Question

An ore contains $1.34\%$ of the mineral argentite, $A{g}_{2}S$, by mass How many gram of this ore would have to be processed in order to obtain $1.00g$ of pure solid silver, $Ag$ ?

• A. 74.6 g
• B. 85.7 g
• C. 107.9 g
• D. 134.0 g

Answer 1

Answer: (B)

$100g$ ore $\equiv 1.34g$ of $A{g}_{2}S$
Molar mass of $A{g}_{2}S=248gmo{l}^{-1}$
$248g$ of $A{g}_{2}S\equiv 2\times 108g$ of $Ag=216g$ of $Ag1.00g$ of $Ag$
$=1.00g$ of $Ag\times \left(\frac{248A{g}_{2}S}{216gAg}\right)\times \left(\frac{100gore}{1.34gA{g}_{2}S}\right)$
$=85.68g$
$=85.7g$