Question

Oxygen gas reacts with powdered aluminum according to the reaction: 4Al(s)+3O2(g)→2Al2O3(s)

Part A What volume of O2 gas (in L), measured at 771 mmHg and 34 ∘C, is required to completely react with 51.9 g of Al?

Answer 1

Given data,

Mass = 51.9 g

Pressure = 771 mmHg

Temperature = 34⁰C

Part A:

Let us consider a reaction,

4Al(s) + 3O₂(g) ----> 2Al₂O₃(s)

Moles of Al = Mass / Molar mass

= 51.9 g / 26.98 g/ mol

= 1.92 mol Al

No. of moles of O2 needed = 1.92 mol x 3 / 4

= 1.44 mol O2

Volume of O2 gas needed = nRT / P

= 1.44 x 62.4 L. mmHg K.mol x ( 34 + 273 ) / 771

= 89.85 x 307 / 771

= 35.77 L