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  4. An ore contains 1.34 % of the mineral argentite, Ag2S, by mass How many gram of this ore would have to be processed in order to obtain 1.00 g of pure solid silver, Ag ?

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Q. An ore contains $1.34\%$ of the mineral argentite, $Ag_2S$, by mass How many gram of this ore would have to be processed in order to obtain $1.00\, g$ of pure solid silver, $Ag$ ?

Some Basic Concepts of Chemistry

Solution:

$100 \,g$ ore $\equiv 1.34\,g$ of $Ag_{2}S$
Molar mass of $Ag_{2}S=248\,g\,mol^{-1}$
$248\, g$ of $Ag_{2}S \equiv 2\times 108\,g$ of $Ag=216\,g$ of $Ag\, 1.00\, g$ of $Ag$
$=1.00\,g $ of $Ag\times\left(\frac{248\,Ag_{2}\,S}{216\, g\,Ag}\right)\times\left(\frac{100\,g ore}{1.34\,g\,Ag_{2} S}\right)$
$=85.68\,g$
$=85.7\,g$