Question

$27$ g of silver was reacted with excess sulfur, according to the following equation:
$2Ag+S\to {Ag}_{2}S$
$25$ g of silver sulfide was collected. What are the theoretical yield, actual yield, and percent yield?

• A. $31$ g, $35$ g, $71$%
• B. $21$ g , $33$ g , $69$%
  
• C. $31$ g, $25$ g, $81$%
• D. $31$ g, $45$ g, $91$%

Answer 1

The correct option is C $31$ g, $25$ g, $81$%

$2Ag+S\to A{g}_{2}S$
The atomic mass of Ag and S are $108$ g/mol and $32$ g/mol respectively.

The Molar mass of $A{g}_{2}S=2\left(108\right)+32=248$ g/mol.
$2$ moles of Ag gives 1 mole of $A{g}_{2}S$.
$27$ g of Ag will give $\frac{27}{108\times 2}\times 248=31$ of $A{g}_{2}S$.
Hence, the theoretical yield is $31$ g.
However only $25$ g of silver sulfide was collected.
Hence, the actual yield is 25 g.
The percentage yield is $\frac{25\times 100}{31}=81$ %.