Question

In the copper – silver nitrate lab Copper medals and silver nitrate solution reacted to produce silver metal and copper (II) nitrate in solution. A student placed a copper wire with a mass of 2.93 g in the reaction test tube. The silver nitrate solution contained 1.41 g of silver nitrate. He obtained .87 g of silver metal. Calculate the percent yield of silver.

Answer 1

Percent Yield = 97.75 %

Step-by-step 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

Second step : Find the limiting Reagent (which is in less amount)

Balanced Chemical equation :

`Cu + 2AgNO_(3) \rightarrow Cu(NO_(3))_(2)+2Ag`

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

`= (170* 2)/(63.54)`

= 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 =

`(107.86)/(170)`

1.41 g of AgNO3 will produce =

`(107.86* 1.41)/(170)`

= 0.89 g

`Yield\ Percent = (Actual\ Yield)/(Theoritical\ Yield)* 100`

Actual yield = 0.87 g

Theoritical yield = 0.89 g

`Yield\ Percent = (0.87)/(0.89)* 100`

Percent Yield = 97.75 %