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

    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 %

law of conservation of matter

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.