How do you balance this chemical equation?
473164
You have to make sure all the elements on each side of the equation are equivalent. \n\n4 PH3 + 8 O2 = P4O10 + 6 H2O\n\n\nThus each side of the equation has total:\n4 Phosphorus\n16 Oxygen\n12 Hydrogen
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!
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
856873
The balanced chemical equation and the molecular weights of the reactants and products.\n\nIn simplest form, assume\n\nXY + AB = XB + AY.\n\nAssume the molecular weight of XY is 50, and of XB 35.\n\nFor every 50 grams (or pounds, tons, or any other measure of mass) of XY you use, you will get 35 grams (or whatever) of XB.\n\nEDIT:\n\nOK. Atomic weights are H=1, S=32, Cl=35.5, Fe=56. Molecular weights are FeS=88, HCl=36.5, H2S=34, FeCl2=127.One gram molecular weight of each is its molecular weight in grams. So if you react 88 grams of FeS with 36.5 x 2 grams of HCl (the balanced equation tells you that you need 2 moles of HCl, so you multiply by 2), you will get 34 grams of H2S and 127 grams of FeCl2. \n\nNotice that the weights balance. You put in 88 + (36.5 X 2) grams, or 161 grams. You got out 127 + 34, or 161 grams.\n\nYou need the mass of any one of those compounds - either reactant or product - to determine all the others. Calculate the moles of the one for which you have the mass, determine the ratio of the moles of that one to the others, multiply by that ratio and then multiply result by molecular weight.
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
727458
2. 2 Na + 2 H2O --> H2 + 2 NaOH\n3. 2 HNO3 + Ca(OH)2 --> Ca(NO3)2 + H2O\n4. C5H12 +  8 O2 --> 6 H2O + 5 CO2
138065
The equilibrium constant of any equation is [C]^c[D]^d/[A]^a[B]^b.\n\nthe equation is already balanced. But you have to know how many moles of each element there are in order to compute the problem
342344
The answer is C.\n\nIn order for the equation to be balanced, you have to make sure there are the same number of each element on both sides of the equation.\n\nIn answer C, the number of each element is equal on both sides. \n\nIn answer A, the number of Sodium (Na) and Hydrogen (H) atoms are not equal on both sides of the equation.\n\nIn answer B, the number of Nitrogen (N) and Hydrogen atoms are not equal on both sides of the equation.\n\nIn answer D, the number of Sodium atoms is equal, but there is one more Sulfur (S) atom on the product (right) side of the equation.
749125
The unbalanced equation is:\nCH3-OH + O2 (+ heat) ---> H2O + CO2\n\nThe balanced equation would be:\n2CH3-OH + 3O2 (+ heat)  ---> 4H2O + 2CO2
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)
539827
(a) 2 NaClO3 -> 2 NaCl + 3 O2\n\n(b) 3 MnCl2 + 2 Al -> 3 Mn + 2 AlCl3\n\n(c) C4H10 + 13/2 O2 -> 5 H20 + 4 CO2 \n                  or\n(c) 2 C4H10 + 13 O2 -> 10 H20 + 8 CO2\n\nBalancing is quite simple for these simple equations. You just have to see the left hand side pick up an element, see how many atoms of that element are there in it and multiply the R.H.S compound containing that element with a number which will be equal to no. of atoms in the l.h.s. Then again repeat the procedure by picking another element.....see it's quite difficult to explain you with writing aneways i hope it helps u :)
866732
do you want to know how to balance chemical equations? It's actually quite simple when you know how\n\nrather than asking for direct answers, it's better if you know how to do it for yourself\n\nI'm not going to answer your questions but show you how to do it for future reference\n\nH2 + O2-->H2O\n\nthe above equation is not balanced because there is only one oxygen atom on the right and two on the left\n\nnow, how do we balance this?\n\nfirst thing we need to do is ensure that there are equal amounts of atoms on the left and on the right\n\nso, to balance that equation\n\nwe will write\n\n2H2+O2-->2H2O\n\nthe equation is now balanced with respect to equal atoms on the left and right\n\nin words\n\n2  molecules of hydrogen  react with 2 oxygen atoms to produce 2 molecules of water\n\nthere are now 4 hydrogen atoms and two oxygen atoms on each side of the equation\n\nthe 2 in front of each compound represents the molecules and the subscripts represent the number of atoms
805565
Ca + 2H2O yields Ca(OH)2 + H2\n\nBasically it's almost balanced already, just add a 2 in front of the H2O.
900316
This is a common redox question easily balanced by the half-reaction method.   The half-reaction method is taught using a few different styles, so you'd best learn the style that your teacher or professor expects you to learn.  \n\nFirst, identify your half reactions and balance them:\n\n(Cr2O7)2- contains 2 Cr6+ ions being reduced to 2Cr3+, gaining  6 electrons:\n\n6e-  +  2Cr6+  --->  2Cr3+    (net charge now adds up to +6 on both sides)\n\nMn2+ is oxidized to Mn4+ (as MnO2) losing 2 electrons, or\n\nMn2+  ---->  MnO2  + 2e-\n\nNet electron transfer must equal zero, so we must triple the oxidizing reaction to yield 6 e- to balance the 6 e- gained, so e- gained now equals e- lost (that is, 6 e' in, 6 out)\n\nThat gives us the clue to multiply the manganese term by 3:\n\n(Cr2O7)2- + H+ + 3Mn2+ --->  2Cr3+ + 3MnO2  + H2O\n\nInspection shows the Cr, Mn, and O terms to be balanced but the left side of the equation to be down by 1 H+, so let's put a 2 by the H+....\n\n(Cr2O7)2- + 2H+ + 3Mn2+ --->  2Cr3+ + 3MnO2  + H2O\n\nWe think we're finished here, so let's make the important two final checks:\n1.  Mass Balance and 2.  Charge Balance\n\nEach side of the reaction has 2Cr, 3Mn, 7O, and 2H\n(Mass balance OK)\n\nNow add up the net charges on both sides:\n-2  +2  +6 = +6 on the left side...\n+6 + 0 + 0 = +6 on the right.  (Charge Balance OK)\n\nReaction is balanced.\n\nAgain, WARNING:  Some methods teach you to balance in ACID or in BASE.  This problem, by virtue of the H+ presence is in acid.  Balancing a redox equation in base has a couple more steps attached.  Again, consult your text or your instructor as to their preferred style.
927756
2H2O2 → 2H20 + 02 is the balanced equation for the above decomposition
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
851400
the first equ is balanced, b/c on the left side u have Pb-1, C-1 O-3 on the right side u have Pb-1 C-1 O-3 so since both side are equal it is balanced\n\nans for question 2 BeCl2 + 3O2------Be(ClO3)2\nBe-1                                  Be-1\nCl-2                                    Cl-2\nO-was 2 now 6                                     O-6