Question

During a laboratory experiment, 46.54 grams of Al2O3 was formed when O2 reacted with aluminum metal at 300.0 K and 1.2 atm. What was the volume of O2 used during the experiment?

Answer 1

Answer:

13.96 L

Explanation:

Answer 2

Answer:

14.04 L.

Explanation:

• The balanced reaction to form Al₂O₃ is:

4Al + 3O₂ → 2Al₂O₃,

4.0 moles of Al react with 3.0 moles O₂ to produce 2.0 moles Al₂O₃.

• Firstly, we need to calculate the no. of moles in (46.54 grams) of Al₂O₃:

n = mass/molar mass = (46.54 g) / (101.96 g/mol) = 0.4565 mol.

Using cross multiplication:

3.0 moles of O₂ produce → 2.0 mole Al₂O₃, from the stichiometry.

??? moles of O₂ produce → 0.4565 mole Al₂O₃.

∴ The no. of moles of O₂ needed to produce 0.4565 mol (46.54 grams) of Al₂O₃ = (3.0 mol)(0.4565 mol)/(2.0 mol) = 0.6847 mol.

• To calculate the volume of O₂ needed to produce 0.4565 mol (46.54 grams) of Al₂O₃, we can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 1.2 atm).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 0.6847 mol).

R is the general gas constant (R = 0.082 L/atm/mol.K),

T is the temperature of the gas in K (T = 300.0 K).

∴ V = nRT/P = (0.6847 mol)(0.082 L/atm/mol.K)(300.0 K)/(1.2 atm) = 14.04 L.