Answer:
Theoretical yield =26.03 g
Percent yield = 87%
Limiting reactant = CaO
Explanation:
Given data:
Mass of CaO = 14.4 g
Mass of CO₂ = 13.8 g
Actual yield of CaCO₃ = 22.6 g
Theoretical yield = ?
Percent yield = ?
Limiting reactant = ?
Solution:
Chemical equation:
CaO + CO₂ → CaCO₃
Number of moles of CaO:
Number of moles = Mass /molar mass
Number of moles = 14.4 g / 56.1 g/mol
Number of moles = 0.26 mol
Number of moles of CO₂:
Number of moles = Mass /molar mass
Number of moles = 13.8 g / 44 g/mol
Number of moles = 0.31 mol
Now we will compare the moles of CO₂ and CaO with CaCO₃ .
CO₂ : CaCO₃
1 : 1
0.31 : 0.31
CaO : CaCO₃
1 : 1
0.26 : 0.26
The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.
Mass of CaCO₃: Theoretical yield
Mass of CaCO₃ = moles × molar mass
Mass of CaCO₃ =0.26 mol × 100.1 g/mol
Mass of CaCO₃ = 26.03 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 22.6 g/ 26.03 g × 100
Percent yield = 0.87× 100
Percent yield = 87%
Limiting reactant:
The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.
