Answer:
1362.9 grams
Step-by-step explanation:
The balanced chemical equation for the reaction of aluminum (Al) with oxygen (O2) to form aluminum oxide (Al2O3) is:
4 Al + 3 O2 → 2 Al2O3
From the balanced equation, we can see that the stoichiometric ratio of Al to Al2O3 is 4:2 or 2:1. This means that for every 2 moles of Al2O3 produced, 4 moles of Al must react.
To determine the amount of Al2O3 that can be produced from 45 moles of Al and 20 moles of O2, we need to determine which reactant is limiting. The limiting reactant is the one that is completely consumed in the reaction, and the amount of product that can be formed is limited by the amount of the limiting reactant.
To do this, we can use the mole ratio between Al and O2 to determine how much Al is needed to react with 20 moles of O2:
4 Al + 3 O2 → 2 Al2O3
4 mol Al / 3 mol O2 = x mol Al / 20 mol O2
x = (4/3) x 20 = 26.67 mol Al (rounded to two decimal places)
Since we have 45 moles of Al, which is more than the amount needed (26.67 moles), Al is not the limiting reactant. Therefore, the amount of Al2O3 that can be produced is limited by the amount of O2.
Using the mole ratio between O2 and Al2O3, we can calculate the theoretical yield of Al2O3:
4 Al + 3 O2 → 2 Al2O3
3 mol O2 / 2 mol Al2O3 = 20 mol O2 / x mol Al2O3
x = (2/3) x 20 = 13.33 mol Al2O3 (rounded to two decimal places)
Therefore, the theoretical yield of Al2O3 is 13.33 moles.
To convert moles of Al2O3 to grams, we can use the molar mass of Al2O3:
Molar mass of Al2O3 = 2 x atomic mass of Al + 3 x atomic mass of O = 2 x 26.98 g/mol + 3 x 16.00 g/mol = 101.96 g/mol
Mass of Al2O3 = moles of Al2O3 x molar mass of Al2O3
Mass of Al2O3 = 13.33 mol x 101.96 g/mol
Mass of Al2O3 = 1362.9 g
Therefore, the mass of Al2O3 that can be produced is approximately 1362.9 grams (to one decimal place).