K2Cr2O7(aq) + CH3CH2OH(aq) + H2SO4(aq) = Cr2(SO4)3(aq) + K2SO4(aq) + CH3COOH(aq) + H2O(l)
215740
1) FIND THE REDOX-REACTION\n\nChromate (+VI) is reduced to 2 Cr(III) meaning this that 6 electrons are needed.\n\nThe corresponding oxidation on the alpha-C-atom in Ethanol goes from oxidation-level (-I) to oxidation-level (+III) in acetic acid, meaning it delivers 4 electrons per molecule.\n\nSo we find a stochiometric electron equilibrium if 12 electrons are exchanged:\n\n2 Cr2O7(2-) + 3 CH3CH2OH  =  4 Cr(3+) + 3 CH3COOH\n\n2) THE PROCESS MUST BE ELECTRICALLY NEUTRAL\n\nWe add the K-Ions having the left side of the equation neutral and the right side with 16 loads positive:\n\n2 K2Cr2O7 + 3 CH3CH2OH  = 4 Cr(3+) + 4 K(+) + 3 CH3COOH\n\nIn order to have the left side also with 16 loads positive, we add 16 protons, meaning this, that the reaction needs acid. Since we use sulfuric acid we also add 8 sulfat-ions on both sides and hereby can neutralize the product side:\n\n2 K2Cr2O7 + 8 H2SO4 + 3 CH3CH2OH  =\n                         2 Cr2(SO4)3 + 2 K2SO4 + 3 CH3COOH\n\n3) STOCHIOMETRIC BALANCE OF THE ELEMENTS H and O\n\nTaking into consideration that unused protons finally end up as water, we add on the right side 11 H2O and hereby is also reached the balance of oxigen:\n\n2 K2Cr2O7 + 8 H2SO4 + 3 CH3CH2OH  =\n                  2 Cr2(SO4)3 + 2 K2SO4 + 3 CH3COOH + 11 H2O\n\nSo now good luck for counting the atoms!
426846
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
378281
2K + I2 -> 2KI\n\nBa + S -> BaS\n\nMgBr2 + MnI6 -> No Reaction\n\nC7H16 + H2O -> No Reaction\n\n2KI + Br2 -> 2KBr + I2
296298
H+Cl = HCL - Hydro Cholric Acid\nK+Br = KBr - Pottasium Bromide\nSr+S = Sb2S3 - Stibnite\nFr+O = FrO - Francium Oxide\nCa+Co = CaCo3 - Calcium Carbonate\nH+OH = Not sure - what is OH - H+O+H - H2O - Water
108468
4 OH-(aq) --->  O2(g) + 2H2O(l) + 4e-
394818
CoCl2(aq) + FeSO4(aq) = CoSO4 (aq) + FeCl2(aq)
105551
4Fe3+ + 9O2 ->  6Fe2O3
402270
a)4 Fe + 3O2--->2 Fe2O3\n\nCH4 + 2O2 ---> CO2 +2 H2O
219453
AgBr-------->Ag+ + Br-\n(a) Ksp=[Ag][Br] solubility=s=[Ag]=[Br]=Ksp^0.5\ns=(6*10^-7)^0.5\n(b) [Br]=0.01 (because all of kBr was dissuciated)\ns=Ksp/[Br]=(6*10^-7)/0.01 mol/lit
440498
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)
448805
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
402503
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
512544
2C18H18 + 45 O2 --->  36 CO2 + 18 H2O\n3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O
484918
44 A) 2AgO --> 4Ag + O2\n     B)NiCO3 --> NiO + CO2\n     C)NH4NO3 --> N2O + 2H2O\n48)A)2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n     B)3Ag2SO4 + 2AlCl3--> 6AgCl + Al2(SO4)3\n     C)&D) I've no idea, sry\n50)A)C4H8 + 6O2 --> 4CO2 + 4H2O\n     B)2C8H18 + 25O2 --> 16CO2 + 18H2O\n     C)2C3H8O3 + 7O2 --> 6CO2 + 8H2O\n     D)2C3H6O + 5O2 --> 6CO2 + 6H2O\n52)A)2Al + 3H2SO4 --> Al2(SO4)3 + 3H2\n     B)2HCl + Ba(OH)2--> BaCl2 + 2H2O\n     C)Au + HCl--> No Reaction(gold is below hydrogen in the metal reactivity series)\n54)A)2Cl2 + 2NaI --> 2NaCl2 + I2\n     B)2NH3--> N2 + 3H2\n     C)4Na + O2--> 2Na2O\n     D)Mg + H2SO4--> MgSO4 + H2\n     E)MgCl2 + Ca(OH)2--> Mg(OH)2 + CaCl2\n     F)H2 + Cl2--> 2HCl
347740
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
369339
4Cr + 3O2 --> 2Cr2O3\nChromium(III) oxide or Chromium oxide.
466328
6HCl +3 I2 + KClO3 --->6 ICl+ KCl + 3H2O\n\n\nMnO4 - +2 NO2- +4 H+ ---> Mn2+2 NO3- +2 H2O