A metal oxide is reduced by heating it in a s | vedclass

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(D) Let $W_{M}$ be the weight of the metal and $W_{O}$ be the weight of oxygen in the oxide.Given: $W_{M} = 1.05 \, g$ and the total weight of the oxi

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The equivalent weight of the metal is $4$

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(D) Let $W_{M}$ be the weight of the metal and $W_{O}$ be the weight of oxygen in the oxide.Given: $W_{M} = 1.05 \, g$ and the total weight of the oxide is $3.15 \, g$.Therefore,the weight of oxygen $W_{O} = 3.15 \, g - 1.05 \, g = 2.10 \, g$.According to the law of equivalence,the equivalent weight of the metal $(E_{M})$ is given by the formula:$E_{M} = \frac{W_{M} 	imes E_{O}}{W_{O}}$,where $E_{O}$ is the equivalent weight of oxygen,which is $8$.Substituting the values:$E_{M} = \frac{1.05 	imes 8}{2.10} = \frac{8.4}{2.1} = 4$.Thus,the equivalent weight of the metal is $4$.

$A$ metal oxide is reduced by heating it in a stream of hydrogen. It is found that after complete reduction,$3.15 \, g$ of the oxide yielded $1.05 \, g$ of the metal. We may deduce that:

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