Antimony is obtained by heating pulverized stibnite (Sb 2 S 3 ) with scrap iron

Question

Antimony is obtained by heating pulverized stibnite (Sb 2 S 3 ) with scrap iron and drawing off the molten antimony from the bottom of the reaction vessel. Sb 2 S 3 + 3Fe 2Sb + 3FeS Suppose that 0.600 kg of stibnite and 0.250 kg of iron turnings are heated together to give 0.200 kg of Sb metal. Determine:

a. The limiting reactant

b. The percentage of excess reactant

c. The degree of completion (fraction)

d. The percent conversion based on Sb 2 S 3

e. The yield in kg Sb produced/kg Sb 2 S 3 fed to the reactor

Explanation / Answer

Sb2S3(s) + 3Fe(s) -----> 2Sb(s) + 3FeS(s)

0.600 kg of stibnite =   600 / 339.715 = 1.76618 Moles

0.25 kg of Iron = 250 / 55.84 = 4.4766 Moles

Limiting reagent is Iron

4.4766 Moles of iron wiil react with 0.0014922 Moles (4.4766/3) of stibinite to give 2.984 Moles (4.4766 x 2) of Sb thoretically.

0.002984 Moles of Sb =    2.984 x 121.76 = 363.3 gm is the theoretical yield

The percent conversion based on Sb2S3 = 506.9 x 100 / 600 = 84.48 %

The percentage of excess reactant = 100 -84.48 = 15.51 %

The yield in kg Sb produced/kg Sb 2 S 3 fed to the reactor = 0.5069 / 0.6 = 0.8448

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