5 pounds of bismuth (MW = 209) is heated along with 1 pound of sulfur to form Bi2S3 (MW = 514) through this rxn: 2Bi + 3S --> Bi2S3. At the end of the reaction, the mass is extracted and the free sulfur recovered is 5 % of the reaction mass. Determine:

a.the limiting reactant

b.the percent excess reactant

c.the percent conversion of sulfur to Bi2S3

Answer:

2 Bi + 3 S ---> Bi₂S₃

a) given , amount of Bi = 5 pounds = 2267.96 g [ as 1 lb = 453.59 g]

hence , the moles of Bi =2267.96 / 209 = 10..85 [ moles = weight / mol. weight]

similarly, amount of S = 1lb =453.59 g

molecular weight of sulphur = 32 g/mol

moles of S = 453.59 /32 = 14.174

2 moles of Bi requires 3 moles S for complete reaction,

hence, 10.85 moles of Bi requires = 10.85 *3 /2 = 16.275 moles of S

but given moles of S are lesser than required amount.

Hence, Sulphur is limiting reactant.

b)

now, we have only 14.174 mole of S

for these moles of S, amount of Bi reacted = 14.174* 2/3 = 9.449

hence, % excess reacctant = (given amount - required amount ) / required amount *100

= (10.85- 9.449 ) /9.449 *100 = 14.82 %

c)

moles of S that will come out of a reactor = 5 % of inital reaction mass 5% of total mass of reactants

= 0.05 *( mass of Bi + mass of S) = 0.05* ( 5+1) =0.3lb = 136.179 g [ 1 lb =453.93g]

moles of S coming out of the reactor = 136.179/32 = 4.225 moles

rconversion = (total initial moles of S fed - total moles of S coming out ) *100 / total initial moles of S fed

= (14.174 -4.225) *100 /14.174

= 69.97 % = 70 %