$8ml=0,008l$
Gọi CTTQ của este là RCOOR', (este không no vì tạo ra andehit => R' ko no)
$n_{KOH}=0,008.2,5=0,02(mol)$
$n_{Ag}=\dfrac{27}{108}=0,025(mol)=2n_{andehit}$
$⇒n_{andehit}=\dfrac{0,025}{2}=0,0125(mol)=n_{este}=n_{muối}$
$⇒n_{KOH}dư=0,02-0,0125=7,5.10^{-3}(mol)$
$M_{este}=\dfrac{1,425}{0,0125}=114(g/mol)$
$m_{muối}=1,645-7,5.10^{-3}.56=1,225(g)$
$M_{muối}=\dfrac{1,225}{0,0125}=98(g/mol)$
$M_R=98-44-39=15(g/mol)⇒R:CH_3-$
$M_{R'}=114-15-44=55(g/mol)$
$⇒R':CH=CH-CH_2-CH_3, CH=C(CH_3)-CH_3$
$CTCT:CH_3COOCH=CH-CH_2-CH_3$
$\text{ hoặc }CH_3COOCH=C(CH_3)-CH_3$