Answer to Question #264035 in Inorganic Chemistry for lilly

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) 2 mol*239.3 g/mol PbS - 2 mol*207.2 g/mol Pb

100 g PbS - x g Pb

x = (100*2*207.2)/(2*239.3) = 86.6 g

(b) n(PbS) = 100 / (2*239.3) = 0.21 mol

n(C) = 4 / (2*12) = 0.17 mol

n(O₂) = 20 / (3*32) = 0.21 mol

C is limiting reactant

(c) 2 mol*12 g/mol C - 2 mol*207.2 g/mol Pb

4 g C - x g Pb

x = (4*2*207.2) / (2*12) = 69.1 g

(d) percent yield = 31.2 g /69.1g * 100% = 45.2%

(e) 3 mol*32 g/mol O₂ - 2 mol*207.2 g/mol Pb

x g O₂ - 500 g Pb

x = (500*3*32) / (2*207.2) = 115.8 g