Excess reagent in the reaction  ${\text{2H}}_2+{\text{O}}_{\text{2}}\to {\text{2H}}_{\text{2}}\text{O}$

In the given equation when 2 moles of Hydrogen produce 2 moles of water while 2 moles of Oxygen produce 4 moles of water .

${\text{2 moles of H}}_2\to {\text{2 moles of H}}_{\text{2}}\text{O}$

${\text{2 moles of O}}_2\to 4{\text{ moles of H}}_{\text{2}}\text{O}$

The reagent which gives larger amount of product is taken as excess reagent...hence excess reagent is ${\text{O}}_2$.