Question

Net Equation: C6H10O + H2O + 2 MNO4- -----------> C6H10O + 2MnO2 + 2OH-

Calculate theoritcal yield and limiting reagent and percent yield:

0.7945grams of KMnO4

0.3642 grams cyclohexanone

water 15ml

Ended with a adipic acid product of 0.02 grams

Answer 1

C6H10O + H2O + 2 KMnO4 -----------> C6H10O + 2MnO2 + 2KOH

158 g of KMnO4 = 1 mole

0.7945 g of KMnO4 = (0.7945/158) = 5.028 * 10^-3 moles

98 g of C6H10O = 1 mole

0.3642 g of C6H10O = (0.3642/98) = 3.716 * 10^-3 moles

1 mole of Cyclohexanone requires 2 moles of KMnO4

3.716 milli moles of Cyclohexanone requires 2 * 3.716 milli moles of KMnO4

                                                                                 = 7.432 milli moles

hence from the above KMnO4 is limiting reagent.

2 mole of KMnO4 produces 1 mole of C6H10O4

5.028 milli moles produces (5.028/2) milli moles of C6H10O4

                                                 = 2.514 milli moles of C6H10O4

mass of C6H10O4 = (2.514 * 10^-3 * 146.14) g

                                    = 0.3673 g

percent yield = (0.02 / 0.3673) *100 = 5.445 %