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ce cantitate de must ce contine 20% glucoza este necesara pentu a obtine 2,07kg tuica de tarie 35% in conditiile unui randament de 90%

Răspuns :

180g. ...........2×40
C6H12O6=2C2H5OH+2CO2
x.....................724.5

2.07 kg = ms (aplicam concentratia)
md = 2.07 ×35/100 = 0.7245 kg = 724.5 g

x=1630.125 g C6H12O6 = mp (aplicam randamentul )
mt = 1630.125 × 100/90 = 1.81125 kg = md
ms = 1.81125 ×100/20= 9.056 kg must