Question

Consider the balanced chemical reaction below and determine the percent yield for iron(III) sulfide if 6.37 moles of iron(III) bromide yielded 1.84 moles of iron(III) sulfide

2FeBr3+3Na2S -> Fe2S3+6NaBr

Answer 1

Percent yield = 57.7 %

Step-by-step 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 %