Answer:
Explanation:
Molecular weight of Al: 27
Molecular weight of O: 16
Percentage composition of Al by weight in Al2O3
= (27*2) / (27*2 + 16*3)
= 0.5294
In 200 grams of Al2O3, there are
= 200*0.5294
= 105.88 grams of Al
77.18 grams of Al was produced in the experiment.
Percent yield for the experiment
= product mass / reactant mass * 100%
= 77.18 / 105.88 * 100%
= 72.89%
