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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

Junaid , 6 Years ago
Grade 12
anser 4 Answers
Arun
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
 
Last Activity: 6 Years ago
Vikas TU
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 
Last Activity: 6 Years ago
Shafi Shaik
12 moles of H2 electron deposit 12/6=6 mole Cu.Hence, mass of copper deposited will be 6×63.5=381gram.
Last Activity: 5 Years ago
Anuranjan
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
Last Activity: 3 Years ago
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