where can i find help balancing chemical equations?
106704
If you go to a search engine, like google, you can just type in "balancing chemical equations" and it will give you tons of info. We were doing this in chemistry a few weeks ago, and thats how I found alot of help online. \n\nGood luck!!
148175
The correct balanced equation is:\n2ZnS + 3O2 --> 2ZnO + 2SO2\nThis is BALANCED\nMr chemistry teacher, you didnt balance your equation,...and i'm in the ninth grade
814943
There is no balanced equation for that question.\n\nYou need to show your products going to reactants for the world to be able to balance the equation.\n\nGood Luck!
1053423
Chemistry rocks!  Try these sites...
329382
c2h4+ 3o2 -> 2co2 + 2h2o is balanced\n\nBefore, there are 2 carbon, 2 hydrogen and 6 oxygen\nAfter is the exact same.   \n\nIs there something else you are asking?
938147
google search chemistry tutorials . there are thousands of sites\nchemtutor.com\nhighschoolace.com\nchemistrycoach.com\nechemistryhelp.com\nare just some of the few that i use.\nthey're really helpful!
753594
It is perfectly permissible to write the products as Br2 + Br2.  It is probably easier to work this way.  After you balance the equations, just add the Br2's together to give the final reaction
775066
You have the products, but now you have to balance the equation. The unbalanced equation is:\n\nLi2CO3 + Cr2(SO4)3 --> Li2SO4 + Cr2(CO3)3\n\nBalanced:\n\n3 Li2CO3 + Cr2(SO4)3 --> 3 Li2SO4 + Cr2(CO3)3
188263
http://chemfinder.cambridgesoft.com/
39364
To understand balancing chemical equations better, it helps to know what it is called. It is called "stoichiometry". Just a quick web search and I found a TON of very accessible web sites on balancing chemical equations. I put a couple of them below. \n\nIt should be noted that in mathematics, there is a very easy way to solve stoichiometry equations, or do chemical balancing. It is much easier than what chemists are usually taught. To look at the mathematical version of the same problem, look for web sites and books on linear algebra. The first linear algebra web site listed below is a simple introduction, but I also included a few links to free books in linear algebra which is probably more detail than you want.\n\nHere is an easy way to start thinking about it, however. It is all about counting how many atoms of each type are on the left side of the equation, and how many are on the right side. You want to adjust the numbers in front of the molecules so that there are as many atoms of each type on the right side as there is on the left side. \n\nSuppose you are considering the burning of carbon C in oxygen O2 to make carbon monoxide, CO. \n\nThere are j C atoms and k O2 molecules that combine to make m CO molecules \n\nSo jC + k O2 ----> m CO\n\nIt is up to you to pick j, k and m so that \n\nthere are the same number of C and O atoms on both sides.\n\nIf m=2, and k=1 and j=2, then \n\n2C+O2----->2CO \n\nand you see there are 2 C atoms on each side, and 2 O atoms on each side, and you are done.\n\nOf course it gets much more complicated than this. That is addressed in the references below.
266694
Balance the equation.\n\nDetermine what the limiting reactant is.\n\nDetermine how much product can be produced from the limiting reactant.\n\nDetermine how much of each reactant is needed to fully react the limiting reactant.  The remainder of each reactant is the excess.
157730
I think FePO4 +O2 >>>>> Fe2O3 + PO2. balance it and solve it . n its simple stoichiometry.
364627
http://www.juliantrubin.com/fairprojects/chemistry/chemicalreactions.html\nhttp://www.scifair.org/\nhttp://physics.usc.edu/~gould/ScienceFairs/
837250
The equation doesn't balance.  There's no hydrogen on the right side.  But, assuming that you ended up with H2 gas on the right (I don't think that reaction works real-world, but whatever), the main thing you need to look at is the chlorine.  It's Cl2 on one side and Cl3 on the other.  That means that you must have a total number of chlorine atoms which is a multiple of 6.  Try balancing the equation with 6 chlorine atoms and see what you get:\n3CuCl2 + 3H2O + 2Al -----> 2AlCl3 + 3 CuO + 3H2
744922
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
188200
I had trouble with the same thing. The best way, and it's worked till now in college, is to take the most complicated element used (i.e. CH3OOH + C2H3O9 ---> etc), figure out O's first since O is used the most. Then work backward using the 2nd most used element. You might get a non-reduced form of balancing, but you can check whats divisible (i.e.  4CH + 2 CH2H3O2 <---> 4 OH + 2 CH3; all divisible by 2 so it'd be 2, 1, 2, 1 respectively). See if this works, it's worked when I've helped other people in the past as well.
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)