Question

In timpul functionarii acumulatorului de plumb se cedeaza 0,018 moli de electroni. Cu cat scade masa anodului de plumb?

Answer 1

la Anod: Pb + HSO4(-) => PbSO4 + H(+) + 2 e(-)

asadar: 
1mol Pb cedeaza 2 mol e(-) (pentru a forma ionul Pb(2+) )
a mol Pb cedeaza  0.018 mol e(-)
a= (0.018 x 1)/2 => a=0.009 mol Pb

m Pb= n Pb x M Pb => 0.009 x 207.2 =1.865 g Pb.
M Pb= 207.2 g/mol
n Pb = a= 0.009 mol Pb