maybe 1+8 (4+548) 4/6 4+4
Answer:
Volume O₂ at STP = 50.4 Liters
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
Nitrogen will be in the column to the right of carbon
