heating an ore of antimony sulfide in the presence of iron gives th element antimony and iron(2) sulfide. when 15.0 antimony sulfide reacts with an excess of iron, 9.84 g antimony is produced. what is the percent yield of this reaction?
You need to calculate the theoretical yield (Y) starting from 15.0 (I presume g) of antimony sulfide. In order to do this you need a balanced equation, so you need to insert symbols and balance:
iron + antimony sulfide -> antimony + iron(II)sulfide.
To calculate the theoretical yield find the number of moles of antimony sulfide in 15.0 g.
15.0 g/(Molar mass of antimony sulfide)
then calculate the theoretical number of moles of antimony formed using the balanced equation.
Y (theoretical yield) is then
Y=moles of antimony x molar mass of antimony.
percentage yield
= (9.84 g x 100)/Y
To calculate the percent yield of a reaction, you need to know the actual yield and the theoretical yield. The actual yield is the amount of product obtained in the experiment, and the theoretical yield is the maximum amount of product that could be obtained based on stoichiometry.
In this case, the reaction is described as follows:
Antimony sulfide (Sb₂S₃) + Iron (Fe) → Antimony (Sb) + Iron(II) sulfide (FeS)
We are given that 15.0 g of antimony sulfide reacts to produce 9.84 g of antimony. We can assume that antimony is the limiting reactant because the amount produced is less than the initial amount of antimony sulfide.
Now, we need to determine the theoretical yield of antimony using stoichiometry:
1 mole of Sb₂S₃ reacts with 3 moles of Sb (according to the balanced equation). We can calculate the number of moles of Sb₂S₃ using its molar mass:
Molar mass of Sb₂S₃ = (2 × atomic mass of Sb) + (3 × atomic mass of S)
= (2 × 121.75 g/mol) + (3 × 32.06 g/mol)
= 339.29 g/mol
Moles of Sb₂S₃ = Mass of Sb₂S₃ / Molar mass of Sb₂S₃
= 15.0 g / 339.29 g/mol
≈ 0.0443 mol
Since the stoichiometric ratio is 1:3, the theoretical yield of antimony (Sb) is:
Theoretical yield of Sb = Moles of Sb₂S₃ × (3 moles of Sb / 1 mole of Sb₂S₃)
= 0.0443 mol × 3
= 0.1329 mol
Now we can calculate the percent yield:
Percent yield = (Actual yield / Theoretical yield) × 100
= (9.84 g / (0.1329 mol × molar mass of Sb)) × 100
Since the molar mass of antimony (Sb) is approximately 121.75 g/mol:
Percent yield = (9.84 g / (0.1329 mol × 121.75 g/mol)) × 100
≈ 60.7%
Therefore, the approximate percent yield of this reaction is 60.7%.