Answer to Question #147008 in Chemistry for Jackens

Question

Mass of lead (II) oxide needed to deposit 20.561g of lead metal

Answer 1

Pb+O₂+PbO₂

n=m/M

n (PbO₂) = n (Pb)

M (Pb) = 207.2 g/mol

M (PbO₂) = 239.2 g/mol

n (Pb) = n (PbO₂) = 20.561/207.2 = 0.1 mol

m (PbO₂) = 0.1 x 239.2 = 24 g