Gọi mol ancol=x molVancol= 60.15%=9ml$C_2H_5OH$ + $O_2$ \rightarrow $CH_3COOH$ + $H_2O$x mol. =>x molMol $CH_3COOH$ =0,009.1=0,009=xm$CH_3COOH$ =0,009.60=0,54(g)
Gọi mol ancol=x molVancol= 60.15%=9ml$C_2H_5OH$ + $O_2$ \rightarrow $CH_3COOH$ + $H_2O$x mol. =>x molMol $CH_3COOH$ =0,009.1=0,009=xm$CH_3COOH$ =0,009.60=0,54g
Gọi mol ancol=x molVancol= 60.15%=9ml$C_2H_5OH$ + $O_2$ \rightarrow $CH_3COOH$ + $H_2O$x mol. =>x molMol $CH_3COOH$ =0,009.1=0,009=xm$CH_3COOH$ =0,009.60=0,54
(g
)