How can we add two chemical equations?
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.
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
424592
H2SO4 + 2NaOH ----> 2H2O + Na2SO4\n\nYOU MUST SEPARATE AND CANCEL THE COMMON IONS (AQUEOUS ONES)
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
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
617758
Actually, all you need is:\n\n2 H2O2 ======> 2 H2O + O2\n\nif you have water on both sides of the equation, they "cancel" out.  Just like in algebra:\n\na + b = a + c  is the same as b = c
849074
K2S + COCl2 --> 2KCl + COS
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=)
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...
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...
795154
2CO(g) + O2(g) --> 2CO2(g)
660043
1.) do your own homework\n2.) this is the computer forum, not high-school chemistry.  Next time post to the right furom.
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2
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.
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.
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
707973
You have to start with a balanced equation.  The subscripts will give you the correct charge and a prefix will give you the right amount of atoms on each side of the equation.  Some elements, like O, F, and I are diatomic and come in pairs.