how can i Calculate those ?? 1)Mass of Cu wire reacted 2)Mass of Ag produced 3)M
ID: 871909 • Letter: H
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
info.
mass of vial + AgNO3 : 13.34 g
mass of vial : 12.16 g
mass of copper wire : 1.12 g
mass of copper wire ( after reaction): 1.05 g
mass of beaker : 56.97 g
mass of beaker + silver produced: 57.78 g
Explanation / Answer
The Balanced equation for above reaction is
Cu + 2AgNO3 --> 2Ag + Cu(NO3)2
From the above equation we can say that, one mole of Cu will react with 2 moles of AgNO3 to produce 2 moles of Silver.
From given data,
Hence
1) Mass of Cu wire reacted: 1.12-1.05 = 0.07 g
2) Mass of Ag produced = 57.78-56.97 = 0.81 g
3) Mass of AgNO3 reacted = 13.34 - 12.16 = 1.18 g
4)Moles of Cu reacted = [Mass of Cu wire reacted] / [molecular weight of Cu]
= 0.07 / 63.5
= 1.1 x 10-3 moles
5)Moles of AgNO3 reacted = [Mass of AgNO3 reacted] / [molecular weight of AgNO3]
= 1.18 / 169.9
= 6.95 x 10-3 moles
6)Moles of Ag produced = [Mass of Ag produced] / [molecular weight of Ag]
= 0.81 / 107.87
= 7.51 x 10-3 moles
7)Ratio of moles Ag to moles AgNO3 = [7.51 x 10-3 moles] / [6.95 x 10-3 moles]
= 1 : 0.93
8)Ratio of moles Ag to moles Cu = [7.51 x 10-3 moles] / [1.1 x 10-3 moles]
= 1: 0.15
9)Number of Cu atoms removed from wire
1 mole Cu = 6.02 x 1023 Cu atoms from Avogadro's number
so 1.1 x 10-3 mole of Cu = 1.1 x 10-3 x 6.02 x 1023
= 6.622 x 1020 atoms
10)Number of Ag atoms produced
1 mole Ag = 6.02 x 1023 Ag atoms from Avogadro's number
so 7.51 x 10-3 mole of Ag = 7.51 x 10-3 x 6.02 x 1023
= 45.21 x 1020 atoms
11)Theoretical yield of Ag
We Know that one moles of AgNO3 will produce one moles of Silver.
Hence fro the above calculations we can say 6.95 x 10-3 moles of AgNO3 reacted hence 6.95 x 10-3 moles
of Ag will produce
12)% Yield of Ag = [7.51 x 10-3 moles] x 100/ [6.95 x 10-3 moles]
= 108.1%
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