Question

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 O_2, 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.

Answer 1

(a) M(PbS) = 239.3 g/mol

n(PbS) $= \frac{100}{239.3} = 0.417 \;mol$

According to the reaction

n(Pb) = n(PbS) = 0.417 mol

M(Pb) = 207.2 g/mol

m(Pb) $= 0.417 \times 207.2 = 86.40 \;g$

(b) M(C) = 12.01 g/mol

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

M(O₂) = 32.0 g/mol

n(O₂) $= \frac{20}{32.0} = 0.625 \;mol$

Theoretical proportion:

PbS : C : O₂

2: 2: 3

Real proportion:

0.417: 0.333: 0.625

So, $0.333<0.417 < \frac{3}{2} \times 0.625$

C is a limiting reactant.

(с) According to the reaction

n(Pb) = n(PS) = 0.333 mol

m(Pb) $= 0.333 \times 207.2 = 68.99 \;g$

(d) Proportion:

68.99 g – 100 %

31.2 g – x

$x = \frac{31.2 \times 100}{68.99} = 45.22 \; \%$

The percentage of yield is 45.22 %

(e) n(Pb) $= \frac{500}{207.2}= 2.41 \;mol$

According to the reaction

n(O₂) $= \frac{3}{2}n(Pb) = \frac{3}{2} \times 2.41 = 3.619 \;mol$

m(O₂) $= 3.619 \times 32.0 = 115.8 \;g$