Oblicz ile kilogramów glukoazy uległo fermentacji alkoholowej, jezeli otrzymano 11,5 kg etanolu, a wydajnosc procesu wynosiła 84%
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C6H12O6 --bakterie--> 2 C2H5OH + 2CO2
1 mol C6H12O6 --- 2 mole C2H5OH
180g C6H12O6 --- 92g C2H5OH
X --- 11,5kg = 11500g C2H5OH
x = 22500g = 22,5kg
22,5kg --- 84%
x --- 100%
x = 26,79 kg