QUESTION 18 Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) 3O 2 (g) 2Al 2O 3 (s) In a particular experiment, the reaction of 2.5 g of Al with 2.5 g of O 2 produced 3.5 g of Al 2O 3. The % yield of the reaction is .

Percent yield = 85.4%

Explanation:

Given data:

Mass of Al = 2.5 g

Mass of oxygen = 2.5 g

Mass of aluminium oxide = 3.5 g

Percent yield = ?

Solution:

Chemical equation:

4Al + 3O₂  → 2Al₂O₃

First of all we will calculate the number of moles of each reactant.

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 oxygen and aluminium.

                         Al            :          Al₂O₃

                           4           :            2

                        0.09        :         2/4×0.09 = 0.045

                          O₂         :          Al₂O₃

                           3           :            2

                        0.08        :         2/4×0.08 = 0.04

The  number of moles of aluminium oxide produced by oxygen are less so it will limiting reactant.

Theoretical yield of aluminium oxide:

Mass = number of moles × molar mass

Mass = 0.04 mol ×  101.96 g/mol

Mass = 4.1 g

Percent yield:

Percent yield = actual yield /theoretical yield × 100

Percent yield = 3.5 g/ 4.1 g × 100

Percent yield = 85.4%

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