I am working on the following lab: OXIDATION OF BENZYL ALCOHOL : for this experi
ID: 575045 • Letter: I
Question
I am working on the following lab: OXIDATION OF BENZYL ALCOHOL : for this experiment which reagent is the limiting reagent?
after I did the math the result showed me the benzyl alcohol as the limiting reagent
Please show work
"Place 150 mL of 0.5 N NaOH into a 500 mL three-neck round-bottom flask. Dilute the solution by adding 50 mL of distilled H2O. Add 2.2 g of benzyl alcohol into the flask and swirl well. Attach a water condenser to the center neck and stopper the other two necks. Attach a 500 mL heating mantle and begin gentle heating (Variac %) Weigh 6.3 g of KMnO4. Add approximately half of the permanganate to the flask through one of the side necks of the round bottom. Potassium permanganate is purple in aqueous solution."
Explanation / Answer
Ans. # Moles of benzyl alcohol (C6H5CH2OH) = Mass / Molar mass
= 2.2 g / (108.14 g/ mol)
= 0.0203 mol
# Mass of KMnO4added = half of 6.3 g = 3.15 g
Moles of KMnO4 added = 3.15 g/ (158.03 g/ mol) = 0.0200 mol
# Oxidation of benzyl alcohol (C6H5CH2OH) into benzoic acid (C6H5COOH) by KMnO4 takes place in the presence of alkaline medium (So, NaOH is the catalyst).
Balanced reaction:
3 C6H5CH2OH + 4 KMnO4 ----> 3 C6H5COOH + 4 MnO2 + H2O + 4 KOH
According to the stoichiometry of balanced reaction, 3 mol C6H5CH2OH reacts with 4 mol KMnO4.
Theoretical molar ratio of reactants = C6H5CH2OH: KMnO4 = 3 : 4
# Theoretical molar ratio of reactants = C6H5CH2OH: KMnO4 =
0.0203 mol : 0.0200 mol
= 3 : 2.99
Comparing the theoretical and experimental molar ration of the reactants, the moles of experimental moles of KMnO4 is less than its theoretical value 4 mol while that of benzyl alcohol is kept constant at 3.
So, KMnO4 is the limiting reactant (when 3.15 g of it is added).
Note: If all the 6.3 g KMnO4 is added to the reaction mixture, it would become the reagent in excess. Then, benzyl alcohol would the limiting reactant.
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