Question

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

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

Answer 1

Answer: (A)

$A{g}_{2}S\to 2Ag+S$

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

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

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

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

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

$1.24gA{g}_{2}S$ is obtained from ore $=100g$

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

$=\frac{24800}{267.84}$

$=92.6g$