4 g of metal oxide `M_xO_y` is reduced by `H_2` and 2.4 g metal is obtained, if the atomic weight of metal is 32, the formula of oxide is

Options

A. `M_2O`
B. `M_3O_4`
C. `M_2O_3`
D. `MO`

Answer

B

Explanation

Given, 32x=2.49 metal :
` therefore x=(2.4)/(32) = 0.75g`
and 32x + 16y = 4.0 = Metal oxide
` rArr32 xx (2.4)/(32) + 16 y = 40`
` rArr 16 y = 4.0 -2.4 = 1.6`
` therefore y = 0.1`
` therefore x :y = 0.75 : 0.1 = 3:4`
Hence, oxide is `M_3O_4`