Answer: 85.27%
Explanation:
CaCO3 (s) --> CaO (s) + CO2 (g)
From the equation, 1 mole of CaCO3 will give 1 mole of CaO and 1 mole of CO2
Number of moles of CaCO3 reacted = 44.53g/100.09g/mol = 0.445 mol
Therefore;
CO2 produced = 0.445 moles (theoretical yield)
Mass of CO2 formed = 0.445mol * 44.01g/mol = 19.584 grams of CO2
Percentage yield = (experimental yield/theoretical yield) x 100 = (16.7/19.584) x 100 = 85.27%
