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8.08g of a metallic oxide on being reduc...

8.08g of a metallic oxide on being reduced by `H_(2),` produces 1.8g of water. Find the quantity of `O_(2)` in the above oxide and the equivalent mass of the metal.

Text Solution

Verified by Experts

Amount of oxygen present in 18g of water=16g
`therefore`Amount of oxygen present in 1.8g of water=1.6g
`therefore` Oxygen content in 8.08 g of the metallic oxide=1.6g
[since all this oxygen comes from the metallic oxide]
Hence, the amount of the metal present in the oxide =(8.08-1.6)=6.48g
`therefore1.6g` of `O_(2)` combine with 6.48g of the metal
or, 8g of `O_(2)` combine with `((6.48)/(1.6)xx8)g` of the metal
Therefore, the equivalent mass of the metal`=(6.48)/(1.6)xx8=32.4`.
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