Answer to Question #204304 in General Chemistry for Joyce Massu

Question #204304

On mélange 8.6g de toluène, C7H8 (Masse molaire = 92g/mol) avec 59.08 g de permanganate de potassium, KMnO4 (Masse molaire = 158g/mol). Cette réaction a un pourcentage de rendement en benzoate de potassium, KC7H502 (Masse molaire = 160g/mol), inférieur à 100%.

C7H8(l) + 2 KMnO4(aq) KC7H502(aq) + 2MnO2(S) + KOH(aq) + H2O(l)

a) Détermine le RL.


1
Expert's answer
2021-06-08T22:57:01-0400

Molar mass C7H8 = 92.14 g/mol

Molar mass KC7H5O2 = 160.21 g/mol

Percent yield = Actual yield/Theoretical yield × 100%

Theoretical yield of KC7H5O2 = actual yield/percent yield × 100% = 11.4g/65% × 100% = 17.54 g KC7H5O2

According to the reaction equation from 1 mol of C7H8 be produced 1 mole of KC7H5O2.

Mass of C7H8 = 17.54g KC7H5O2/160.21g/mol KC7H5O2 × 92.14g/mol C7H8 = 10.09 g C7H8


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