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$29.$

$a.$

$H_2NC_nH_{2n}COOH+HCl \to ClH_3NC_nH_{2n}COOH $

$b.$

Bảo toàn khối lượng:

$n_X=\dfrac{m_{muối}-m_X}{M_{HCl}}=\dfrac{12.55-8.9}{36.5}=0.1(mol)$

`->`$M_{X}=89(g/mol)$

`->`$CH_3CH(NH_2)COOH$

$30.$

$a.$$(C_{17}H_{35}COO)_3C_3H_5+3NaOH \xrightarrow{heat} 3C_{17}H_{35}COOH+C_3H_5(OH)_3$

$b.$ $C_6H_{12}O_6 \xrightarrow[35-45^oC]{enzim}2CO_2+2C_2H_5OH$

$c.$ $H_2NCH_2COOH+HCl \to ClH_3NCH_2COOH$

$d.$$C_6H_5NH_2+HCl \to C_6H_5NH_3Cl$

$31.$

$n_{ester}=\dfrac{55.5}{74}=0.75(mol)$

`-` Mà chỉ toàn ester đơn chức, mạch hở và no.

`->`$n_{Y}=n_{Z}=n_{NaOH}=n_{ester}=0.75(mol)$

`->`$m_{Y}=n_{ete}+\dfrac{1}{2}n_Y \times 18 $

$=21.45+\dfrac{18}{2} \times 0.75=28.2(gam)$

`-` Bảo toàn khối lượng:

$m_Z=m_{ester}+m_{NaOH}-m_{Y}$

$=55.5+40 \times 0.75-28.2=57.3(gam)$

$32.$

`-` Hỗn hợp gồm:

$C_6H_{12}O_6=(CH_2O)_6$

$CH_2O$

$CH_3COOH=(CH_2O)_2$

$HCOOCH_3=(CH_2O)_2$

`-` Quy về :

$CH_2O:a(mol)$

$n_{O_2}=0.2(mol)$

`-` Bảo toàn $e:$

$(4+2 -2) \times a=4n_{O_2}=0.8$

`=>`$a=0.2(mol)$

`-` Bảo toàn nguyên tố $carbon:$

$n_{CaCO_3}=n_{CO_2}=0.2$

`->`$m=0.2 \times 100=20(gam)$