UGRENT! Please help showing all work

a. Oxygen

b. 20.39g of Al₂O₃

c. 87%

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%