Question

1)calculati volumul de oxigen necesar pentru arderea de 100 g de propan
2)calculati masa alcoolului etilic si volumui dioxidului de carbon care se formeaza la fermentatia alcoolica a glucozei C6H12O6 cu masa de 18 kg
varog e urgent

Answer 1

1) MC3H8=44g/mol

n=m/M=100/44=2.27 moli C3H8

1mol.....5mol
C3H8 + 5O2 --> 3CO2 + 4H2O + Q
2.27mol...x=11.35 moli O2

n=V/Vmolar==>V=n*Vmolar=11.35*22.4=254.24 L O2

2) MCO2=44g/mol=44kg/kmol
MCH3-CH2-OH=46g/mol=46kg/kmol
MC6H12O6=180g/mol=180kg/kmol

n=m/M=18/180=0.1 kmol=100 moli

1kmol............2kmol.................2kmol
C6H12O6 --> 2CH3-CH2-OH + 2CO2
0.1kmol.......x=0.2kmol............y=0.2kmol

n=m/M==>m=n*M=0.2*46=9.2 kg CH3-CH2-OH

n=V/Vmolar==>V=n*Vmolar=0.2*22.4=4.48m3 CO2