Question

Amount of oxygen required for complete combustion of 27 g Al is:

• A. 24g
• B. 12g
• C. 20g
• D. 6 g

Answer 1

Answer: (A)

24g

$4Al+3O_2\to 2A{l}_2{O}_3$
The atomic mass of Al is 27 g/mol. Hence, 27 g of Al corresponds to 1 mole of Al. 4 moles of Al requires 3 moles of oxygen for complete combustion. 1 moles of Al will require $\frac{3}{4}\times 1=0.75$ moles of oxygen for complete combustion. The molar mass of oxygen is 32 g/mol. The mass of oxygen required will be $0.75\times 32=24$ g.