Mass of lead deposited of a current 3.7 ampere was passer through a solution of lead 2 chloride for 1 hour 30 minutes

Answer 1

Lead(II) chloride = PbCl₂

Pb²⁺ (aq) + 2e- ==> Pb(s)

3.7 amp x 1 Coulomb/s / amp = 3.7 C/s

1 hour 30 min = 90 min and 90 min x 60 sec / min = 5400 sec

5400 sec x 3.7 C / sec = 19,980 Coulombs passed

Converting this to moles of electrons, we have ...

19,980 C x 1 mol e- / 96,485 C = 0.207 mols of e-

Converting this to moles of Pb, we have ...

0.207 mol e- x 1 mol Pb / 2 mol e- = 0.1035 mol Pb

Converting this to mass (g) of lead, we have ...

0.1035 mol Pb x 207 g Pb / mol = 21.4 g Pb