Lead metal is obtained from lead sulfide through the following reaction:
2PbS(s) + 2C(s) + 3O2(g) → 2Pb(s) +2CO(g) + 2SO2(g)
(a)If 100g of PbS is reacted with excess C and O2, calculate the mass of lead metal obtained. ( 5 )
(b)Suppose 100 g of PbS is reacted with 4g of C and 20g O2.
Find the limiting reactant for this reaction.
( 4 )
(c)For question (b) above, calculate the mass of lead metal obtained. ( 4 )
(d)By referring to the question (c) above, If 31.2g Pb is obtained as the yield. Calculate the percentage of yield. ( 2 )
(e)Calculate mass of oxygen is needed to produce 500g of lead metal.
Molar mass of PbS = 239.3
= 0.418 moles
Mole ratio = 2:2
= 0.418 moles of Pb
Molar mass of Pb = 207.2