\(n_{CH_3COOCH_3}=\dfrac{1,48}{74}=0,02\left(mol\right)\)
PTHH: CH3OH + CH3COOH --> CH3COOCH3 + H2O
_______0,02<--------------------------0,02
=> \(m_{CH_3OH\left(pthh\right)}=0,02.32=0,64\left(g\right)\)
=> \(m_{CH_3OH\left(thựctế\right)}=\dfrac{0,64.100}{25}=2,56\left(g\right)\)