Asample of 0.53 g of carbon dioxide was obtained by heating 1.31 g of calcium carbonate. what is the percent yield for this reaction? caco3(s) ⟶ cao(s) + co2(s)

CaCO3(s) ⟶ CaO(s)+CO2(s) 

moles CaCO3: 1.31 g/100 g/mole CaCO3= 0.0131 

From stoichiometry, 1 mole of CO2 is formed per 1 mole CaCO3, therefore 0.0131 moles CO2 should also be formed. 
0.0131 moles CO2 x 44 g/mole CO2 = 0.576 g CO2 

Therefore:
% Yield: 0.53/.576 x100= 92 percent yield

Answer : The percent yield for this reaction is, 91.9 %

Solution : Given,

Mass of carbon dioxide = 0.53 g

Mass of calcium carbonate = 1.31 g

Molar mass of carbon dioxide = 44 g/mole

Molar mass of calcium carbonate = 100 g/mole

First we have to calculate the moles of .

Now we have to calculate the mass of carbon dioxide.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 1 mole of react to give 1 mole of

So, 0.0131 mole of react to give 0.0131 mole of

Now we have to calculate the mass of carbon dioxide.

Now we have to calculate the percent yield for this reaction.

Therefore, the percent yield for this reaction is, 91.9 %