How many moles of oxygen are required to produce 436 L of water vapor?

The reaction for formation of water vapour is as follows;

2H2 + O2 - > 2H2O

Water vapour is in the gaseous form, at STP, 1 mol of any gas occupies 22.4 L volume

If 22.4 L occupied by 1 mol

Then amount of moles in 436 L - 1/22.4 x 436 = 19.5 mol

Stoichiometry of O2 to H2O is 1:2

Amount of O2 moles reacted to formed H2O is - 19.5 mol / 2 = 9.75 mol