Answer:
33.9 g
Step-by-step explanation:
We can use stoichiometry to determine the amount of AI2O3 formed from the given amount of O2.
First, we need to calculate the amount of O2 in moles
n(O2) = m(O2) / M(O2) = 16 g / 32 g/mol = 0.5 mol
According to the balanced equation, 4 moles of AI react with 3 moles of O2 to produce 2 moles of AI2O3. This means that the mole ratio of O2 to AI2O3 is 3:2.
Since we have an excess of AI, all of the O2 will react with AI to form AI2O3. Therefore, we can use the mole ratio to calculate the amount of AI2O3 formed
n(AI2O3) = n(O2) x (2/3) = 0.5 mol x (2/3) = 0.333 mol
Finally, we can calculate the mass of AI2O3 formed using the molar mass of AI2O3
m(AI2O3) = n(AI2O3) x M(AI2O3) = 0.333 mol x 102 g/mol = 33.9 g
Therefore, 33.9 g of AI2O3 will form from 16 g of O2 and excess AI.