Question

5.6 l of oxygen at NTP equivalent

Answer 1

Hello!

22.4 L of Oxygen at NTP = $\sf 6.022 \:x\: 10^{23}$ O-molecules

Then,

5.6 L of Oxygen at NTP contains $\dfrac{\sf 5.6}{\sf 22.4} \:x\: \sf 6.022 \:x\: 10^{23}$

⇒ $\sf 1.505 \:x\: 10^{23}$ O-molecules

Cheers!

Answer 2

22.4l = 1 mole o2 at NTP

1l= 1÷22.4 mole o2
5.6l = (1÷22.4)×5.6
=0.25 eq