See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

    3 x 3 2x x X g 2Y g Z g    1 1 2 2 2 1 r P M r P M  1 3 x 4 2 2x 16   1 3 x 1 2 2x 2    X = 2 mole Y = 2 mole Z = 1 mole Total moles = 5 PV = nRT 5