A metal oxide has the formula `M_(2)O_(3)`. It can be reduced by `H_(2)` to free metal and water. 0.1596g of `M_(2)O_(3)` required 6 mg of `H_(2)` for complete reduction. The atomic mass of the metal is:

Options

A. 27.9
B. 79.8
C. 55.8
D. 159.8

Answer

C

Explanation

`M_(2)O_(3)+3H_(2)rarr2M+3H_(2)O`
`(2x+48)g 6g`
x=Atomic mass of metal
`because` 0.006 g `H_(2)` reduces 0.1596g `M_(2)_(3)`.
`therefore 6g H_(2)` will reduce `(0.1596)/(0.006)xx6 g M_(2)O_(3)=159.6M_(2)O_(3)`
`2x+48=159.6`
`2x=11.6`
`x=55.8`.