Cann u tell me how to "balance chemical equations" or a website that shows me cause this stuff is hard and i..
154085
the purpose of balancing chemical equations is to have the same amount of elements that you have in the left side in your right side...i'd explain it to you but i hard to type the subscripts...i looked up a website that it explaings it pretty good...if u still don't get it, get at me...ok
952049
just browse over formulas which are tough for u... and review how to do steps to problems that have been tough. otherwise u should b ok. good luck
1040448
ask your chem prof or tutor ..it's hard to explain over the internet!
683249
Oh, come on, it doesn't get any easier than this!\n\nSnO2 + 2H2 --> Sn + 2H2O\n\nAbove is balanced. 1 Sn on both sides, 2 O on both sides and 4 H on both sides.\nYou need to work a little harder on these yourself.
777428
well actually me and you are in the same situation. im trying to find out how to write wrod equations into a reaction. lol but neway i just learned how to balance equations its very simple.\nFe2O3+CO--Fe+CO2\n:draw a box around each set of elements:Fe2O3 would b in a box CO would be in a box and so forth.\n:then write out a diagram of how many atoms each element has:\nEx: Element/ Before/After\n      Fe          2         1\n      O            4         1\n      C            1          1\n:Then Write numbers in front of each of the boxes until the inventory for each element is the same both before and after the reaction. Now, what happens when we put a number in front of a formula? Basically, anything in that box is multiplied by that number, because we're saying that we have that many of that kind of molecule. \n:Now do this until the diagram is equal on both sides\nAND YOU HAVE YOUR BALANCED EQUATION
755565
1.) Cr(NO2)2+(NH4)2SO4\n---->CrSO4+(2)NH4NO2\n2.) (2)KNO3+H2CO3\n----->K2CO3+(2)HNO3\n3.) (3)KBr+Fe(OH)3\n----->(3)KOH+FeBr3\n4.) B2Br6+(6)HNO3\n----->(2)B(NO3)3+(6)HBr\n\nHope this helps=)\nTip: Start by balancing those more isolated atoms first, especially when you are balancing more complex equations (they may have atoms that appear in more than 1 products).\nAlso, feel free to leave numbers in fractions while you balance ur equation. You can always multiply all the numbers later to make them all whole numbers. You will find this useful later when you have to balance long equations. Itd definitely more tedious to alter every number in ur mind and writing them all out at one go. Do it progressively=)
914017
I will suggest that first you write the equation in a compact mode, balance it and then add all the bells and whistles.\n\nCuSO4 +K3PO4 ---> Cu3(PO4)2 + K2SO4\n\n3CuSO4 +2K3PO4 ---> Cu3(PO4)2 + 3K2SO4\n\n3CuSO4(aq) +2K3PO4(s) ---> Cu3(PO4)2(s) + 3K2SO4(aq)\n\nNow that it is balanced it is easier to break down:\n\n3Cu(+2) + 3SO4(-2) + 2K3PO4(s) --->  Cu3(PO4)2(s) + 6K(+) + 3SO4(-2)
413972
man you can't balance an equation if there is none and HCL is not an equation
158900
Yup. They are not simple...\nChemistry...\nhttp://www.chm.davidson.edu/ChemistryApplets/equilibria/EquilibriumConstant.html\nhttp://en.wikipedia.org/wiki/Chemical_equilibrium\nThey use the equilibrium constant expression (Kc) here...\nhttp://www.cofc.edu/genchemlab/equilibrium.htm
119628
So, usually the best way to do these is to just fiddle with them until you get them right, but for the sake of it, let's go with a systematic approach using algebra.  First we assign a variable in front of each of the terms on the left and right like:\n\nx Pb  +  y H2O  +  z O2  -->  a Pb(OH)2\n\nSo now we'll write down equations to make sure that we have the same numbers of both sides of each *element*.  I'll do this by writing down the variable in front of each compound in your equation (so x, y z and a) and then multiplying those by the number of times an element exists in that compound.  (So when I'm writing the equation for hydrogen, the H20 term will be 2*y, because y is the coefficient in front of H20 and H occurs twice in that compound).\n\nRemember, the number of any element on the reactants (left) side equals the number of that element on the products (right) side.\n\nFor Pb: x = a, because Pb shows up only once on each side with a subscript of 1, those two compounds will have to be present in the same molar amounts.\n\nFor Hydrogen: 2y = 2a\n\nFor Oxygen: y + 2*z = 2 * a\n\nSo now we have three equations:\n\nx = y\ny = a\ny + 2z = 2a\n\nBut four unknowns, which is fine because we'll just solve for them in terms of one of the variables and then simply choose the last variable to make all the numbers nice.  So solving the equations in terms of a:\n\nx = a\ny = a\nz = (2a - y)/2 = (2a - a)/2 = a/2\n\nSo now we have x, y and z in terms of a free parameter, a.  We'll just choose a = 2, for example, to make x, y and z all turn out as integers.  So now we've chosen:\n\nx = 2\ny = 2\nz = 1\na = 2\n\nSo the balanced equation is:\n\n2 Pb + 2 H2O + O2 --> 2 Pb(OH)2\n\nMake sense?
777578
The same way that you balance them in chemistry.  Start with the most irregular element and go from there.
691513
Check that both sides of equation match. Here is an example:\n\n3 CH3CH2OH + 2 Cr2O7 2- + 16 H+ --> 3 CH3COOH + 4 Cr3+ + 11 H2O\n\nThere should be an equal amount of H, C, O, Cr in both sides of the equation. \nIt is sometimes difficult to know where to start balancing an equation. In such case you should try to find an element/molecule that does not appear in every molecule on the other side of the equation. Eg. when balancing this equation above, I first balance the number of Cr because it is on its own on the right side. Sometimes you just need to try and play with numbers for a while to find the right answer. Hope this helped...
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.
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.
223710
Start by balancing Al, then Cl. Al is already  balanced, so when you balance Cl everything is easy from their. \nYou final amswer will be:\n3HCl + Al(OH)3 ----> 3H2O + AlCl3
747348
Hi Kitty!  Dr. J to the rescue with a full answer.\n\n1) 4Li + O2 ---> 2Li2O\n\n2) 4HF + SiO2 ---> SiF4 + 2H2O\n\n3) It is simpler if we eliminate the K and the sodium and write the equation ionic because it is a redox reaction and we need to balance.\n\nThe overall redox equation will be:\n\nCr2O7(=) + SO3(=) + H(+) ---> Cr(+3) + SO4(=).\n\nNow you balance by the half reaction method.  The two reactions are:\n\n6e(-) + Cr2O7(=) + 14H(+) ---> 2Cr(+3) + 7H2O\n3SO3(=)  + 3H2O --->  3SO4(=) + 6H(+) + 6e(-)\n\nEliminating the electrons:\n\nCr2O7(=) + 3SO3(=) + 8H(+) ---> 2Cr(+3) + 3SO4(=) + 4H2O\n\nThis one was tough!!!  Thumbs up!\n\nIf you want you could put the K and the Na back.\n\nBest answer?
660043
1.) do your own homework\n2.) this is the computer forum, not high-school chemistry.  Next time post to the right furom.