Limiting reagent question

Question

In aqueous solution, ferric sulfate [Fe2(SO4)3] can be reduced to ferrous sulfate [FeSO4] by the action of sulfur dioxide [SO2] gas (as depicted in the following unbalanced chemical equation) = Fe2(SO4)3 + H20 + SO2 ? FeSO4 + H2SO4. How many moles of sulfur dioxide are required for the treatment of 12.40 mL of 0.4813 M Fe2 (SO4)3 solution in order to produce 0.7300 g of FeSO4?

 

A. 5.968 x 10^-3 mol

B. 2.403 x 10^-3 mol

C. 4.805 x 10^-3 mol

D. This is not possible because Fe2(SO4)3 is the limiting reagent

Answer 1

Seung,

The easiest way to solve this type of problem is dimensional analysis, which in chemistry is lumped in with stoichiometry. The amount of ferric sulfate put in will be converted to moles. Then based on the "balanced" reaction, we'll know the number of moles of sulfur dioxide that react with it as well as the moles of ferrous sulfate that are produced. So first we balance the equation.

 

               Fe₂(SO₄)₃ + 2H₂O + SO₂ → 2FeSO₄ + 2H₂SO₄

 

Now you can see that one mole of ferric sulfate reacts with one mole of sulfur dioxide to produce two moles of ferrous sulfate. For brevity, I'm going to abbreviate ferric sulfate as FC and ferrous sulfate as FS in the dimensional analysis.  We'll need to know the molar mass of the ferrous sulfate.

FeSO₄

Fe   55.847 * 1 = 55.847

S     32.06  * 1 = 32.06

O    15.9994*4 = 63.996

------------------------------

molar mass     = 151.903 g

 

 

12.4 mL FC          L        .4813 moles FC   2 moles FS    151.903 g FS

-------------- x ------------ x ------------------- x -------------- x ----------------   = 1.81 g FS

         1            1000 mL             L            1 mole FC       mole FS

 

This shows us that the amount of ferric sulfate given, if the sulfur dioxide is equal or greater, will produce more than the desired .73 g of ferrous sulfate. So we redo the computation based on the desired amount of ferrous sulfate.

 

 

.7300 g FS     mole FS      1 mole SO₂

------------- x -------------- x --------------  =  2.403 x 10^-3  moles SO₂

        1       151.903 g FS  2 moles FS

 

I hope that helps.