Does anyone know how to balance equations, for chemistry??HELP!!?
151775
giving some sample equations would help!\n\nyou have to make the left hand side balance wilh the right hand side...\n\nsuppose you have the unbalanced equation\n\nC4H10 + O2   =    CO2  +  H20\n\nin this equations the easiest thing to balance first would be the carbon so on the right hand side you put \n\nC4H10  + O2   =   4CO2  +  H2O\n\nThen the hydrogen,you have 10 Hydrogens on the left so you need 10 on the right\n\nC4H10  +  O2  =  4CO2  + 5H2O\n\nthus you will need 13 oxegen atoms on the left side.. since oxegen exists as O2 and 13 is an odd number you multiply everything by a factor of 2 to make it just right... so your final equation looks like....\n\n2C4H10  +  13O2  =  8CO2   +  10H2O
789547
first you need to balance the equation\n\nC3H8(g) + O2(g) -------> CO2(g) + H2O(l) (unbalanced)\n\nCompare carbon atoms on both and you will make three CO2 for every one C3H8.  That leaves 8 Hydrogen atoms to make water which has 2 hydrogen atoms per molecule so you would need  4 H2O to balance that out. The equations starts to look balanced like this\n\nC3H8(g) + O2(g) -------> 3CO2(g) + 4H2O(l) (unbalanced)\n\nnow we have to figure out how much oxygen is needed to balance the whole thing.  There are 2 Oxygens in CO2 and 1 in H2O so we have a total of (3x2)+(4x1)=10 oxygen on the right side of the equation. since O2 has 2 oxygen we need 5 of them to get 10 so the equation looks like this.\n\nC3H8(g) + 5O2(g) -------> 3CO2(g) + 4H2O(l) (balanced)\n\nsince oxygen has an atomic mass of 16.000 amu then O2 has a mass of twice that or 32.000 amu=32g/mol\n\nfrom our equation we get that there is 1 mole of C3H8 to every 5 moles of O2. \n\n5molx32g/mol=160g
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
129663
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
930806
When you want to balance ANY chemical equation, always follow these steps:\n\n1. Balance metals\n2. Balance non-metals\n3. Balance hydrogen\n4. Balance oxygen\n\nLet's do it\n\nFirst let's balance K (metal):\n\n"2" KHCO3 + __H3PO4 yields __K2HPO4 + __H2O + __CO2\n\ndone. now let's balance P (non metal): if you check, it is balanced already.\n\nnow C (also non metal):\n\n2 KHCO3 + __H3PO4 yields __K2HPO4 + __H2O + "2" CO2\n\nnow hydrogen:\n\n2 KHCO3 + __H3PO4 yields __K2HPO4 + "2" H2O + 2 CO2\n\nfinally, oxygen (its already balanced!!)\n\nSo you end with\n\n2 KHCO3 + H3PO4 yields K2HPO4 + 2 H2O + 2 CO2
881744
First, balance the equation.\n\n2LiOH + CO2------>Li2CO3  +  H2O\n\nSo, for every 2 LiOH, there will be 1 Li2CO3 and 1 H2O\n\nSo, for every 5.4 LiOH, there will be 2.7 Li2CO3 and 2.7 H20
973794
moles = concentration* volume\n\nfind the moles of the substance with given concentration and volume\n\n\nwrite and balance the equation. determine the moles equivalence of every reactant then use the equation above
577084
Acid + base = salt + water.\n\nUnbalanced eqn is: Al(OH)3 + H2CO3 --> Al2(CO3)3 + H2O
723776
Walk through these things slowly... First off, lets correct your first equation (?)\n\nLi + OH- + K2(2e-) <--> Li + 2e- + 2K + OH- \n\nThat doesn't make since... (is that your initial equation?)\n - Li are ballanced\n - OH- are ballanced\n - K2(2e-) goes to 2e- and 2K which is balanced\n\nWhat's the question again? I think there was a typo somewhere in your original expression. Are you going from LiOH to KOH? If so, your expression would be more like this...\n\nLiOH + K2(-2) <--> Li(+) + 2e- + 2KOH\n\nStart by balancing the OH and then the Li to get\n\n2LiOH + K2(-2) <--> 2Li(+) + 2e- + 2KOH\n\nAnd, there you go.\n - Li, K and OH are balanced
1040448
ask your chem prof or tutor ..it's hard to explain over the internet!
