Câu 3:A
$\%C=\dfrac{12n}{14n+32}.100=40\%$
=> $n=2$
=> $C_2H_4O_2$
có 1 đồng phân este : $HCOOCH_3$
Câu 4:D
$\%O=\dfrac{32}{14n+32}.100=31,37\%$
=> $n=5$
=> $C_5H_10O_2$
có 4 đồng phân có phản ứng tráng gương:
$HCOOCH_2CH_2CH_2CH_3$
$HCOOCH(CH_3)CH_2CH_3$
$HCOOCH_2CH(CH_3)_2$
$HCOOCH(CH_3)_3$
Câu 5:C
$n_{NaOH}=1.0,13=0,13(mol)=n_{este}$
$M_{este}=\dfrac{11,44}{0,13}=88(g/mol)$
=> $C_4H_8O_2$
$HCOOCH_2CH_2CH_3$
$HCOOCH(CH_3)_2$
$CH_3COOC_2H_5$
$C_2H_5COOCH_3$