Answer:
• Molecular mass of Iron (III) tetraoxide
[ molar masses: Fe → 56, O → 16 ]
The oxidation of at least two atoms should change
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.
Th actual yield of the reaction is 24.86 g
We'll begin by calculating the theoretical yield of the reaction.
2Na + Cl₂ → 2NaCl
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl from the balanced = 2 × 58.5 = 117 g
From the balanced equation above,
46 g of Na reacted to produce 117 g of NaCl.
Therefore,
11.5 g of Na will react to produce = (11.5 × 117) / 46 = 29.25 g of NaCl.
Thus, the theoretical yield of NaCl is 29.25 g.
Finally, we shall determine the actual yield of NaCl.
- Theoretical yield = 29.25 g
Actual yield = Percent yield × Theoretical yield
Actual yield = 85% × 29.25
Actual yield = 0.85 × 29.25 g
Actual yield = 24.86 g
Learn more about stoichiometry: brainly.com/question/25899385
