Use stoichiometry and the balanced equation to determine the mass of rust (Fe2O3) that will react with 25 mL of 0.1 M oxalic acid (H3C2O4)
It can be seen in the balanced equation that 1 mole of Fe2O3 reacts with 6 moles of H3C2O4
Recall that molarity can be calculated as:
One of the uses of oxalic acid (H2C2O4) is rust removal. It reacts with rust (Fe2O3) according to the equation
Fe2O3(aq) + 6 H3C2O4(aq) → 2 H3Fe2(C2O4)3(aq) + 3 H2O(l)
Calculate the number of grams of rust (159.7 g/mol) that can be removed by 25.0 mL of a 0.100 M solution of oxalic acid.
(A) 0.0665 grams
(B) 2.66 grams
(C) 0.200 grams
(D) 0.400 grams
(E) 0.00250 grams
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