Question

nu stiu cum e cu raportul ala

Answer 1

n1 moli  CH2=CH2 + [O] + H2O ⇒ HO-CH2-CH2-OH
n2 moli  CH2=CH-CH3 + [O] + H2O ⇒ HO-ch2-CH(OH)-CH3
2KMnO4 + H2O ⇒ 2KOH + 2MnO2↓ + 3[O]
n1 + n2 = 134,4/22,4 = 6 mmoli
n1/n2 = 1/2    (n1 +n2)/n2 = 3/2    6/n2 = 3/2   n2 = 4mmoli propena
n1 = 2mmoli etena
nKMnO4 = 3(n1 + n2)/2 = 9mmoli    Vs = n/cM = 9/0,02 = 450ml
c) nMnO2 = nKMnO4 = 9      m = 9·87 = 783 mg  = 0,783g