A metal oxide has the formul `Z_(2)O_(3)`. It can be reduced by hydrogen to give free metal and water. `0.2 g` of the metal oxide requires `12 mg` of hydrogen for complete reduction. The atomic weight of the metal is

Options

A. 52
B. 104
C. 26
D. 78

Answer

C

Explanation

`{:(Z_(2)O_(3),+,3H_(2),rarr,2Z,+,3H_(2)O),([2g = 200 mg],,(12mg)/(2mg),,,,),(,,"= 6 m mol",,,,):}`
Since `H_(2)` used is 6 m mol, `Z_(2)O_(3)` used should be 2 mmol.
Mol of `Z_(2)O_(3) = ("Weight")/(Mw) = (200m g xx 10^(-3)g)/(Mw) = 2 xx 10^(-3) "moles"`
`:. Mw (Z_(2)O_(3)) = 100 :. 2Z + 16 xx 3 = 100 rArr Z = (100 - 48)/(2) = 26`