing to do with chemistry. But looks very seriously chemically :)
You may generate such equations by yourself, just follow description from my lecture on balancing failures (http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-failure) :)
-
Another interesting one ;)
Pb(NO3)2 + Cr(MnO4)2 = Pb3O4 + Cr2O3 + MnO2 + NO
-
Apart from Borek's monster-equations ;) , This is one that I had to balance at school yesterday:
P + KOH + H2O ? PH3 + KH2PO4