See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

3N of H2O2 = 3  5.6 = 16.8 volume of O2 Volume of O2 produced by H2O2 = 200  16.8 = 3360 mL Same volume will be produced by KMnO4 = 3360 mL Total volume of O2 = 6720 mL = 6.72 L AITS-FT-VIII (Paper-2)-PCM(Sol.)-JEE(Advanced)/2021 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 8 Mathematics PART – III SECTION – A