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%