Answer to Question #70287 in General Chemistry for Princess

Question

If you were to use 414.4g of Pb, how many g of S would you need and how many g of Pb(II) sulfide would you produce? I need the g of Sulfur and Lead. I am having a hard time figuring out how to set this up.

Answer 1

Pb + S = PbS
Mr(Pb) = 207.4 g/mol
Mr(S) = 32 g/mol
Mr(PbS) = 239.2 g/mol
m(Pb) = 414.4 g
m(S) = 414.4*32/207.2 = 64 g
m(PbS) = 414.4*239.2/207.2 = 478.4 g