Calculating Moles of Base

Question

For some reason I am really struggling with a titration lab to figure out the moles of base in

            a) 675 mg CaCO₃

            b) 135 mg Mg(OH)₂

This is the actual amount of pertinent ingredients in the antacid tablet I back-titrated.

I back-titrated this antacid and calculated the experimental value to be 4.65 mmol base based on the equivalence volume found on the titration curve. I am struggling to find the theoretical yield to calculate percent error.

I know that when both are titrated with HCl, there are 2 moles HCl neutralized for every 1 mole CaCO₃ and for every 1 mole Mg(OH)₂ based on the balanced equations.

From (a) and (b) I did stoich. to find 0.0067 mole CaCO₃ and .0023 moleMg(OH)₂.

While doing the lab, I added 25mL of 0.1 M HCl to the antacid before titration. I used this information to calculate 0.0025 Moles HCl added.

It seems too easy to do this from stoich. alone so I need guidance on where to go next or if I have really overcomplicated this for myself. Thank you!

Answer 1

Adding 0.0025 moles HCl wouldn’t be enough to neutralize the CaCO3 + Mg(OH)2 and also have excess to be back titrated. Maybe you could add more information so that we might be better able to assist you. Why are you looking for theoretical yield when you haven’t performed a synthesis? You don’t get a theoretical yield when you do a titration. I’m not sure what you’re trying to do. Sorry.