Calculate the theoretical yield and percent yield?

Question

The following chemical reaction was carried out in a hood in a university laboratory. Calculate the theoretical yield of calcium carbonate, identify the limiting reactant, and given an actual yield of 6.35g, calculate the percent yield.

The chemical reaction is:

Ca(OH)2 (aq) + CO2 (g) -----> CaCO3 (s) + H2O (l)

  (6.50g)           (6.00g)              (?g)

Answer 1

moles of Ca(OH)₂: 6.50 g x 1 mole/74.092 g = 0.0877 moles

 

moles of CO₂: 6.00 g x 1 mole/44.009g = 0.136 moles

 

Since the reaction gives a ratio of 1:1 for Ca(OH)₂ interacting with CO₂, the limiting reagent would be Ca(OH)₂ because it is the lesser amount.

 

 

The theoretical yield would use this limiting amount of Ca(OH)₂:

 

0.0877 mole Ca(OH)₂ x 1 mole CaCO₃/1 mole Ca(OH)₂ x 100.086 g/1 mole CaCO₃ = 8.78 g

 

The actual yield of 6.36 g will produce a percent yield of 6.35 / 8.78 x 100% =  72.3%