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

Ramesh V
70 Points
14 years ago

consider +1 and +2 oxidation states for metals

case 1 :

M2O + H2  -- > 2M  +H2O

2x+16                 2x

so, its , 2*3.15/(2x+16) = 1.05/x

or x= 4 i.e., the compound may be Helium(but helium being inert gas doesn't for oxide)

case 2 :

MO + H2  -- > M  +H2O

x+16                 x

so, its , 3.15/(x+16) = 1.05/x

or x = 8 (but there is no element with atomic weight of 8)

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Ramesh

sakshi
11 Points
6 years ago
actually it can be much more easier . we all know that there is no metal with atomic mass of 8 and helium doesn’t form oxide . so the answer can be calculated by the following method:
mass of metal/massof oxygen(both given in ques.)= eq. mass of metal (E) /eq.mass of oxygen (i.e.8)
1.05 / (3.15-1.05) =E/8
E=4

Atul Patel
15 Points
5 years ago
Simply apply the relationship between equivalent wt. And mass of element i.e.Mass of element/eq.of element=mass of oxid