The percent yield : 4. 84.58%
Reaction
CaCO₃ ⇄ CaO+CO₂
mass CaCO₃ = 2.3 × 10³ g
mol CaCO₃ (MW=100.0869 g/mol) :
[tex]\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{2.3\times 10^3}{100,0869}\\\\mol=22.98[/tex]
From the equation, mol CaCO₃ : CaO = 1 : 1, so mol CaO=22.98
mass CaO(MW=56.0774 g/mol)⇒ (theoretical) :
[tex]\tt mass=mol\times MW\\\\mass=22.98\times 56,0774\\\\mass=1288.659~g[/tex]
The percent yield :
[tex]\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{1090}{1288.659}\times 100\%\\\\\5yield=84.58\%[/tex]
