Answer:
a. Oxygen
b. 20.39g of Al₂O₃
c. 87%
Step-by-step explanation:
In order to produce 2moles of Al₂O₃, we need to make react 4 moles of Al, to 3 moles of O₂.
We determine the limiting reactant with the moles of each:
12.98g / 26.98g/mol = 0.481 moles of Al
9.6g / 32g/mol = 0.300 moles of oxygen
4 moles of Al react to 3 moles of O₂
Then, 0.481 mol may react to (0.481 . 3)/4 = 0.361 moles
We only have 0.300 moles of O₂ and we need 0.361. Oxygen is the limiting reactant.
So 3 moles of O₂ can produce 2 moles of Al₂O₃
Our 0.300 moles may produce (0.300 . 2) / 3 = 0.200 mol
Our theoretical yield is: 0.2 mol . 101.96g/ mol = 20.39 g
To determine the percent yield we calculate:
(Produced yield/ Thoeretical yield) . 100 → (17.74 /20.39) . 100 = 87%