8 g of a metal reacted with oxygen to form 9 | vedclass

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(D) $1$. Calculate the mass of oxygen reacted: $Mass_{oxide} - Mass_{metal} = 9 \ g - 8 \ g = 1 \ g$ of oxygen.$2$. Equivalent weight of metal $(E_M)$

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$512$

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(D) $1$. Calculate the mass of oxygen reacted: $Mass_{oxide} - Mass_{metal} = 9 \ g - 8 \ g = 1 \ g$ of oxygen.$2$. Equivalent weight of metal $(E_M)$ is given by: $E_M = \frac{Mass_{metal}}{Mass_{oxygen}} 	imes 8 = \frac{8 \ g}{1 \ g} 	imes 8 = 64$.$3$. The equivalent weight of hydrogen is $1$.$4$. To react with $8 \ g$ of hydrogen,the mass of metal required is: $Mass_{metal} = E_M 	imes Mass_{hydrogen} = 64 	imes 8 = 512 \ g$.

$8 \ g$ of a metal reacted with oxygen to form $9 \ g$ of its oxide. What is the weight (in $g$) of the metal required to react with $8 \ g$ of hydrogen? $(H=1.0)$

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