Question

How many moles of electrons are required to produce 57.1 g of lead metal from a solution of aqueous lead(IV)nitrate

Answer 1

Pb(NO3)4 ------> Pb4+ + 4NO3-

Pb4+ + 4e- -----> Pb

We see that to convert 1 Pb4+ ion to Pb, 4 electrons are needed

Given that mass of Pb required = 57.1 g

We know that atomic mass of Pb = 207.2 g / mol

=> Moles of Pb = 57.1 / 207.2 = 0.2756 moles

We know that 1 mole of Pb contains 6.022 x 10^23 (avagadro number) of atoms

=> Number of atoms in 0.2756 moles Pb = 0.2756 x 6.022 x 10^23 = 1.66 x 10^23 atoms

Numer of electrons needed = 1.66 x 10^23 x 4 = 6.638 x 10^23

=> Moles of electrons needed = 6.638 x 10^23 / 6.022 x 10^23 = 1.1 moles