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dane V=10dm3 etanolu, d=0,87g/cm3 = 0,87kg/dm3
........masa etanolu m=d*V = 10dm3*0,87kg/dm3 = 8,7 kg etanolu
1.utlenianie do etanalu
C2H5OH + 1/2 O2------>CH3-CHO + H2O
2.utlenianie do kwasu octowego
C2H5OH + O2----->CH3-COOH + H2O
3.utlenianie do CO2 i H2O czyli do całkowitego spalenia
C2H5OH + 3 O2----->2CO2 + 3H2O
C2H5OH.......... +............. n O2-------->
46kg.............................n*22,4m3
8,7kg..............................x
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x = n*22,4m3*8,7kg /46kg = 4,2365n m3 tlenu = 4 236,5n dm3
Do utlenienia zadanej ilości etanolu potrzeba :
1.gdy utlenianie do etanalu /n=0,5 /
V1 = 4 236,5 *0,5 dm3 = 2118,25 dm3 O2
2.gdy utlenianie do kwasu octowego
V2 = 4 236,5*1 dm3 = 4 236,5 dm3 O2
3.gdy utlenianie do CO2 i H2O
V3 = 4 236,5*3 dm3 = 12 709,5 dm3 O2
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