2.
Lead metal is obtained from lead sulphide 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.
(b) Suppose 100 g of PbS is reacted with 4g of C and 20g O2.
Find the limiting reactant for this reaction.
(c) For question (b) above, calculate the mass of lead metal obtained.
(d) If 31.2g Pb is obtained as the yield. calculate the percentage of yield.
(e) Calculate mass of oxygen is needed to produce 500g of lead metal.
Molar mass of PbS = 239.3
= 100/239.3
= 0.418 moles
Mole ratio = 2:2
= 0.418 moles of Pb
Molar mass of Pb = 207.2
= 207.2×0.418
= 86.59g
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