Question

An ore contains $1.34$% of the mineral argentite, ${Ag}_{2}S$ 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. $74.6g$
• B. $85.7g$
• C. $107.9g$
• D. $134.0g$

Answer 1

The correct option is B $85.7g$

Argentite will give $Ag$ as:

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

Molecular wt of $A{g}_{2}S=248g$ and $Ag=108g$

Therefore $248g$ of $A{g}_{2}S$ gives $2\times 108=216g$ of $Ag$

Amount of $A{g}_{2}S$ required to obtain $1gAg$ is:

$248/216g=1.148gA{g}_{2}S$

Given that $1.34gA{g}_{2}S$ is obtained from $100g$ ore.

Therefore for $1.148gA{g}_{2}S$, amount of ore required is:

$=\frac{100\times 1.148}{1.34}g$=$85.67g$

Thus option B is correct.