Ecuația reacției de ardere
x 0.5
CₙH₂ₙ₊₂O + (3n+1)/2 O₂ --------> n CO₂ + (n+1) H₂O + Q
1 n
Ecuația reacției cu sodiul:
0.75 x
C₂H₆O + Na --------> C₂H₆O⁻Na⁺ + 1/2 H₂
1 1
sau
H₃C - CH₂ - OH + Na --------> H₃C - CH₂ - O⁻Na⁺ + 1/2 H₂
Rezolvarea problemei:
[tex]m_{CO_2}=22g\\M_{CO_2}=12+(2*16)=12+32=44\ g/mol\\\\n=\frac{m}{M}=\frac{22}{44}=0.5\ moli\ CO_2\\\\n_{C_nH_{2n+2}O}=x=\frac{0.5}{n}\ moli\\m_{C_nH_{2n+2}O}=11.5\ g\\\\n=\frac{m}{M}\implies M=\frac{m}{n}=\frac{11.5}{\frac{0.5}{n}}=11.5*\frac{n}{0.5}=23n\ g/mol\\\\M_{C_nH_{2n+2}O}=nA_C+(2n+2)A_H+A_O=12n+2n+2+16=(14n+18)\ g/mol\\\\14n+18=23n\\23n-14n=18\\ 9n=18\implies n=2\implies\ C_2H_6O\ (alc.\ etilic/etanol)[/tex]
[tex]m_{C_2H_6O}=34.5\ g\\M_{C_2H_6O}=2*12+6*1+16=24+6+16=46\ g/mol\\\\n=\frac{m}{M}=\frac{34.5}{46}=0.75\ moli\\\\n_{Na}=x=0.75\ moli\\M_{Na}=23\ g/mol\\n=\frac{m}{M}\implies m=n*M=0.75*23=17.25\ g\ Na[/tex]