17656
Best technique is to look for where it is unbalanced and change it so it is balanced. Repeat until it is balanced.\n\neg. We need 2 hydrogen here on the left, so go for:\n\nMg + 2HCl -> MgCl + H2\n\nNow we need another 2 as we have 2 Cl on the left and one of the right.\n\nMg + 2HCl -> 2MgCl + H2\n\nNow we need to have 2 Mg on the left as we have 2 Mg on the right. So we get:\n\n2Mg + 2HCl -> 2MgCl + H2\n\nAll balanced so that is the answer.
701763
First, notice you have two S atoms on right hand side.\nTo get two S atoms on left, you must have two H2SO4 molecules.\nYour equation is now Cu + 2H2SO4 ---> CuSO4 + SO2 + ??\nIf you subtract right from left, you are left with 4H + 2O atoms, for 2H2O\nSo final eq is Cu + 2H2SO4 ---> CuSO4 + SO2 + 2H2O\n\nHope this helps you to approach these equation-balancing problems methodically.
926891
just count how many of each element you have, and make sure they add up\n\n\nC2H6 + 5O2 ---> 2CO2 + 3H2O\n2 carbon 6 hydrogen 2 oxygen----> 1 carbon 2 oxygen 2 hydrogen 1 oxygen\n\n\nso to even it out it would be...\n\nC2H6 + 7O2--->2CO2 + 3H2O\n2 carbon 6 hydrogen + 7oxygen--->2 carbon 4 oxygen + 6 hydrogen 3 oxygen\n\niff you add up each element they are the same on each side
268703
Use the overall rate equation: rate = k[A]^x[B]^y...\n1) define for rate1 and rate 2 with the appropriate concentrations.\n2) divide both sides (e.g. rate1/rate2 = ...\nNote that this cancels k\n3) Solve for x, y, ...  These are your respective orders.\n4) Add x, y, ...  This is your overall order.\n\nYou can get the value of k by:\n5) Plug in values of x, y, ... to either rate1 or rate2 equation.\n5) Solve for k.
967368
Just treat all other letters as constants. You can multiply, divide, add and subtract both sides of an equation and maintain the equality.
157730
I think FePO4 +O2 >>>>> Fe2O3 + PO2. balance it and solve it . n its simple stoichiometry.
822606
CuSO4 + NH4S = CuS + NH4SO4\n\n1)Write a rough equation without balancing anything. The 2 after copper, which I presume is to be written as copper (II), indicates that copper has a charge of +2. It is a transition metal and can have a differing charge.\n\n2)Because ammonium has a charge of +1 and the sulfate has a charge of -2, there needs to be 2 ammoniums. SAme thing is for the left side. The ammonium is +1 and the sulfide is -2. There needs to be two ammoniums on the left side. \nCuSO4 + 2NH4S = CuS + 2NH4SO4\n\n3) Because there are two sulfides on the left side, there needs to be two sulfides on the right side. \n\nCuSO4 + 2NH4S = 2CuS + 2NH4SO4\n\n4) Now there are two sulfates on the right side, but not on the left. So we place a two in front of copper (II) sulfate.\n\n2CuSO4 + 2NH4S = 2CuS + 2NH4SO4\n\n5) I believe that we are done with balancing so let's check our work.\n\n6) 2 coppers are on the left and right side.\n2 sulfates are on the left and right side.\n2 ammoniums are on the left and right side. \n2 sulfides are on the left and right side.\n\n7) In most cases, not all compounds will have the same number on each side. It's just a coincidence that there are all 2 of each. If you need anymore help, email me ^_^