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

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