In an experiment, 4 g of M2Ox oxide was reduced to 2.8 g of the metal. If the atomic mass of the metal is 56 g mol–1, the number of O atoms in the oxide is (AFMC 2010)

Given, The weight of metal oxide ( 2 x M O ) = 4 g Weight of reduced metal = 2.8 g The molecular mass of given metal = 1 56 gm ol The balanced chemical reaction of given metal reduction is 2 2 2 2 x x M O M O → Mass of metal oxide Mass of reduced metal 4 g 2.8 g (given) 2 56 112 g x 2 56 16 112 16 x x g x (as per balanced equation) Solving for x, we get 112 4 448 112 16 160 2.8 2.8 112 16 160 16 160 112 16 48 48 3 16 x x x x x x Therefore, the number of oxygen atoms in the oxide is