Se ard 44,8 cm cubi C2H6. Calculeaza:
- volumul de aer (cu 20% O2 in volume), masurat in conditii normale, consumat;
- numarul de moli de CO2 degajat;
- masa de apa obtinuta in urma reactiei de ardere.

n=V/Vm=44,8/22,4=2 mmoli C2H6

C2H6 + 7/2O2 = 2CO2 + 3H2O + Q

1 mmol C2H6...........3,5 mmoli O2
2 mmoli C2H6..........x=7 mmoli O2

VO2=22,4*7=156,8mL O2

Vaer=5*VO2=5*156,8=784mL=0,784L aer

1 mmol C2H6...................2 mmoli CO2
2 mmoli C2H6.................x=4 mmoli CO2

1 mmol C2H6........................3*18mg H2O
2 mmoli C2H6.........................x= Calculeaza !!! (masa in mg)

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Raoull Raoull · Answer 2 · 2016-09-26T20:16:03+03:00

_0.02moli_    _x=0.07moli_ _y=0.04moli_ _z=0.06moli_
   C₂H₆            +        7/2 O₂   ---------> 2CO₂    +    3H₂O
   1mol 3.5moli 2moli 3moli

[tex]V_{C_2H_6}=44.8 cm^3=0.448dm^3[/tex]

n C₂H₆ = [tex]\frac{V}{22.4}=\frac{0.448}{22.4}=0.02moli[/tex]

a) n O₂ = x = 3.5*0.02 = 0.07moli

n=[tex]\frac{V}{22.4}[/tex] ⇒ V = n*22.4 = 0.07*22.4 = 1.568 dm³ O₂

V aer = 5* V O₂ = 5*1.568 = 7.84 dm³

b) n CO₂ = y = 2*0.02 = 0.04moli

c) n H₂O = z = 3*0.02 = 0.06moli

n=[tex]\frac{m}M[/tex] ⇒ m = n*M = 0.06*18 = 1.08 g H₂O

*M H₂O = 2+16 = 18g/mol
__________________________________________________________________
n-----> nr. de moli
M----> masa molara
V-----> volum