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 %

gas

explanation:

smoke is the matter of a gas