Answer:
Percent yield = 57.7 %
Explanation:
Given Data:
moles of iron(III) bromide = 6.37 mol
actual yield of iron(III) sulfide = 1.84 mole
percent yield for iron(III) sulfide = ?
Reaction Given
2 FeBr₃ + 3 Na₂S ------------> Fe₂S₃+ 6 NaBr
Solution:
First we have to know the theoretical yield
We Know the Information given in the reaction
2 FeBr₃ + 3 Na₂S ------------> Fe₂S₃+ 6 NaBr
2 mol 3 mol 1 mol
Now,
if two mole of iron(III) bromide (FeBr₃) give 1 mole of iron(III) sulfide (Fe₂S₃)
Then how many moles of Fe₂S₃ will be produced if 6.37 moles of FeBr₃ will be used
Apply the unity formula
2 mol of FeBr₃ ≅ 1 mol of Fe₂S₃
6.37 mol of FeBr₃ ≅ ? mol of Fe₂S₃
By doing cross multiplication
no. of mol of Fe₂S₃ = 1 mol x 6.37 mol/ 2 mol
no. of mol of Fe₂S₃ = 3.185 mol
So,
The theoretical yield is 3.19 mol
Formula Used to find Percent yield
Percent yield = Actual yield/ theoretical yield x 100%
Now put all values in above equation
Percent yield = 1.84 mol / 3.19 yield x 100 %
Percent yield = 0.577x 100%
Percent yield = 57.7 %
So the Percent yield of Iron(III) Sulfide = 57.7 %
