In the reaction shown above, 2.00 x 10^3 g caco3 produce 1.05 x 10^3 g of cao, what is the percent yield?

First, we have to get the theoretical yield of CaO:

the balanced equation for the reaction is:

CaCO3 (s) →CaO (s) + CO2 (g)

covert mass to moles:

moles CaCO3 = mass of CaCO3 / molar mass of CaCO3

= 2x10^3 / 100 = 20 moles

the molar ratio between CaCO3 : CaO = 1:1

∴moles of CaO = 1 * 20 = 20 moles

∴mass of CaO = moles of CaO * molar mass of CaO

= 20 * 56 = 1120 g

∴the theoritical yield = 1120 g and we have the actual yield = 1.05X10^3

∴Percent yield = actual yield / theoritical yield * 100

= (1.05x10^3) / 1120 * 100

= 94 %