Question

Prin combustia a 1.08 g de hidrocarbura gazoasa, care are densitatea fata de oxigen 2,25, s-au obtinut 1,68 dm3 de CO2. Sa se determine hidrocarbura si volumul de aer consumat pentru arderea unui mol de hidrocarbura.

Answer 1

d=masa molara hidrocarbura/masa molara oxigen
=> m.m.h = 32*2,25 = 72 g/mol
22,4 l CO2..........12 g C
1.68 l CO2..............x               
x = 0,9 g C
m H = 1,08 - m C = 1.08 - 0,9 = 0,18 g H
0,9/12 = 0,075
0,18/1 = 0.18
=> FB: CH2.4
FM: (CH2.4)n
M=72 => 12n+2,4n=72=>14,4n=72=> n=5 => FM: C5H12
C5H12 + 8 O2 = 5 CO2 + 6 H2O
1 mol hidrocarbura consuma 8 moli de oxigen
in aer se gaseste aproximativ 20% oxigen
=> V aer = 100/20 * V oxigen
V oxigen = 8*22,4=179,2 l oxigen
V aer = 5*179.2 = 896 l aer

Answer 2

dgaz=M/Mgaz==>M=d*M=2.25*32=72g/mol

m/M=V/Vm==>m=M*V/Vm=44*1.68/22.4=3.3g C

℅C=300a/11s=300*3.3/11*1.08=990/11.88=83.33
℅H=100-℅C=16.66

nC=℅C*M/AC*100=83.33*72/1200=5
nH=℅H*M/AH*100=16.66*72/100=12

F.m=C5H12 => n-pentan

C5H12 + 8O2 ---> 5CO2 + 6H2O + Q

nO2=8 moli

n=V/Vm==>V=8*22.4=179.2 L O2

Vaer=5*VO2=179.2*5=896 L ..