Fe2O3 + 6H2C2O4 -----> 2Fe(C204)3^3- + 3H20 +6H What maaa of rust can be removed by 540 g of oxalic acid (H2C2O4)?

1 Answer
Al E.
Apr 29, 2018

Consider the reaction when you bathe ferric oxide in oxalic acid,

#Fe_2O_3 + 6H_2C_2O_4 to 2Fe(C_2O_4)_3^(3-) +3H_2O + 6H^(+)#

The "rust" is the ferric oxide on the left side of your equation.

An easy approximation to make for this relatively trivial question is that the given molar quantity of oxalic acid will "remove" a stoichiometrically equivalent molar quantity of ferric oxide.

Hence,

#540"g" * (H_2C_2O_4)/(90"g") * (Fe_2O_3)/(6H_2C_2O_4) * (159.6"g")/(Fe_2O_3) approx 159.6"g"#

of rust can be removed by the given mass of oxalic acid.

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