How many mL of 0.122 M Na2CO3 would be needed to precipitate all of the copper ions in 23.7 mL of 0.167 M CuSO4? How many grams of CuCO3 could be recovered?

Answer 1

CO3^2- + Cu^2+ ==> CuCO3
millimols CuSO4 = mL x M = 23.7*0.167 = approx 4 but that's an estimate so you need to redo all of these calculations.
So you will need approx 4 mmols NaCO3
and M Na2CO3 = mmols/mL or
mL = mmols/M = approx 4/0.122 = ? mL.

g CuCO3 = mols CuCO3 x molar mass CuCO3 or g = approx 0.004 mols x molar mass CuCO3