need some help please?

Quantity of electricity = It = (4.79 A) × (1.10 × 3600 s) = 18970 C

1 mole of e⁻ carries an electricity of 96500 C
No. of moles of e⁻ passed = (18970 C) / (96500 C/mol) = 0.1966 mol

Pb²⁺(aq) + 2e⁻ ⇌ Pb(s)
Mole ratio e⁻ : Pb = 2 : 1
No. of moles of Pb plated = (0.1966 mol) × (1/2) = 0.0983 mol

Molar mass of Pb = 207.2 g/mol
Mass of Pb plated = (0.0983 mol) × (207.2 g/mol) = 20.4 g (3 sig. fig.)

4.79A = 4.79 C/s;
one coulomb is 6.24 x 10^18 electrons.
It takes two electrons transferred, to neutralize one lead(2) ion,
so each amp plates out 3.12 x 10^18 lead atoms per second.
Then 4.79A plates out 1.494 x 10^19 atoms per second,
or 5.918 x 10^22 atoms in 3960 seconds.

Finally, that's 0.09827 moles, or
(0.09827 mol)(207.2 g/mol) = 20.4 grams of lead.