During a laboratory experiment, 36.12 grams
of al2o3 was formed when
o2 reacted with aluminum metal at 280.0 k and 1.4 atm. what wa
volume of o2 used during the experiment?
302 + 4a1 - 2a1203

V = 9.86 L

Explanation:

Given data:

Mass of Al₂O₃ produced = 36.12 g

Temperature  = 280.0 K

Pressure = 1.4 atm

Volume of O₂ 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 O₂:

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