Am obtinut 13,2g de acetaldehida cu r=60% => a ramas etanol neoxidat, deci aducem masa de acetaldehida la r=100, cat s-ar obtine teoretic, daca s-ar oxida tot alcoolul
60/100=13,2/mteoretica => mteoretica=13,2*100/60=22g etanal
........................1mol.....44g
CH3-CH2-OH + [O] -> CH3-CH=O + H2O
.......................0,5moli..22g
1mol..............................3moli
K2Cr2O7 + 4H2SO4 -> 3[O] + K2SO4 + Cr2(SO4)3 + 4H2O
0,66moli.......................0,2moli
Cm=n/Vs => Vs=n/Cm=0,33/0,66=0,5L K2Cr2O7