balanced equation
posted by christine .
? MnO4– + ?H+ + ?C2O42– > ?Mn2+ + ?H2O + ?CO2
what's the balanced equation? correct coefficients?
is it 1MnO4 + 2H+ + 1C2O42 > 1Mn2+ + 1H20 + 2CO2 ?
c2042..? are you missing parenthesis somewhere?
oh i see, no i believe this is correct:
MnO4 + 8h+ + C2042 > Mn2+ + 4H2O + 2CO2
o sorry
_?(MnO4) + _? (H)+ + _?(C2O4)2 >
_?(Mn)2+ + _?(H2O) +_? (CO2)
The two half equations are:
MnO4^ + 8H^+ 5e ==> Mn^+2 + 4H2O
C2O4^= ==> 2CO2 + 2e
multiply equation 1 by 2 and equation 2 by 5 and add. Cancel common species if any.
are you balancing in general or blancing in a basic/acidic solution? if so then i think you would add electrons like DrBob said, what year chemistry are you? cause i didn't learn the electron thing til second year.
The H^+ tell you it is balanced in acid solution. Besides, permanganate is always used to oxidize oxalate ion in an acid solution.
Here another solution if your algebra is upto speed!
? MnO4– + ?H+ + ?(C2O4)2– > ?Mn2+ + ?H2O + ?CO2
MnO4– + xH+ + y(C2O4)2– > Mn2+ + aH2O + bCO2
We have 4 unknowns and can then generate 4 equations.
Charge
x2y1=2
x2y=3(1
H atoms
X=2a(2
O atoms
4+4y=a+2b(3
C atoms
2y=b(4
Substitute 2x(4 in (3
4+2b=a+2b
Hence a=4
From (2
x=8
from (1
82y=3
y=5/2
and from (4
b=5
substitute the values back into the equation.
MnO4– + 8H+ + 5/2(C2O4)2– > Mn2+ + 4H2O + 5CO2
Or
2MnO4– + 16H+ + 5(C2O4)2– > 2Mn2+ + 8H2O + 10CO2
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correct, that is the answer I got 2, 16, 5, 2, 8, 10

16H(+) + 2MnO4() + 5C2O4(2) → 2Mn(+2) + 4H2O + 5CO2

c2o4 has a charge of 2
so it's only + 1e, not 2 
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