Oxygen gas reacts with powdered aluminum according to the following reaction:

4Al (s) + 3O2 (g) →2Al2O3 (s)

What volume of O2 gas, measured at 787 mmHg and 21 ∘C, is required to completely react with 55.0 g of Al?

Express the volume in liters to three significant figures.

35.5L of O2

Explanation:

4Al (s) + 3O2 (g) → 2Al2O3 (s)

First, let us calculate the number of mole O2 that reacted with 55g of Al. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al = 55g

Number of mole = Mass / Molar Mass

Number of mole of Al = 55/27 = 2.04moles

From the equation,

4moles of Al reacted completely with 3 moles of O2.

Therefore, 2.04moles of Al will react completely with = (2.04 x 3) / 4 = 1.53moles of O2

Now let us obtain the volume of O2 occupied by 1.53moles of O2 measured at 787 mmHg and 21°C. This can be achieved by doing the following:

n = 1.53moles

P = 787mmHg

Recall: 760mmHg = 1atm

787mmHg = 787/760 = 1.04atm

T = 21°C = 21 + 273 = 294K

R = 0.082atm. L/Kmol

V = ?

Using the Ideal gas equation PV = nRT, we can obtain the volume as follows:

PV = nRT

V = nRT/P

V = (1.53 x 0.082 x 294) / 1.04

V = 35.5L

Therefore, 35.5L of O2 required to react with 55.0g of Al