Question

on reduction with hydrogen 3.6 grams of an oxide of metal 3.2 grams of metal left if vapour density of metal is 32 the simplest formula of the oxide would be​

Answer 1

Answer:

M₂O

Explanation:

The question also contains 4 options where the metal is denoted by M and Oxygen by O.

Given, vapour density of metal = 3.2 grams

Also we know that

Vapour Density = (Molecular Weight)/2

or, Molecular Weight = 2 × Vapour Density

∴ Molecular Weight of the metal = 2 × 32 = 64

On reduction with hydrogen, the amount of metal left will be the same as that of in the Oxide

∴ Weight of metal = 3.2 gram

Mole ratio of the metal = Weight/Molecular Weight

                                      = 3.2/64 = 0.05

Weight of Oxygen in the metal = 3.6 - 3.2 = 0.4 gram

Mole ration of the Oxygen = 0.4/16  (∵ Molecular Weight of Oxygen = 16)

                                             = .025

Thus, the formula of the metal can be $M_{0.05}O_{.025}$  

Or, the simplest formula is M₂O

Hope this helps.