I am struggling with this problem on molarity

Question

33.0 mL of aqueous 0.878M potassium oxide solution is mixed with a 25.0mL solution of 0.699M iron( III) nitrate, producing a precipitate.

 

How do I:

 

A. Write a balanced net equation for the reaction

 

B. Calculate the concentration of the potassim ions after the two solutions have been mixed

 

C. Determine the limiting reagent

 

D. Calculate the concentration of the excess ion after the reaction has completed

 

E. Determine the mass of the precipitate

 

 

Answer 1

This problem involves a rather long, complex series of calculations.  I'm quite sure the approach is correct, but you may want to check my calculations, as I am tired and there may be an error or two.  After all, I'm only human.

 

A.  3K2O(aq) + 2Fe(NO3)3(aq)  ===> 6KNO3(aq) + Fe2O3(s) = balanced equation

Net ionic:  2Fe³⁺(aq) + 3O^₂-(aq) ===> Fe₂O₃(s)

 

B.  moles K₂O present = 0.033 L x 0.878 mole/L = 0.03897 moles

moles Fe(NO₃)₃ added = 0.025 L x 0.699 mole/L = 0.01748 moles

From balanced eq. 3 moles K2O react with 2 moles Fe2(NO3)2, thus moles K2O reacted = 0.03897 moles K2O x 2 moles Fe2(NO3)3/3 moles K2O = 0.02598 moles K2O that reacted.

Concentration of K+ ions after mixing = 0.03897 mol K2O - 0.02598 moles = 0.01299 moles K2O remaining

0.01299 moles K2O = 0.02598 moles K+

Final volume = 33.0 ml + 25.0 ml  = 58.0 ml = 0.058 L

Final Concentration of K+ ions = 0.02598 moles K+/0.058 L = 0.4479 M

 

C. Limiting reagent is Fe(NO₃₎₃ because from the balanced eq. it runs out first.

 

D.  The excess ion will be K⁺ and O^2-, and the [K⁺] was answered in (B) above. The [O^2-] will be1/2 that.  Not really sure what part D is asking.

 

E.  Mass of precipitate, Fe₂O₃:  0.01748 moles Fe(NO₃)₃ x 1 mole Fe₂O₃/2 moles Fe(NO₃)₃ = 0.00874 moles Fe₂O₃

0.00874 moles Fe2O3 x 160 g/mole = 1.398 g