When 10.51 g CaCO3 reacts with excess hydrochloric acid, as below, 3.29 g of CO2 is produced.

CaCO3 (s) + HCl (aq) → CO2 (g) + CaCl2 (aq) + H2O (l)

What is the percent yield of CO2?

71.2 %

Explanation:

This is a problem we can solve by making use of the stoichiometry of the balanced (very important!) chemical reaction:

CaCO3 (s) + 2HCl (aq) → CO2 (g) + CaCl2 (aq) + H2O (l)

Now determine the moles of CaCO3 and CO2 and perform the calculations

# mol CaCO3 = 10.51 g / 100.09 g/mol = 0.105 mol

From the stoichiometry of the reaction, we know 1 mol CO2 is produced per mol of CaCO3, thus whe should expect

0.105 mol CaCO3 x 1 mol CO2 / mol CaCO3 = 0.105 mol

lets compare this theoretical value with the one obtained:

mol CO2 obtained = 3.29 g / 44 g/mol = 0.075

Therefore our yield is

(0.075mol / 0.105 mol) x 100 = 71.2 %