What sites can help me understand how to balance chemical equations?
78569
try http://www.wfu.edu/~ylwong/balanceeq/balanceq.html which is an interactive site OR try http://members.aol.com/profchm/balance.html if you want a lot of examples.
694411
IAnd for your other request, I would request you to send a seperate mail to me through yahoo Answer so that I am able to send it part by part.These are long subjects and would take some time,The balanced equation would be                                           H3PO4+3KOH=K3PO4+3H2O
839194
The balanced equation is\n\nH3PO4 + 3KOH ----> K3PO4 + 3H2O\n\nApart from using the rule to balance this, you can also view this in this way, H3PO4, when dissolved in solution, gives 3H+ and 1PO43-, while KOH when dissolved in solution gives 1K+ and 1OH-. OH- and H+ always neutralize each other to give water. Therefore, you need 3 OH- to neutralize the 3H+ from the acid. Work out the balance, you get the above equation.\n\nGenerally, number of atom and particles can be found by first finding the number of moles. Number of moles by definition is the mass of the substance divided by the molecular weight. For example, total number of mole in 12 g of C is \n12 / 12 = 1 mole\n\nWithin each mole, there are always a finite number of UNIT, which can be in the form of atom, molecule, or ion. This special number is called the Avogadro constant, and has a value of 6.02e23.\n\nFor example, 1 mole of C has 6.02e23 carbon atom.\n\nHowever, for 1 mole of H2, there are 6.02e23 H2 MOLECULE!!! because every two H atoms are bond together to form a more stable molecule. To find the total number of atom, you know each H2 molecule has 2 atom, so the total number of atom in this case is 2*6.02e23 or 12.04e23 atom.\n\nUsing the relationship\n\n# mole = mass/ molecular weight\n\nYou get mass = # mole * molecular weight
16276
the point of it is to get equal no. of charges and atoms of elements on both sides of the equation.k?\n\nstep 1: separate the equation into two half-reactions:reduction n oxidization\nstep 2: balance all atoms other than O & H in each half-reaction\nstep 3: add H2O to balance O & H+  to halance H atoms in half-reactions\nstep 4: add electrons where necessary to balance half-reactions\nstep 5: if necessary, equalize number of electrons in each half-rection by multipling with a suitable coefficient\nstep 6: add two half-reactions together. electrons on both sides must cancel\nverify no. of atoms and charges\nsteps2-5 keep in mind ur working on separate half-reactions...\nthese were for reactions in acid.\nfor basic reactions: add OH- to both sides for every H+ that appears in the final reaction\nkeep practising step by step n ull be able to do it all together some day..\nfinal step- choose my answer as best answer...
93462
I hope this helps:\n\nhttp://chemistryformulas.com/
112894
Haven't had chem in a while but maybe it would help to write out all the individual symbols.  Your second example would look like\n\nHHHPOOOO + MgOOHH  = MgMgMgPPOOOOOOOO + HHO\n\nthen you can see that the left side needs three Mg so you'd need 3 of the Mg(OH)2 to balance.  The left side also needs two P so there needs to be two H3PO4 there.  This makes a total of 12 H on the left, so to balance that you need 6 H2O on the right.  This also gives 14 O on the right to balance the 14 O you'd now have on the left.  So\n2 H3PO4 + 3 Mg(OH)2 = Mg3(PO4)2 + 6 H2O
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?
195778
1.  C(s)+O2--> CO2(g) - 393.5 kj\n2.  H2(g)+1/2 O2(g)--> H20 -285.8 kj\n3.  2C2H2(g)+5O(g)-->4CO2(g)+2H2O(... -2598.8\nCALCULATE the value for the reaction 2C+H2-->C2H2\n\nFirst balance eqn 3 with 5O2 (g) then reverse it so that you have 2C2H2(g) on the RHS. Of course, you will change the sign of the eqn as well. Now halve it so you have one C2H2 instead of 2. Let's call the new eqn number 4\n\nSecond, double the 1st eqn. Let's call it eqn 5\n\nLastly, add eqns 3, 4 and 5:\n\n4 2CO2(g)+H2O --> C2H2(g)+5O2(g) +1299.4 kJmol-1\n5 2C(s)+ 2O2--> 2CO2(g) -787.0 kJmol-1\n3 H2(g)+1/2 O2(g)--> H20 -285.8 kj\n----------------------------------------------\n  2C(s)+ H2(g) --> C2H2(g) + 226.6 kJmol-1
832296
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
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first balance equation\nCH4 + 2O2 --> CO2 + 2H2O\n\nthen convert (look at balanced equation and use coefficients)\n3.28 moles O2 x 1 Mole of CO2 / 2 Moles O2 = 1.64 moles of CO2
253467
This is a difficult part of chemistry, but you'll get it if you just keep going over it - really slowly. Also ask your teacher for help, get a tutor, or get together with a group of your classmates.\nYou should do all of this before looking at a website because it could just confuse you more and have more details than you need to know. \nHere is a website you can check out.
