Answer to Question #227559 in Inorganic Chemistry for thasha

Question #227559

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.

Expert's answer

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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