Question

how can i Calculate those ??

1)Mass of Cu wire reacted

2)Mass of Ag produced

3)Mass of AgNO3 reacted

4)Moles of Cu reacted

5)Moles of AgNO3 reacted

6)Moles of Ag produced

7)Ratio of moles Ag to moles AgNO3

8)Ratio of moles Ag to moles Cu

9)Number of Cu atoms removed from wire

10)Number of Ag atoms produced

11)Theoretical yield of Ag

12)% Yield of Ag

Answer 1

1.Mass of Cu wire reacted = mass of copper wire - mass of copper wire ( after reaction) = 1.12 g - 1.05g = 0.07g

2.Mass of Ag produced = mass of beaker + silver produced - mass of beaker = 57.78g - 56.97 g = 0.81g

3.Mass of AgNO3 reacted = mass of vial + AgNO3 - mass of vial = 13.34g - 12.16 g = 1.18g

4.Moles of Cu reacted = (Mass of Cu wire reacted) / (gram atomic mass of Cu) = 0.07g / 63.5gmol^-1 = 0.0011mol

5.Moles of AgNO3 reacted = (Mass of AgNO3 reacted ) / (molecular mass of AgNO3) = 1.18g / 170gmol^-1 = 0.0069 mol

6.Moles of Ag produced = (Mass of Ag produced ) / (gram atomic mass of Ag) = 0.81g / 108gmol^-1 = 0.0075mol

7. Ratio of moles Ag to moles AgNO3 = 0.0075mol / 0.0069 mol = 1.078

8)Ratio of moles Ag to moles Cu = 0.0075 mol /  0.0011g = 6.82

9)Number of Cu atoms removed from wire:

Number of Cu atoms present in 1 mol of Cu = 6.023x10²³ atomsmol^-1

Hence number of Cu atoms that would present in 0.0011mol of Cu = (6.023x10²³ atomsmol^-1)x0.0011mol = 6.63x10²⁰ atoms

10)Number of Ag atoms produced:

Number of Ag atoms present in 1 mol of Ag = 6.023x10²³ atomsmol^-1

Hence number of Ag atoms that would present in 0.0075mol of Ag = (6.023x10²³ atomsmol^-1)x0.0075mol = 4.5x10²¹ atoms

11)Theoretical yield of Ag:

The chemical reaction can be written as

Cu + 2AgNO3 --------------> 2Ag + Cu(NO3)2

1mol 2 mol 2 mol

2 mol of AgNO3 gives = 2 mol of Ag

=> 0.0069 mol of AgNO3 will produce mol of Ag = (2/2)x(0.0069 mol ) = 0.0069 mo

12)% Yield of Ag = Actual yield / theritical yield = 0.0075 / 0.0069 = 107.8 %