The correct option is D 0.65 g
4KO2+2CO2→2K2CO3+3O2
Moles of KO2=given massmolar mass=2.4571=0.034 molMoles of CO2=given massMolar mass=4.4444=0.100 mol
Finding the limiting reagent:
For KO2=given molesstoichiometric coefficient=0.0085 molFor CO2=0.12=0.05 mol
So, KO2 is the limiting reagent.
4 moles of KO2 produces 3 moles of O2
0.034 mol will form =34×0.034=0.0255 mol
Mass of oxygen formed = 0.0255 mol×32 g/mol=0.816 g
But, the yield of the reaction is 80%
Actual amount of oxygen formed = 0.816 g ×80100=0.65 g