Question

Imi puteti rezolva problema 4?

Answer 1

M CO2=44  M H2O=18
44gCO2-----------12gC
6,6g----------------------X
X=1,8gC
18gH2O-----------2g H
3,6g-----------------Y
Y=0,4gH
2,2g subst.-------------1,8gC----------------0,4gH
100g-----------------------a-----------------------------b
a=81,81%C    b=18,18% H
C 81,18/12=6,81
H 18,18/1=18,18
C 6,81/6,81=1
H 18,18/6,81=2,66
⇒FB (CH2,66)n
12n+2,66n=44
n=3 ⇒FM C3H8(propan)
C3H8 +5O2⇒3CO2+4H2O