Question

Oxygen gas reacts with powdered aluminum according to the following reaction: 4Al(s)+3O2(g)→2Al2O3(s)..What volume of O2 gas (in L), measured at 795 mmHg and 16 ∘C, is required to completely react with 53.3 g of Al?

Answer 1

Oxygen gas reacts with aluminum as follow:

4Al(s) + 3O₂(g)----> Al₂O₃(s)

Therefore, 4 mol of Al reacts with 3 mol of O₂ .

no. moles of Al = mass/mol.wt
               = 53.3 g/ 26.982g/mol
               = 1.975 mol
Therefore, no. of mol of O₂ = no. moles of Al x 3/4
                           = 1.975 mol x 3/4
                           = 1.4815 mol
                          
Now we need to find the volume of O₂ required at 795 mmHg and temperature T = 16 ^oC

P = 795 mmHg = 795 mmHg x (1 atm/760 mmHg) = 1.046 atm ;
T = 16 ^oC = 16+273 = 289 K
n = 1.4815 mol

From Ideal gas equation;

PV = nRT
Volume V = nRT/P
       = (1.4815 mol x 0.08205 L.atm.mol^-1·K^-1 x 289 K)/1.046 atm
       = 35.13 L
      
The volume of O₂ gas required = 35.13 L