Question

Same electric charge is passed through aq solution of Hcl&cuso4. If 12g of H2 is liberated, find the mass of copper deposited

Answer 1

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