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 %
