A student mixes 20mL of 0.19M Ca(NO3)2 and 12 mL of 0.15M Na2CO3. If the student's dry precipitate weighs 0.169 g, what is the percent yield?

Answer 1

First, write the correctly balanced equation for the reaction taking place:

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

Next, since the amounts of BOTH reactants are given, we must determine which reactant is present in limiting supply. One easy way to do this is to simply divide the moles of each by the corresponding coefficient in the balanced equation, and whichever value is less represents the limiting reactant.

For Ca(NO₃)₂: 20 ml x 1 L /1000 ml x 0.19 mol / L = 0.0038 mols (÷1->0.0038)

For Na₂CO₃: 12 ml x 1 L / 1000 ml x 0.15 mol / L = 0.0018 mols (÷1->0.0018)

Since 0.0018 is less than 0.0038, Ca(NO₃)₂ is the limiting reactant and will determine yield of product.

Final step is to determine the mass of precipitate, CaCO₃, and then the % yield:

theoretical yield of CaCO₃ = 0.0018 mols Ca(NO₃)₂ x 1 mol CaCO₃ / mol Ca(NO₃)₂ = 0.0018 mol CaCO₃

0.0018 mols CaCO₃ x 100 g / mol = 0.18 g CaCO₃

% yield = actual / theoretical (x100%) = 0.169 g / 0.180 g (x100%) = 94% yield (2 sig.figs.)