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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 isA. 52B. 104C. 26D. 78 |
Answer» Correct Answer - C `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` |
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