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 dа ta:

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