mọi người giả giúp e vs

Đáp án: 

Câu 1:

$HCOOCH_3$: metyl fomat 

$CH_3COOCH_3$: metyl axetat 

$(C_{17}H_{33}COO)_3C_3H_5$: triolein 

$HCOOC_2H_5$: etyl fomat 

$CH_3COOC_2H_5$: etyl axetat 

$(C_{17}H_{35}COO)_3C_3H_5$: tristearin 

Câu 2:

$HCOOCH_3+H_2O\buildrel{{H^+, t^o}}\over\rightleftharpoons HCOOH+CH_3OH$ 

$HCOOCH_3+NaOH\xrightarrow{{t^o}} HCOONa+CH_3OH$ 

$CH_3COOH+C_2H_5OH\buildrel{{H_2SO_4 đ; t^o}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O$ 

$(C_{17}H_{33}COO)_3C_3H_5+3NaOH\xrightarrow{{t^o}} 3C_{17}H_{33}COONa+C_3H_5(OH)_3$ 

$(C_{17}H_{33}COO)_3C_3H_5+3H_2\xrightarrow{{Ni, t^o}} (C_{17}H_{35}COO)_3C_3H_5$

Câu 3: 

$n_{NaOH}=0,1.1=0,1$ mol 

Đặt công thức este là $RCOOR'$ 

$RCOOR'+NaOH\to RCOONa+R'OH$ 

$\Rightarrow n_A=n_{RCOONa}=0,1$ mol 

$M_{RCOONa}=\dfrac{6,8}{0,1}=68=M_R+44+23$ 

$\Rightarrow M_R=1(H)$ 

$M_{RCOOR'}=\dfrac{7,4}{0,1}=74$ 

$\Rightarrow M_{R}+44+M_{R'}=74$ 

$\Rightarrow M_{R'}=29(C_2H_5)$ 

Vậy $A$ là $HCOOC_2H_5$