Question

How many liters of oxygen at 0 ∘C and 760 mmHg (STP) are required to completely react with 2.7 g of aluminum?

solve for V

Answer 1

4Al(s)+3O2(g)→2Al2O3(s)

The coefficients in a balanced equation determine the ratio of moles of reactants and moles of products. In this reaction 4 moles of Al react with 3 moles of O2 to produce 2 moles of Al2O3. The ratio of Al to O2 is 4:3. Use the following proportion to determine the number of moles of O2.

4 ÷ 3 = Moles of Al ÷ Moles of O2


To determine the number of moles of Al divide the mass by the mass of one mole.

N = 2.7 ÷ 27
This is approximately 0.1 moles.


4 ÷ 3 = (2.7 ÷ 27) ÷ Moles of O2

Moles of O2 = 0.75 * (2.7 ÷ 27) = 5.175 ÷ 27


This is approximately 0.075 moles. Since the temperature is at 0 ∘C and the pressure is 760 mm Hg, the volume of one mole is 22.4 liters.

V = 22.4 * 2.025 ÷ 27 = 45.36 ÷ 27
V = 1.68 liters.