5. If 0.9g of oxalic acid are used in the reaction, how many grams of calcium oxalate can theoretically be produced? The steps of the analysis for this experiment are in the manual on page 22 and shown below. (Note you are only calculating the theoretical yield before the experiment. You will calculate the % yield after the experiment.) 1.) Oxalate, C2O42, is the limiting reactant in this reaction. Calculate the number of moles of oxalate that you added by dividing the 0.9 grams of oxalic acid by 90g/mol which is the molar mass. (0.5pt) 2.) The mole ratio in the reaction shows that for every one mole of oxalate ion used, one mole of calcium oxalate can be produced. Therefore moles of oxalate ion = moles of calcium oxalate. (0.5pt) 3.) Multiply the moles from step 2 by 128g/mole to calculate the theoretical yield of calcium oxalate in grams. (The molar mass of calcium oxalate is 128g/mol ) (1pt)