Question

ce cantitate de precipitat se obtine prin oxidarea blanda a 58,8 grame 1-butena,in proportie de 90%?

Answer 1

3CH2=CH-CH2-CH3 +2KMnO4 +4H2O---> 2MnO2 ↓ +2KOH+
                                              CH2OH-CHOH-CH2-CH3   (DIOL)
M,= (4X12gC+8gH)/mol= 56g/mol  1 butena
M= (55gMn+32gO)/mol= 87g/mol MnO2
-CALCULEZ masa de butena ce se oxideaza:  8758,8x90/100=53g
-din ecuatie
3x56g butena......2x87g MnO2
53g..........................m=55g