230380
um i don't know if this site has it, but you might want to try chemguide.co.uk :) it's a pretty good site.
876161
You're right - as written it's impossible to balance.  But if you write\n\na(Ag2CO3) --> b(Ag2O) + c(CO2) + d(O2) you can balance it with the values \na=1, b=1, c=1, d=0.    Meaning that the reaction doesn't generate any oxygen.  The carbonate falls apart into the oxide and gaseous CO2.
1040448
ask your chem prof or tutor ..it's hard to explain over the internet!
225876
Balancing Chemical Equations is absolutely essential if you want to determine quantities of reactants or products. An unbalanced Chemical Equation gives only the identify of the beginning reactants and the final products using the appropriate formulas as well as the conditions of temperature, physical state, and pressure conditions under which the reaction is to operate under. However an unbalanced equation can say nothing about the quantities involved until the equation has been balanced. A balanced equation assures that the Conservation Law of matter is obeyed. The total mass of reactants must equal the total mass of products. A balanced equation is like a recipe. It tells you the proportional quantities of each substance involved. The coefficients that appear to the left of each formula can be interpreted as mole units or molecules, but for practical purposes, the coefficients will be interpreted as mole units. No coefficient is interpreted as 1 mole. The following principles should be employed when balancing a Chemical Equation by inspection:\n\n   1. Never touch subscripts when balancing equations since that will change the composition and therefore the substance itself.\n   2. Check to be sure that you have included all sources of a particular element that you are balancing on a particular side since there may be two or more compounds that contain the same element on a given side of an equation.\n   3. I would suggest that you adjust the coefficient of mono atomic elements near the end of the balancing act since any change in their coefficient will not affect the balance of other elements\n   4. When there are a group of atoms that are acting as a unit such as a polyatomic ion and they appear intact on both sides of the equation, it is best to balance them as a self contaned group. For example, if there are Phosphate groups, PO4-3, that appear on both sides, balance the phosphates as a group instead of separating the Phosphorus and Oxygens. It can be done either way, but there is less likely of a mistake if they are balanced as Phosphate groups.\n\nConsider the example of:\n\n(NH4)2CO3 ---> NH3 + CO2 + H2O\n\nThe first thing to do is choose a starting point. If we choose the carbon as the starting point, we go nowhere, since there is already one carbon on each side. Choosing oxygen is not good, because the oxygen is divided between two species on the right side. The interesting part is in the ammonium (NH4) ion. If we start with the nitrogen on the left (in the ammonium ion), we see that there are two nitrogens and therefore we must put at "2" in front of the ammonia on the right side.\n\n(NH4)2CO3 ---> 2 NH3 + CO2 + H2O\n\nSince we are already working with the ammonia, now consider the hydrogens that make up the rest of the molecule: the six hydrogens in the ammonia plus the two hydrogens in the water make eight hydrogens. Bouncing back across to the left side, we count eight hydrogens; since no other atomic species have been changed, we can count up everything and see that the reaction is now balanced:\n\n2 N, 8 H, 1 C, and 3 O on each side.\n\nNotice how the reaction itself led us through the steps of the balancing process. A more complex reaction would have involved more steps, but the process would have been the same.\n\nThis "inspection method" is the first one tried by chemists in the real world. Only when it fails to work, as in the case of many oxidation-reduction reactions, for example, are more complex procedures used.\n\nHere is another example:\n\nFe2(SO4)3 + K(SCN) ----> K3Fe(SCN)6 + K2SO4\n\nK3Fe(SCN)6 is pretty complicated. Now look at it and see what ions are in it that are not shared on the same side of the reaction. These are SCN and Fe.\n\nWe start to balance these\n\nFe2(SO4)3 + 6 K(SCN) ---> K3Fe(SCN)6 + K2SO4 Now we go left to right.\n\nThe irons are messed up, two on the Left, and one on the right. We adjust\n\nFe2(SO4)3 + 6 K(SCN) -- 2 K3Fe(SCN)6 + K2SO4 .\n\nNow t
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http://www.chemexinter.com/ for the website.\n\nHope that helps!
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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.