In reference to item 13, suppose synthesis of al2o3 is done in the lab using 20.50 g of al and excess of o2. upon purification, 30.33 grams of al2o3 is produced. what is the percent yield of al2o3?

The balanced equation for the above reaction is as follows;

4Al + 3O₂ - - > 2Al₂O₃

stoichiometry of Al to Al₂O₃ is 4:2

the number of Al moles reacted - 20.50 g / 27 g/mol = 0.76 mol

we have been told that O₂ was provided in excess, therefore Al is the limiting reactant.

Amount of product formed depends on amount of limiting reactant present.

number of Al₂O₃ moles formed are - 0.76 / 2 = 0.38 mol

therefore mass of Al₂O₃ formed is - 0.38 mol x 102 g/mol = 38.76 g

theoretical yield is 38.76 g

but actual yield is 30.33 g

the percentage yield = actual yield / theoretical yield x 100 %

percentage yield = 30.33 g / 38.76 x 100% = 78.3 %

percentage yield = 78.3 %

The answer is 78.31%

Solution:

Aluminum reacts with oxygen to form aluminum oxide as shown in the chemical equation

4Al + 3O2 → 2Al2O3

We can see that 4 moles of aluminum is required to produce 2 moles aluminum oxide. Therefore we can calculate for the mass of aluminum oxide using the molar masses of aluminum and aluminum oxide:

mass of Al2O3 = 20.50 g Al (1 mol Al / 26.982 g Al) (2 mol Al2O3 / 4 mol Al) (101.96 g Al2O3 / 1 mol Al2O3)

= 38.733 g Al2O3

If only 30.33 grams of aluminum oxide formed under the conditions the percentage yield is

percent yield = (actual mass/theoretical mass) x 100

= (30.33 / 38.733) x 100

= 78.31%