How can I balance LiCO3+CrSO4--->Li2(SO4)+Cr2(CO3)3?
814313
First you need to check if your formulas are correct.  Lithium carbonate should be Li2CO3 and chromium (II) carbonate should be CrCO3.  With the correct formulas, the chemical equation is already balanced.
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.
799504
FeS + 2HBr --> FeBr2 + H2S\nFe2+ + S2- + 2H+ + 2Br- --> Fe2+ + 2Br- + H2S\n2H+ + S2- --> H2S\n\nK2CO3 + CuCl2 --> 2KCl + CuCO3\n2K+ + CO32- + Cu2+ + 2Cl- --> 2K+ + 2Cl- + CuCO3\nCu2+ + CO32- --> CuCO3\n\npeace\nvixklen
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
598514
Anything, at this point... once you add something reactive to the cobalt, the reaction will turn reactants (cobalt and ____) into products, shifting the equillibrium.\n\nCobalt two has a positive-two charge.  You should add any ion with a negative-two charge: carbonate (CO3), chromate (CrO4), dichromate (Cr2O7), an oxide (O), sulfate (SO4), sulfide (S), or sulfite (SO3).
862596
K2SO4\nCa(NO3)2\n\npotassium has a +1 charge and sulfate has a -2 so therefore you need 2 potassiums so that you can have a net zero charge.\ncalcium has a +2 charge and nitrate has a -1 charge so you need two nitrates to balance the charges
205870
CaCO3(s) + 2HC2H3O2(aq)-> Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
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 ^_^
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.
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...
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!!
170129
CuSO4 (s)  --H2O-->  Cu+2 (aq) + SO4-2 (aq)
394069
3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n                                        +11H2O\n\nFirst you must calculate the (average)oxidation number of central atom of main compounds in reactants.\nEx;Cr oxidation no=+6 in K2Cr2O7\n   C  oxidation no=-2 in C2H5OH\nThen in the products.\nEx;Cr oxidation no=+3 in Cr2(SO4)3\n   C  oxidation no=0 in CH3COOH\n\nThen you must evalute how many electrons must be transferred.\nEx;Oxidation no of Cr is changing from +6 to +3 \n   So it is reduced.\n   SO each atom of Cr must receive 3 electrons.\n   So totally 6 electrons must be received by K2Cr2O7\n   \n   Oxidation no of C is changing from -2 to 0\n   SO each atom of C must donate 2 electrons.\n   So totally 4 electrons must be donated by C2H5OH\n\nThen you must write now a explantion reaction.\nEx;you must write oppsite (simple)propotion of electrons in front of each compound\n  3C2H5OH + 2K2Cr2O7\n \nThen you can calculate how many Cr,C(,K) atoms should be there\n So, 3C2H5OH + 2K2Cr2O7=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n\nThen you can calculate the medium(SO42-) should be there\n i think 8 SO4 should be there\n\nSo,3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n\nFinally you must equilize H,O atoms(balanced by water)\nSo,3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n                                        +11H2O\n\nSecond one ................\nC2H5OH+4I2+6NaOH=CHI3+HCOONa\n                  +5NaI+5H2O
589750
SO4 2-\nCO3 2-\nOH -\nPO4 3-
767683
1. First, balance the equation:\n\n2LiOH + CO2 → Li2CO3 + H2O \n\n- here, you'll see that for every 2 moles of LiOH, one mole of  Li2CO3 is produced.\n\n2. If there are 5.36 moles of LiOH, divide this by 2 to get the number of moles of Li2CO3. That is 2.68 moles of Li2CO3.
24810
Cu + 2H2SO4 --> 2H2O + SO2 + CuSO4\nDouble the H2SO4 and the H2O.
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