Question

calculeaza masa alcoolului etilic si volumul dioxidului de carbon care se formeaza la fermentarea alcoolica a glucozei C6H12O6 cu masa de 18 kg
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Answer 1

MC6H12O6=180g/mol
MC2H6O=46g/mol
MCO2=44g/mol

180Kg................2*46Kg
C6H12O6 ---> 2CH3-CH2-OH + 2CO2
18Kg..................x=9.2Kg

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

1kmol glucoza......................2kmoli CO2
0.1kmol................................x=0.2kmoli CO2

n=V/Vmolar ==>V=n*Vmolar=0.2*22.4=4.48 m3 CO2=4480L CO2