What mass of calcium carbonate is precipitated out when 2.20 g of ammonium carbonate is added to 25.7 mL of a 4.00 M solution of calcium nitrate?

Answer 1

This is a limiting reagent problem; I know that because amounts for BOTH reactants are given.
(NH4)2CO3 + Ca(NO3)2 ==> CaCO3 + 2NH4NO3

moles (NH4)2CO3 = 2.20g/molar mass (NH4)2CO3 = approximately 0.02 mole.

moles Ca(NO3)2 = M x L = approx 0.1 mol.

Now convert each, separately, to moles CaCO3.
moles CaCO3 = moles (NH4)2CO3 x [1 mole CaCO3/1 mole NH4)2CO3] = about 0.02 x 1/1 = about 0.02 mole CaCO3.

moles CaCO3 = moles Ca(NO3)2 x [1 mole CaCO3/1 mole Ca(NO3)2] = about 0.1 x (1/1) = about 0.1 mol CaCO3.

Both answers can't be right; the correct value in limiting reagent problems is ALWAYS the smaller one and the reagent producing that number is the limiting reagent.
Now convert to grams by g = moles x molar mass.