Answer:
Percent Yield = 97.75 %
Explanation:
1 MOLE = It is equal to the molar mass of the substance
1 mole of Cu = 63.54 g (Molar Mass of Cu = 63.54 g/mole)
1 mole of AgNO3 = 170 g (Molar Mass of AgNO3 = 170 g/mol)
Given Mass of AgNO3 = 1.41 g
Given Mass of Cu = 2.93 g
<em><u>Second step : Find the limiting Reagent (which is in less amount) </u></em>
Balanced Chemical equation :
![Cu + 2AgNO_{3} \rightarrow Cu(NO_{3})_{2}+2Ag](https://tex.z-dn.net/?f=Cu%20%2B%202AgNO_%7B3%7D%20%5Crightarrow%20Cu%28NO_%7B3%7D%29_%7B2%7D%2B2Ag)
This means
1 mole of Cu will react with = 2 mole of AgNO3
63.54 g of Cu reacts with = 2 x 170 g of AgNO3
1 g of Cu reacts with = (2 x 170)/63.54 of AgNO3
![= \frac{170\times 2}{63.54}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B170%5Ctimes%202%7D%7B63.54%7D)
= 5.35 g of AgNO3
2.93 g should reacts with = 2.93 x 5.35 = 15.67 g of AgNO3
Available AgNO3 = 1.41 g
So , AgNO3 is less than required = limiting reagent
Now the reaction occur 1.41 g of AgNO3
Now, Limiting reagent will decide How much Silver(Ag) Metal will form
2 mole of AgNO3 will produce = 2 mole of Ag
1 mole of AgNO3 will produce = 1 mole of Ag
170 g AgNO3 will produce = 107.86 mole of Ag(Molar mass of Ag = 107.86)
1 g AgNO3 will produce =
![\frac{107.86}{170}](https://tex.z-dn.net/?f=%20%5Cfrac%7B107.86%7D%7B170%7D)
1.41 g of AgNO3 will produce =
![\frac{107.86\times 1.41}{170}](https://tex.z-dn.net/?f=%20%5Cfrac%7B107.86%5Ctimes%201.41%7D%7B170%7D)
= 0.89 g
![Yield\ Percent = \frac{Actual\ Yield}{Theoritical\ Yield}\times 100](https://tex.z-dn.net/?f=Yield%5C%20Percent%20%3D%20%5Cfrac%7BActual%5C%20Yield%7D%7BTheoritical%5C%20Yield%7D%5Ctimes%20100)
Actual yield = 0.87 g
Theoritical yield = 0.89 g
![Yield\ Percent = \frac{0.87}{0.89}\times 100](https://tex.z-dn.net/?f=Yield%5C%20Percent%20%3D%20%5Cfrac%7B0.87%7D%7B0.89%7D%5Ctimes%20100)
Percent Yield = 97.75 %