what is the balanced equation for Cr2(SO4)3+Cr(OH)3+K2SO4?
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!
775066
You have the products, but now you have to balance the equation. The unbalanced equation is:\n\nLi2CO3 + Cr2(SO4)3 --> Li2SO4 + Cr2(CO3)3\n\nBalanced:\n\n3 Li2CO3 + Cr2(SO4)3 --> 3 Li2SO4 + Cr2(CO3)3
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
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.
918374
Fe2O3 -- Fe+O2\n\nthis is the balanced form\n\n2(Fe2O3)--- 4Fe+3O2\nwithout the parentheses
906651
2KOH=>K(2)O+H(2)O\n\nFeO+C=>CO+Fe\n\nCa+2AlCl=>CaCl(2)+2Al\n\nP(2)O(5)+3H(2)O=>2H(3)PO(4)\n\n2KI+Pb(NO3)2=>2KNO(3)+PbI(2)
764101
In all reactions one uses that 2 OH -> H2O + O, the split of oxygen goes to create the ion. Finally balance the equation:\n \nSO3 + 2 NaOH -> Na2SO4 + H2O\n\nCO2 + 2 KOH -> K2CO3 + H2O \n\nSO2 + Ba(OH)2 -> BaSO3 + H2O
606075
Cu(s) + 4HNO3 (aq)--> Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)
849074
K2S + COCl2 --> 2KCl + COS
814611
CdSo4+NH4OH ---->Cd(OH)2 + (NH4)2SO4
832296
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
927706
Al2(SO4)3 ==> Al2O3 + 3SO3
744922
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
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
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
729557
_NaOH(aq) + _Fe(ClO4)3(aq) => _Fe(OH)3(s) + _NaClO4(aq)\n\n3  NaOH(aq) + 1  Fe(ClO4)3(aq) => 1  Fe(OH)3(s) + 3  NaClO4(aq)\nThe coefficients of one could be excluded but I put one which is also equivalent.\n\n_((NH4)2)SO4(aq) + _Pb(NO3)2(aq) => _PbSO4(s) + _NH4NO3(aq)\n\n1  ((NH4)2)SO4(aq) + 1  Pb(NO3)2(aq) => 1  PbSO4(s) + 2  NH4NO3(aq)\n\nThese are only the balanced equations.\n\nThe following is the ionic equations:\n(Na)+ (aq) + (OH)- (aq) + (Fe)3+ (aq) + (ClO4)- (aq) => Fe(OH)3(s) + (Na)+ (aq) + (ClO4)- (aq)\nand\n(NH4)+ (aq) + (SO4)2- (aq) + (Pb)2+ (aq) + (NO3)- (aq) => PbSO4(s) + (NH4)+ (aq) + (NO3)- (aq)\n\nThe ionic equations are already balanced.  It is simply the breaking down of the equations into ions.  The solid remains intact while the compounds in aqueous solution become ions.\n\nThe + and - signs are the charge of the ions.  I can't find any other way to write them so I separated the compound with parentheses.  Sorry if it's confusing.\n\nThe net ionic equations are "simplified" ionic equations.  Basically, the things that are repeated on both sides get crossed off.  What is remaining is simply the net ionic equation.\n\n(OH)- (aq) + (Fe)3+ (aq) => Fe(OH)3(s)\nand\n(SO4)2- (aq) + (Pb)2+ (aq) => PbSO4(s)
412530
1) H3COCH3 (or C2H6O) + 3O2 (g) --> 2CO2 (g) + 3H2O (g)\n2) HNO2 (aq) + NaOH (aq) <--> NaNO2 (aq) + H2O (l)\n3) 2K+ + Cr2O7(2-) + Fe(NO3)2 + 2H+ <--> CrO4(-) + Fe3+ + 2K+ + 2NO3(-) \n4) Na3PO4 + NH4NO3 = No reaction, all soluble salts.\n5) 2HCl (aq) + Na2S (s) --> 2NaCl (aq) + H2S (g)