1.87 grams Looking at the balanced equation, it indicates that for every mole of rust as a reactant, you need 6 moles of oxalic acid. So let's see how many moles of oxalic acid we have by multiplying the molarity of the solution by the quantity. 7.00x10^2 ml / 1000 ml/l * 0.100 M = 7.00x10^-2 mol So the number of moles of rules that can be converted is 7.00x10^-2 mol / 6 = 1.17x10^-2 mol Now determine the molar mass of Fe2O3. Atomic weight iron = 55.845 Atomic weight oxygen = 15.999 Molar mass Fe2O3 = 2 * 55.845 + 3 * 15.999 = 159.687 g/mol Mass Fe2O3 = 159.687 g/mol * 1.17x10^-2 mol = 1.87 g So the amount of rust that can be removed is 1.87 grams.