during a laboratory experiment, 44.18 grams of al2o3 was formed when o2 reacted with aluminum metal at 290.0 k and 1.3 atm. what was the volume of o2 used during the experiment?

3o2 + 4al → 2al2o3

a. 10.10 liters
b. 11.90 liters
c.12.51 liters
d. 15.55 liters

b. 11.90 Liters

Explanation:

3O₂ + 4Al → 2Al₂O₃,

It is clear that 3.0 moles of O₂ react with 4.0 moles of Al to produce 2.0 Al₂O₃.

n = mass/molar mass = (44.18 g)/(101.96 g/mol) = 0.4333 mol.

using cross multiplication:

3.0 mol of O₂ produces → 2.0 mol of Al₂O₃.

??? mol of O₂ produces → 0.4333 mol of Al₂O₃.

∴ The no. of moles of O₂ needed to produce 36.12 grams of Al₂O₃ = (3.0 mol)(0.4333 mol)/(2.0 mol) = 0.65 mol.

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 1.3 atm).

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

n is the no. of moles of the gas in mol (n = 0.65 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 290 K).

∴ V = nRT/P = (0.65 mol)(0.0821 L.atm/mol.K)(290 K)/(1.3 atm) = 11.9 L.

So, the right choice is: b. 11.90 Liters.

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