How do you balance this equation: $F e_{2} {(C_{2} O_{4})}_{3} \to F e C_{2} O_{4} + C O_{2}$ $Fe_2(C_2O_4)_3 -> FeC_2O_4 + CO_2$ ?

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Answer 1 · 2017-03-16T15:13:39

This is a formal redox reaction...........

$2 F e^{3 +} + 3 C_{2} O_{4}^{2 -} \to 2 F e (C_{2} O_{4}) + 2 C O_{2} (g)$ $2Fe^(3+) +3C_2O_4^(2-)rarr2Fe(C_2O_4) +2CO_2(g)$

So we approach the equation by the method of half equations:

$F e (I I I)$ $Fe(III)$ is reduced to $F e (I I)$ $Fe(II)$ :

$F e^{3 +} + e^{-} \to F e^{2 +}$ $Fe^(3+) + e^(-)rarrFe^(2+)$ $(i)$ $(i)$

And $oxalate ion$ $"oxalate ion"$ is oxidized to $C O_{2}$ $CO_2$ :

$C_{2} O_{4}^{2 -} \to 2 C O_{2} + 2 e^{-}$ $C_2O_4^(2-) rarr 2CO_2 + 2e^-$ $(i i)$ $(ii)$

Both half equations conserve mass and conserve charge, as they must if they are to reflect chemical reality.

And we add the individual half reactions together so as to remove the electrons: i.e. $2 \times (i) + (i i) :$ $2xx(i) + (ii):$

$2Fe^(3+) + cancel(2e^(-)) +C_2O_4^(2-)rarr2Fe^(2+) +2CO_2(g)+ cancel(2e^-)$

To give (almost finally):

$2Fe^(3+) +C_2O_4^(2-)rarr2Fe^(2+) +2CO_2(g)$

And we can add TWO $"oxalato ligands"$ to EACH side:

$2Fe^(3+) +3C_2O_4^(2-)rarr2Fe(C_2O_4) +2CO_2(g)$

Are mass and charge balanced? If they don't, what must you do?

How do you balance this equation: Fe_2(C_2O_4)_3 -> FeC_2O_4 + CO_2Fe2(C2O4)3→FeC2O4+CO2Fe_2(C_2O_4)_3 -> FeC_2O_4 + CO_2?