Solution:
Step 1: Calculate the molar mass of CaCO₃ and CaO.
For CaCO₃
molar mass = (40.08 g/mol × 1) + (12.01 g/mol × 1) + (16.00 g/mol × 3)
molar mass = 100.09 g/mol
For CaO
molar mass = (40.08 g/mol × 1) + (16.00 g/mol × 1)
molar mass = 56.08 g/mol
Step 2: Determine the mole ratio needed.
mole ratio = 1 mol CaCO₃ : 1 mol CaO
Step 3: Calculate the theoretical yield of the reaction (mass of CaO produced).
[tex]\text{theoretical yield = 20.7 g CaCO₃} × \frac{\text{1 mol CaCO₃}}{\text{100.09 g CaCO₃}} × \frac{\text{1 mol CaO}}{\text{1 mol CaCO₃}} × \frac{\text{56.08 g CaO}}{\text{1 mol CaO}}[/tex]
[tex]\text{theoretical yield = 11.6 g}[/tex]
Step 4: Calculate the percent yield of the reaction.
[tex]\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} × 100[/tex]
[tex]\text{percent yield} = \frac{\text{6.81 g}}{\text{11.6 g}} × 100[/tex]
[tex]\boxed{\text{percent yield = 58.7\%}}[/tex]
#CarryOnLearning
