Antimony is obtained by heating pulverized stibnite (Sb2S3) with scrap iron and drawing of the molten antimony from the bottom of the reaction vessel?

Sb2O3 + 3 Fe → 2Sb + 3FeS
Suppose 0.06 kg of stibnite and 0.25 kg of iron turnings are heated together to give 0.2 kg of Sb metal. Calculate:
1. The limiting reagent.
2. The percentage of excess reactant.
3. The fraction of iron reacted.
4. The percentage conversion of stibnite.

1 Answer
Ernest Z.
Dec 21, 2017

Warning! Long Answer. Here's what I got.

Explanation:

  1. Stibnite is the excess reactant.
  2. There is 400 % excess of iron.
  3. The fraction of iron reacted is 0.1.
  4. The stibnite was 500 % (!) converted to #"Sb"#.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

Step 1. Gather all the information in one place with molar masses above the formulas and everything else below the formulas.

#M_r:color(white)(mmmml) 339.72color(white)(ml)55.84color(white)(m)121.76#
#color(white)(mmmmmmll)"Sb"_2"S"_3 +color(white)(l) "3Fe" color(white)(l)→ "2Sb" + "FeS"#
#"Mass/g":color(white)(mmml)60color(white)(mmll)250#
#"Amt/mol:"color(white)(mml)0.18color(white)(mml)4.5#
#"Divide by:"color(white)(mmm)1color(white)(mmmll )3#
#"Moles rxn:"color(white)(mm)0.18color(white)(mml)1.5#

#"Moles of Sb"_2"S"_3 = 60color(red)(cancel(color(black)("g Sb"_2"S"_3))) × ("1 mol Sb"_2"S"_3)/(339.72 color(red)(cancel(color(black)("g Sb"_2"S"_3)))) = "0.18 mol Sb"_2"S"_3#

#"Moles of Fe" = 250 color(red)(cancel(color(black)("g Fe"))) × ("1 mol Fe")/(55.84 color(red)(cancel(color(black)("g Fe")))) = "4.5 mol Fe"#

Step 2. Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

#"Sb"_2"S"_3# is the limiting reactant because it gives fewer moles of reaction.

Step 3. Calculate the percentage of excess reactant

#"Fe"# is the excess reactant.

#"Moles of Fe used" = 0.18 color(red)(cancel(color(black)("mol Sb"_2"S"_3))) × ("3 mol Fe")/(1 color(red)(cancel(color(black)("mol Sb"_2"S"_3)))) = "0.53 mol Fe"#

#"Excess moles of Fe = (4.5 - 0.53) mol = 3.9 mol"#

#"% excess" = (3.9 color(red)(cancel(color(black)("mol"))))/(0.53 color(red)(cancel(color(black)("mol")))) × 100 % = 400 %"# (1 significant figure)

Step 4. Calculate the fraction of #"Fe"# reacted

We use 0.53 mol of the 4.5 mol of #"Fe"# provided.

#"Fraction of Fe reacted" = (0.53 color(red)(cancel(color(black)("mol"))))/(4.5 color(red)(cancel(color(black)("mol")))) = 0.1# (1 significant figure)

Step 5. Calculate the percent conversion of stibnite

#"Theoretical yield of Sb" = 0.18 color(red)(cancel(color(black)("mol Sb"_2"S"_3))) × (2 color(red)(cancel(color(black)("mol Sb"))))/(1 color(red)(cancel(color(black)("mol Sb"_2"S"_3)))) × "121.76 g Sb"/(1 color(red)(cancel(color(black)("mol Sb")))) = "43 g Sb"#

#"Actual yield = 200 g Sb"#

#"% conversion" = "actual"/"theoretical" × 100 % = (200 color(red)(cancel(color(black)("g"))))/(43 color(red)(cancel(color(black)("g")))) × 100 %#
#= 500 %# (1 significant figure)

You are one great chemist, to get a 500 % yield!

Note: The answers can have only one significant figure, because that is all you gave for the masses of stibnite and antimony.

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