Balancing chemical reactions... How do you do it step by step?
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
744922
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
957775
Ca(OH)2 + CO2 --> C(OH) + CaO\n\nIs unbalanced..
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!!
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.
979796
2H3PO4+6NaOH-------->6H2O+2Na3PO4\n\nThere has to be two moles of phosphoric acid and six moles of sodium hydroxide for it to balance out
105737
Exchange electrons and protons to get balanced ? It's just my idea, I'm not sure at all though
824141
Start off with the original equation\n\nCaCO3(s) + HCl(aq) --> CaCl2(aq) + CO2(g) + H2O\n\n\nStart off by making the Ca equal on both sides, which is fortunately already done\n\nThen make the Cl equal on both sides by putting a 2 in front of HCL\nCaCO3(s) + 2HCl --> CaCl2(aq) + CO2(g) + H2O\n\nThen make the C's on both sides equal, which again, is already done\n\nNext make the O's equal on both sides, which yet again is already done.\n\nFinally you make the H's equal on both sides, which is done also. \n\nSo the answer to this whole question is to put a 2 in front of the HCl. (I'm guessing you meant to put CaCl2 instead of CaC12 in the products)\n\nThat was easy.
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.
367815
Try finding the standard reduction potential for Cu and NO first then use those poteintial rxn equ to form the above full reaction afterwhich simply add or substract the poteinital u find to get E for the above reaction. \n\nSince E is proportional to the rate constate, u can use the above concentration as well as partial pressure to calculate the rate constant and hence E. Taking care that rate constant is the concentration/partial pressure of the reactant over the product!
301956
The answer is B.\n\nIn your beginning reaction CO2 + C --> CO, you have 2 carbons and 2 oxygens on the reactants side, and you have 1 carbon and 1 oxygen on the products side.\n\nIn order to make the equation balanced, you have to make sure there are the same number of each element on both sides of the equation.\n\nSo to make this chemical equation balanced all you have to do is put 2 as the coefficient of the CO on the products side, making the final equation:  CO2 + C --> 2CO\nThis will give you 2 carbons and 2 oxygens on both sides.
116069
The balance is :\n\n     H2O2(aq) -> O2(g) + 2H+(ag) + 2e-\n MnO2(s) + 4H+(aq)  +  2e-  ->  Mn2+(aq)  + 2H2O\n \n Combine the two half-reactions and cancel on both sides yield the     complete balance equation.\n\n MnO2(s) + H2O2(aq) + 2H+(aq) -> Mn2+(aq) + O2(g) + 2H2O\n\n this is an oxidation-reduction reaction\ntype  decomposition reaction.
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
523189
3NH4ClO4 + 3Al -> Al2O3 + AlCl3 + 6H2O + 3NO\n\nHow's that ?????????\n\nThe trick is to concentrate on element like Oxygen first...........
83087
It depends.  Are you looking at reactions in series (where the products of the first become the reactants of the second), or reactions in parallel (where the products of each are all different from the reactants of each)?\n\nJust take all the reactants of all the equations and put them on the LHS (left-hand side) of the reaction, and all the products of all the equations and put them on the RHS of the equation.  If the reactions are in parallel, then there should be no species on both sides of the equation, so you're done.  If the reactions are in series, then you want to try to eliminate terms on both sides.  Usually, this means you have to pick a single thing to eliminate.\n\nLet's take the following theoretical reaction:\n\nA + 2B --> 2C + 2D\n\n3C + 4D --> 3F + G\n\nIn this case, you can eliminate either C or D, BUT NOT BOTH.  So you pick which one you want.  If you want to get rid of C, you look at the coefficients in front of C (2 and 3) and then identify the least common multiple (LCM) of those two numbers (6).  Now you want to multiply both sides of each equation by whatever number will make the coefficient 6 for C.  That means you mulitiply the first equation by 3 and the second by 2:\n\nA + 2B --> 2C + 2D\nx3\n____________________\n3A + 6B --> 6C + 6D\n\n3C + 4D --> 3F + G\nx2\n___________________\n6C + 8D --> 6F + 2G\n\nAdd these two resultant equations together as before:\n\n3A + 6B + 6C + 8D --> 6C + 6D + 6F + 2G\n\nBecause 6C is on both sides, we cancel it from the equation.\n\n3A + 6B + 8D --> 6D + 6F + 2G\n\nSince D is on both sides, we can subtract the lesser number from each side, but we can't eliminate it entirely from both sides:\n\n3A + 6B + 2D --> 6F + 2G\n\nAnd that's you're final answer.\n\nIf we'd wanted to eliminate D instead of C, we'd find the LCM of the coeffiecients of D (2 and 4, so the LCM is 4) and then multiply each equation by whatever is needed to make both coefficients 4.  Since the second equation already has 4D, we leave it alone and instead multiply the first equation by 2:\n\nA + 2B --> 2C + 2D\nx2\n__________________\n2A + 4B --> 4C + 4D\n\n3C + 4D --> 3F + G\n\nNow add those two and get:\n\n2A + 4B + 3C + 4D --> 4C + 4D + 3F + G\n\nCancel the 4D from both sides:\n\n2A + 4B + 3C --> 4C + 3F + G\n\nSubtract 3C from each side, and your final equation is\n\n2A + 4B --> C + 3F + G\n\nEither of these equations technically correct.  If you have only one species that is on both sides of parallel reactions, you don't have to make a decision.  In this case, if you're looking at series reactions that use a mixture of A and B (without and C or D in the initial mixture), we'd probably choose the second approach, which gives you a final equation that has only A and B on the RHS of the equation.
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
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.