Three stoichiometry questions

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%