Question

In urma arderii unui amestec echimolecular de metanol si etanol se obțin 6, 72 dm3 de CO2. Calculează compoziția amestecului in procente de masa si volum de aer (20%O2), măsurat in condiții normale, necesar arderii amestecului de alcooli.

Answer 1

x=y
MCH3-OH=12+3+16+1=32g/mol
MCH3-CH2-OH=46g/mol

n=V/Vmolar=6.72/22.4=0.3 moli CO2 ==>0.15 moli CH3-OH si 0.15 moli CH3-CH2-OH

n=m/M==>m=n*M=0.15*32=4.8g CH3-OH
n=m/M==>m=n*M=0.075*46=3.45g CH3-CH2-OH

mamestec=4.8+3.45=8.25g

%CH3-OH=m/mamestec*100=4.8/*100=58.18
%CH3-CH2-OH=m/mamestec*100=3.45/8.25*100=41.81

1mol..........................1mol
CH3-OH + 3/2O2 --> CO2 + 2H2O + Q
x=0.15mol.................0.15mol

1mol.................................2mol
CH3-CH2-OH + 3O2 --> 2CO2 + 3H2O + Q
x=0.075mol.....................0.15mol

1mol..........1.5mol
CH3-OH + 3/2O2 --> CO2 + 2H2O + Q
0.15mol.....x=0.225mol

n=V/Vmolar==>V=n*Vmolar=0.225*22.4=5.04 L O2

Vaer=5*VO2=5*5.04=25.2L aer

1mol..................3mol
CH3-CH2-OH + 3O2 --> 2CO2 + 3H2O + Q
0.075mol..........x=0.225mol

n=V/Vmolar==>V=n*Vmolar=0.225*22.4=5.04 L O2

Vaer=5*VO2=25.2 L aer

Volumul de aer necesar arderii amestecului de alcooli este ==> V1+V2=25.2+25.2=50.4 L aer