A metal oxide is reduced by heating it in a stream of hydrogen. It is found that after complete reduction, 3.15g of the oxide have yielded 1.05 g of the metal. We may deduce that

Options

A. The eq. weight of the metal is 8
B. The atomic weight of the metal is 8
C. The atomic weight of the metal is 4
D. The eq. weight of the metal is 4

Answer

D

Explanation

Method I :
Consider +1 and +2 oxidation states for metals
Case I : With oxidation state +1
`M_(2)O + H_(2) to 2M + H_(2)O`
`2x + 16 " " 2x`
`therefore (2 xx 3.15)/(2x + 16) = (1.05)/(x)`
`therefore` x = 4 i.e., the compound may be helium (but helium being inert gas doesn’t form oxide)
Case II : With oxidation state +2
`MO + H_(2) to M + H_(2)O`
`x + 16 " " x`
`therefore (3.15)/(x + 16) = (1.05)/(x)`
` x = 8`
(but there is no element with atomic weight of 8).
As helium doesn’t form oxide and there is no metal with atomic mass 8 therefore the preferable method is method 2.
Method II :
`("Mass of metal")/("Mass of oxygen") = ("Equivalent mass of metal")/("Equivalent mass of oxygen")`
` (1.05)/((3.15 - 1.05)) = (E)/(8)`
`E = 4 `