Answer:
V = 9.86 L
Explanation:
Given data:
Mass of Al₂O₃ produced = 36.12 g
Temperature  = 280.0 K
Pressure = 1.4 atm
Volume of Oâ‚‚ used = ?
Solution:
Chemical equation:
4Al + 3O₂  →  2Al₂O₃
Number of moles of Al₂O₃ :
Number of moles = mass/ molar mass
Number of moles = 36.12 g/ 101.96 g/mol
Number of moles = 0.4 mol
Now we will compare the moles of O₂ and Al₂O₃ .
             Al₂O₃       :       O₂
               2         :       3
              0.4        :      3/2 ×0.4 = 0.6 mol
Volume of Oâ‚‚:
PV = nRT
V = nRT/P
V = 0.6 mol × 0.0821 atm. L/mol.K × 280.0 k / 1.4 atm
V = 13.8 atm.L / 1.4 atm
V = 9.86 L
