See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

    40% 2 5 2 2 3 N O g O g 2NO g O g        60% 2 2 3 NO g O g NO g O g     2 5 N O n 40  then moles of O2 consumed in 1st reaction = 40  2 NO 40 n 40 2 32 mole 100       3 O 40 n 40 16 mole 100    Now, for 2nd reaction, NO2 is L.R. 2 O consumed n 32          2 O left n 85 40 32 13 mole          3 O formed 32 n 60