Answer : The percent yield is, 83.51 %
Solution : Given,
Mass of Zn = 5.00 g
Mass of
= 25.00 g
Molar mass of Zn = 65.38 g/mole
Molar mass of
= 168.97 g/mole
Molar mass of Ag = 107.87 g/mole
First we have to calculate the moles of Zn and
.


Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,

From the balanced reaction we conclude that
As, 2 mole of
react with 1 mole of 
So, 0.1479 moles of
react with
moles of 
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of 
From the reaction, we conclude that
As, 2 mole of
react to give 2 mole of 
So, 0.1479 moles of
react to give 0.1479 moles of 
Now we have to calculate the mass of 


Theoretical yield of
= 15.95 g
Experimental yield of
= 13.32 g
Now we have to calculate the percent yield.


Therefore, the percent yield is, 83.51 %