Question

What mass of Al2O3 is produced when 13.1 L of O2 gas react with Al at 482 K and 3.40 atm. 

4Al(s)+3O2(g)⟶2Al2O3(g)

 

First, use the ideal gas equation to calculate the moles of O2. 

Then, calculate the mass of Al2O3 from the moles of O2. Show the conversions required to solve this problem.

Answer 1

pV = nRT

R = 0.082 L atm K⁻¹ mol⁻¹

n = pV / RT = (3.40 x 13.1) / (0.082 x 482) = 1.1 mol

According to the equation, n (Al₂O₃) = 2/3 x n (O₂) =2/3 x 1.1 = 0.75 mol

n = m / M

M (Al₂O₃) = 101.96 g/mol

m (Al₂O₃) = n x M = 0.75 x 101.96 = 76.6 g