Answer:
Percent yield = 57.8 %
Theoretical yield  = 11.781 g
Explanation:
Given data:
Mass of CaO produced = 6.81 g
Mass of CaCO₃ react = 20.7 g
Theoretical yield = ?
Percent yield = ?
Solution:
Chemical equation:
CaCO₃    →    CaO + CO₂
Number of moles of CaCO₃ :
Number of moles = mass/molar mass
Number of moles = 20.7 g/ 100.1 g/mol
Number of moles = 0.21 mol
Now we will compare the moles of CaCO₃ with CaO.
         CaCO₃     :      CaO
            1        :        1
           0.21     :      0.21
Theoretical yield of CaO: Â
Mass = number of moles × molar mass
Mass = 0.21 mol × 56.1 g/mol
Mass = 11.781 g
Percent yield:
Percent yield = ( actual yield / theoretical yield ) ×  100
Percent yield = (6.81 g/ 11.781 g) × 100
Percent yield = 57.8 %