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4 g of metal oxide MxOy is reduced by H2...

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

A

`M_2O`

B

`M_3O_4`

C

`M_2O_3`

D

`MO`

Text Solution

Verified by Experts

The correct Answer is:
B

Given, 32x = 2.4 g metal
`therefore x=(2.4)/32=0.75g`
and 32x + 16y = 4.0 = Metal oxide
`rArr32x(2.4)/32+16y=4.0`
`rArr16y=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`
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