Calculate the number of oxygen atoms required to combine with 7.0 g of N2 to form N2O3 if  80% of N2 is converted into productsN2+32O2⟶N2O3

2N2 + 3O2 → 2N2O3, (% yield = 80%)Initial mol of N2 = 7.028=0.25 molMoles of N2 converted = 0.25×80100=0.22 mol N2 = 3 mol O2 (1 mol O2 = 2 oxygen atom) = 2 x 0.3 mol O atom =2 x 0.3 x 6.02 x 1023 = 3.6 x 1023