Giải giúp em bài hoá với ạ

$1)$

$a)$$\text{3-methylbutan-2-ol}$

$b)$

$CH_3CH(OH)CH(CH_3)_2\xrightarrow[170^oC]{H_2SO_4}\left[\begin{matrix} CH_3=CHCH(CH_3)_2(spc)\\ CH_3CH=C(CH_3)_2(spp)\end{matrix}\right. +H_2O$

$(spc)$ $3-methyl-1-butene$

$(spp)$ $2-methylbut-2-ene$ 

$c)$

$CH_3CH(OH)CH(CH_3)_2+CuO\xrightarrow{\Delta} CH_3C=OCH(CH_3)_2+Cu+H_2O$

Tên sản phẩm : $\text{ Methyl isopropyl ketone}$ hoặc $\text{3-methyl-2-butanone}$

$d)$

$n_{H_2}=\dfrac{1}{2}n_{alcohol}=1(mol)$

`=>`$V_{H_2}=24.79(L)$

$2)$

$(C_6H_{10}O_5)_n \xrightarrow nC_6H_{12}O_6 \to 2nC_2H_5OH+2nCO_2$

$n_{C_6H_{10}O_5}=\dfrac{m_{tb}}{165n}(mol)$

$n_{C_2H_5OH}=2n \times n_{C_6H_{10}O_5}$

$=2n \times \dfrac{m_{tb}}{165n}$

$=\dfrac{m_{tb}}{82.5}(mol)$

$m_{C_2H_5OH}=46 \times \dfrac{m_{tb}}{82.5}=\dfrac{92m_{tb}}{165}(gam)$

Giả sử $D_r=0.78(g/cm^3)$

$V_{C_2H_5OH}=\dfrac{1}{0.78} \times m_{C_2H_5OH}$

$=\dfrac{1}{0.78} \times \dfrac{92m_{tb}}{165}$

$=\dfrac{920m_{tb}}{1287}(cm^3)$

$V_r=\dfrac{\dfrac{920m_{tb}}{1287}}{30} \times 100$

$=\dfrac{\dfrac{920 \times 10 \times 10^3}{1287}}{30} \times 100$

$=23828(cm^3)$

$=23.828(L)$