Answer:
Explanation:
Given data:
Mass of calcium oxide = 14.4 g
Mass of carbon dioxide = 13.8 g
Actual yield of calcium carbonate = 19.4 g
Mass of calcium carbonate produced = ?
Limiting reactant = ?
Percent yield = ?
Chemical equation:
CaO + CO₂ → CaCO₃
Number of moles of CaO:
Number of moles of CaO = Mass /molar mass
Number of moles of CaO = 14.4 g / 56.1g/mol
Number of moles of CaO = 0.26 mol
Number of moles of CO₂:
Number of moles of CO₂= Mass /molar mass
Number of moles of CO₂ = 13.8 g / 44 g/mol
Number of moles of CO₂ = 0.31 mol
Now we will compare the moles of CaCO₃ with CO₂ and CaO.
CaO : CaCO₃
1 : 1
0.26 : 0.26
CO₂ : CaCO₃
1 : 1
0.31 : 0.31
The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.
Limiting reactant:
CaO
Theoretical yield:
Mass of CaCO₃ = moles × molar mass
Mass of CaCO₃ = 0.26 mol × 100 g/mol
Mass of CaCO₃ = 26 g
Percent yield:
Percent yield = Actual yield / theoretical yield × 100
Percent yield = 19.4 g/ 26 g× 100
Percent yield = 74.6 %
