Answer:
Explanation:
7)
Given data:
Mass of aluminium = 2.5 g
Mass of oxygen = 2.5 g
Mass of aluminium oxide = 3.5 g
Percent yield = ?
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Number of moles of Al:
Number of moles = mass/ molar mass
Number of moles = 2.5 g/ 27 g/mol
Number of moles = 0.09 mol
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 2.5 g/ 32 g/mol
Number of moles = 0.08 mol
Now we will compare the moles of aluminium oxide with aluminium and oxygen.
Al ; Al₂O₃
4 : 2
0.09 : 2/4×0.09 = 0.045
O₂ : Al₂O₃
3 : 2
0.08 : 2/3 ×0.08 = 0.053
The number of moles of aluminium oxide produced by Al are less so it will limiting reactant.
Mass of aluminium oxide:
Mass = number of moles × molar mass
Mass = 0.045 × 101.96 g/mol
Mass = 4.6 g
Percent yield:
Percent yield = actual yield / theoretical yield ×100
Percent yield = 3.5 g / 4.6 ×100
Percent yield = 76.1%
8)
Given data:
Mass of copper produced = 3.47 g
Mass of aluminium = 1.87 g
Percent yield = ?
Solution:
Chemical equation:
2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu
Number of moles of Al:
Number of moles = mass/ molar mass
Number of moles = 1.87 g/ 27 g/mol
Number of moles = 0.07 mol
Now we will compare the moles of copper with aluminium.
Al ; Cu
2 : 3
0.07 : 3/2×0.09 = 0.105
Mass of copper:
Mass = number of moles × molar mass
Mass = 0.105 × 63.55 g/mol
Mass = 6.67 g
Percent yield:
Percent yield = actual yield / theoretical yield ×100
Percent yield = 3.47 g / 6.67 × 100
Percent yield = 52%
