Question

How many litres of $O_2$ gas at NTP is needed to react completely with 27 gm of $Al$?

• A. 5.6 litres
• B. 11.2 litres
• C. 22.4 litres
• D. 16.8 litres

Answer 1

The correct option is D 16.8 litres

$4Al+3O_2⟶2A{l}_2{O}_3$

Moles of Al $=\frac{27}{27}=1mole$

4 moles of aluminium reacts with 3 moles of oxygen.

So, 1 mole of aluminium reacts with 0.75 moles of oxygen.

At NTP, 1 mole of a gas is equal to 22.4 litres.

Volume of oxygen used is $0.75\times 22.4$ litres = 16.8 litres.