Question

6.7 și 6.10
Dau coroana!!

Answer 1

6.7. CnH2n-2

24n/4n-4 = 7,5

6n = 30 => n = 5 ; C5H8

1-pentina si 2-pentina

b) 1-pentina + Na => acetilura monosodica de pentina + 1/2 H2

n = m/M = 69/23 = 3 moli Na

n Na = n 1-pentina = 3

m pentina = 3×68 = 204g
m 2-pentina = 272-204 = 68g

n = m/M = 68/68 = 1

n1/n2 = 3:1

c) In ambele reactii reactioneaza 2 moli de H2

n alchine total×n moli H2 = 4×4 = 16 moli H2

VH2 = 16×22,4 = 358,4L


6.10. CnH2n+2

%C = mC/M×100
81,81 = 12n/14n+2×100

Dupa aflare n = 3, C3H8

C3H8 -H2 -----> C3H6

nC3H6 ----> -[-H2C-CH-CH3-]-n


b) MC3H8 = 44g/mol
n =m/M = 1650/44 = 37,5 moli alcan (teoretic)
practic : 0,8×0,9×37,5 = 27 moli

n alcan = n polimer = 27 moli

m polimer = 27×42 = 1134g
( nu sunt sigur daca e asa)