Question

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?

Answer 1

71.2 %

Step-by-step 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 %