Question

An ore contains $1.24 \%$ of mineral argentate, $Ag _{2} S$ by mass. How many grams of this ore would have to be processed in order to obtain $1 g$ of pure solid silver?

• A. 92.6 g
• B. 88.1 g
• C. 101.11 g
• D. 107.25 g

Answer 1

Answer: (A)

$Ag _{2} S \rightarrow 2 Ag + S$

To obtain 2 mole $Ag$, mole of $Ag _{2} S$ required $=1$

To obtain $\frac{1}{108}$ mole $Ag ,$ mole of $Ag _{2} S$ required

$=\frac{1}{2} \times \frac{1}{108}=\frac{1}{216} mole$

grams of $Ag _{2} S$ required $=\frac{1}{216} \times 248 \,g$

$Ag _{2} S$ required $=\frac{248}{216} \,g$

$1.24\, g\, Ag _{2} S$ is obtained from ore $=100 g$

$ \frac{248}{216} g Ag _{2} S$ is obtained from ore $=\frac{100}{1.24} \times \frac{248}{216}$

$=\frac{24800}{267.84}$

$=92.6\, g$