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

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 %