Question

se ard 1,53 l alcan gazos ,masurat la 100*C si o atm, cu aer in exces.Gazele rezultate dupa condensarea apei sunt trecute prin 200ml sol. de NaOH 12% (q=1,14g/mol), reducand concentratia procentuala a sol. de NaOH la 6,54%(se formeaza sarea neutra)
Deter. f. m. a alcanului supus analizei si scrieti ecuatiile reactiilor care au loc

Answer 1

pV=nRT => n=pV/RT=1*1,53/(0,082*373)=0,05moli hidrocarbura
ms=d*Vs=1,14*200=228g solutie NaOH 12%
12/100=md/228 => md=27,36g NaOH
consideram ca se consuma x g NaOH
80g...........1mol.........106g........18g
2NaOH + CO2 -> Na2CO3 + H2O
x g..........a=x/80....1,325x....0,225x
mdfinal=27,36-x
msfinal=228-x+0,225x+1,325x=228+0,55x
6,54/100=(27,36-x)/(228+0,55x)
2736-100x=1491,12+3,597x
103,597x=1244,88 => x=12g NaOH consumati
a=12/80=0,15moli CO2
1mol......................................n moli
CnH2n+2 + (3n+1)/2 O2 -> nCO2 + (n+1)H2O
0,05moli...............................0,15moli
0,05n=0,15 => n=3
f.m: C3H8 - propanul