First balance the equation to find that so the moles of ##O_2## released is equal to one-half the moles of ##Na_2O_2## added:
##2 Na_2O_2 + 2 H_2O rarr 4 NaOH + O_2##
The molar mass of ##Na_2O_2## is 77.98 g/mol, so the moles of ##Na_2O_2## is
##(1.15g)/(77.98 g/(mol))=0.01456## mol
(carries 1 extra sig fig for intermediate calculation)
The molar mass of ##O_2## is 32.00 g/mol, so the mass of ##O_2## released is
##1/2 times 0.01456 mol times 32.00 g/(mol)=0.233##g