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Same electric charge is passed through aqua solution of HCL and CuSo4. if 12 gram of H2 is liberated find the mass of copper deposited 1. 402 g 2. 360 g 3. 381 g 4. 369 g

Same electric charge is passed through aqua solution of HCL and CuSo4. if 12 gram of H2 is liberated find the mass of copper deposited
1.    402 g
2.    360 g
3.    381 g 
4.    369 g

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Grade:12

4 Answers

Arun
25750 Points
4 years ago
W1 = (A1 X I X t) ÷ (n1 x 96500)
W1 is mass of H2
12 = (1 X I X t) ÷ (2 x 96500)
12 x 193000 = I X t ---------------- equation 1
W2 = (A X I X t) ÷ (n2 x 96500)
W2 is the mass of Cu
W2 = (63.5 X 12 x 193000) ÷ 193000 = 762g/2 = 381
 
Vikas TU
14149 Points
4 years ago
2 mole of electron will liberate 1 mole of H2 gas 
Mole of H2 liberated = 12/2 = 6 
To liberate 6 mole of e we require 12 mole of e 
To liberate 2 mole of e we need 1 mole of cu 
12 mole of e will deposit 6 mole of cu 
Mass of cu = 6* 63.5 = 381 
Shafi Shaik
15 Points
3 years ago
12 moles of H2 electron deposit 12/6=6 mole Cu.Hence, mass of copper deposited will be 6×63.5=381gram.
Anuranjan
13 Points
one year ago
Amount of H2 liberated = 12g 
Moles of electron required to liberate 12g of H2 = given mass/molar mass = 12/2=6
Now , 6 mole of H2 require 12 mole of electron. 
For  2 mole of electron we require 1 mole of copper. 
Hence, 12 mole of electron will deposit 6 mole of copper. 
So finally , mass of copper deposited = 6× 63.5=381

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