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14 years ago
guywithalotoflifeexperience

Balance the following equation. For a balanced eq. aA + bB => cC + dD enter your answer as an integer : abcd. C2H4 (g) + MnO4- (aq) + H+ (aq) --> CO2 (g) + Mn2+ (aq) + H2O (l)
14 years ago
ChemistryMom
first assign oxidation numbers and write the half reactions C in C2H4 is (-2), C in CO2 is +4 C(-2) --->C(+4) + 6e- 2C(-2) ---> 2C(+4) + 12 e- Mn(+7) + 5e- ---> Mn(+2) balance electrons 10C(-2) ---> 10C(+4) + 60e- 12Mn(+7) + 60e- ---> 12 Mn(+2) 5C2H4 ---> 10CO2 + 60e- 12MnO4(-) + 60e- ---> 12Mn(+2) add equations 5C2H4 + 12MnO4(-) ---> 10CO2 + 12Mn(+2) the left side has a charge of -12, the right side of +24 - balance charge using H+ 5C2H4 + 12MnO4(-) + 36H+ ---> 10CO2 + 12Mn(+2) balance O and H using H2O 5C2H4 + 12MnO4(-) + 36H(+) ---> 10CO2 + 12Mn(+2) + 28H2O double check by counting each atom to make sure there are the same number on both sides
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14 years ago
Anonymous
Break the equation off into 2 half equations and balance the equation... ie. C2H4 ---> 2CO2 + 4e- + 4H+ and MnO4- + 5e- +8H+ ---> Mn2+ +4H2O then times the first equation by 5 and the second equation by 4, to cross out the e- ie 5C2H4 ----> 10CO2 + 20e- + 20H+ 4MnO4- +20e- + 42H+ ---> 4Mn2+ + 16H2O then put the 2 equations together and cross off the electrons and hydrogen ions 5C2H4 + 4 MnO4- + 22H+ ---> 10CO2 + 4Mn2+ 16H2O
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14 years ago
ok_ok361
C2H4 (g) + 2 MnO4- (aq) + 4 H+ (aq) --> 2 CO2 (g) + 2 Mn2+ (aq) + 4 H2O (l)
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