**3. How many liters of fluorine gas can be produced when 0.67 L of HF reacts with excess o, at STP
(standard temperature and pressure; normal conditions)?

0.336L

Explanation:

4HF + O2 —> 2F2 + 2H2O

1mole of HF occupies 22.4L at stp.

Therefore Xmol will occupy 0.67L i.e

Xmol of HF = 0.67/22.4 = 0.03mol

From the equation,

4moles of HF 2moles of F2.

Therefore, 0.03mol of HF will produce = (0.03x2)/4 = 0.015mol of F2.

1mol of F2 occupies 22.4L at stp.

Therefore, 0.015mol will occupy = 0.015 x 22.4L = 0.336L