Question

1.Prin arderea a 22,4 mL (c.n) dintr-un alcan gazos se obtin 0,132 g CO2. Determina formula moleculara a hidrocarburii.

2.Prin arderea a 0,58 g alcan se formeaza 0,896 L CO2.Stabileste formula moleculara a alcanului.

Va rog ,urgent!Multumesc anticipat!

Answer 1

1.
1mol.................................x mol
CnH2n+2 + 3n+2/2 O2 --> nCO2 + (n+1) H2O + Q
0.001mol...............................0.003mol
0.003=0.001x==>x=0.003=3 moli

C3H8-propan

n=V/Vmolar=22.4/22.4=1milimoli=0.001 moli alcan

n=m/M=0.132/44=0.003 moli CO2

2.

14n+2g...............................n*44g
CnH2n+2 + 3n+2/2 O2 --> nCO2 + (n+1) H2O + Q
0.58g..................................1.76g

n=(14n+2)*1.76=44n+0.58==>24.64n+3.52=25.52n==>0.88n=3.52==>n=4

C4H10-butan


n=V/Vmolar=0.896/22.4=0.04 moli

n=m/M==>m=n*M=0.04*44=1.76g

MCO2=44g/mol