Question

Question 2(Multiple Choice Worth 5 points) (06.04 HC) 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 was the volume of O2 used during the experiment? 3O2 + 4Al → 2Al2O3 8.70 liters 9.22 liters 10.13 liters 12.81 liters

Answer 1

Answer:

8.70 liters.

Explanation:

• The balanced equation for the mentioned reaction is:

3O₂ + 4Al → 2Al₂O₃,

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

• Firstly, we need to calculate the no. of moles (n) of 36.12 g of Al₂O₃:

n = mass/molar mass = (36.12 g)/(101.96 g/mol) = 0.3543 mol.

using cross multiplication:

3.0 mol of O₂ produces → 2.0 mol of Al₂O₃.

??? mol of O₂ produces → 0.3543 mol of Al₂O₃.

∴ The no. of moles of O₂ needed to produce 36.12 grams of Al₂O₃ = (3.0 mol)(0.3543 mol)/(2.0 mol) = 0.5314 mol.

• Now, we can find the volume of O₂ used during the experiment:

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

where, P is the pressure of the gas in atm (P = 1.4 atm).

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

n is the no. of moles of the gas in mol (n = 0.5314 mol).

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

T is the temperature of the gas in K (T = 280 K).

∴ V = nRT/P = (0.5314 mol)(0.0821 L.atm/mol.K)(280 K)/(1.4 atm) = 8.726 L ≅ 8.70 L.

So, the right choice is: 8.70 liters.

Answer 2

Answer:

8.70 liters

Explanation: