Question

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.Suppose 100 g of PbS is reacted with 4g of C and 20g O₂.

Find the limiting reactant for this reaction

b.For question (b) above, calculate the mass of lead metal obtained.

Answer 1

a. M(PbS) = 239.22 g/mol

M(C) = 12.01 g/mol

M(O₂) = 32.0 g/mol

n(PbS) $= \frac{100}{239.22} = 0.418 \;mol$

n(C) $= \frac{4}{12.01} = 0.333 \;mol$

n(O₂) $= \frac{20}{32.0}=0.625 \;mol$

According to the reaction for two moles of PbS and two moles of C we need 3 moles of O₂. So,

theoretical proportion:

2 : 2 : 3 = 1 : 1 : 1.5

Real proportion:

0.418 : 0.333 : 0.625 = $\frac{0.418}{0.418} : \frac{0.333}{0.418} : \frac{0.625}{0.418}$ = 1 : 0.796 : 1.495

So, C is limiting reactant.

b. According to the reaction n(Pb) = n(C) = 0.333 mol

M(Pb) = 207.19 g/mol

m(Pb) $= 0.333 \times 207.19 = 68.99 \;g$

Answer: 67 g