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

8.70 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 = (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.

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.

8.70 liters

Explanation:

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2