Question

How many liters of oxygen are needed to react completely with 81.0 g of aluminum at STP?

A.) 25.2 L
B.) 72.0L
C.) 50.4 L
D.) 67.2 L

Answer 1

I think is C
Al= 108g have Vo2= 67.2L
Al= 81.0g have Vo2= ?
What you need to do is:
67.2 multiply with 81.0 and divided by 108 you will see the answer which is 50.4 L

Answer 2

Volume O₂ at STP = 50.4 Liters

Step-by-step explanation:

4Al(s) + 3O₂(g) => 2Al₂O₃(s) at STP conditions

81g Al(s) = 81g/27g/mole = 3mole Al

moles O₂ consumed = 4/3(3)moles O₂ = 2.25 moles O₂ consumed

Volume O₂ at STP = 2.25moles x 22.4L/mole = 50.4 Liters O₂