A student mixes 20mL of 0.19M Ca(NO3)2 and 12 mL of 0.15M Na2CO3. What is the theoretical yield of CaCO3?

Answer 1

Ca(NO₃)₂ + Na₂CO₃ ==> CaCO₃ + 2NaNO₃ .. balanced equation

moles Ca(NO₃)₂ present = 20 ml x 1 L / 1000 ml x 0.19 mol / L = 0.0038 moles

moles Na₂CO₃ present = 12 ml x 1 L / 1000 ml x 0.15 mol / L = 0.0018 moles

Since the mole ratio in the balance equation is 1:1, the Na₂CO₃ will be present in limiting supply and will run out, leaving excess Ca(NO₃)₂ left over. Thus, the amount of product is determined by the amount (moles) of Na₂CO₃.

Theoretical yield of CaCO₃ = 0.0018 mols Na₂CO₃ x 1 mol CaCO₃ / 1 mol Na₂CO₃ = 0.0018 mols CaCO₃

If you want this in grams, just multiply 0.0018 mols CaCO₃ by the molar mass of CaCO₃