$\bullet$ Cu đứng sau Hydrogen trong dãy hoạt động hoá học nên Cu không phản ứng với HCl.

Ta có: $\rm n_{H_2}=\dfrac{3,36}{22,4}=0,15\ (mol)$

Phản ứng xảy ra:

$\rm\quad\quad\;\;\; 2Al+6HCl\to 2AlCl_3+3H_2\\(mol)\;\;0,1\hspace{1,6cm}\leftarrow\hspace{2cm}0,15$

Vậy $\rm\%m_{Cu}=\dfrac{2,75}{10}×100\%=27,5\%$

$\rm\quad\;\;\%m_{Al}=\dfrac{0,1×27}{10}×100\%=27\%$

$\rm\quad\;\;\%m_{Al_2O_3}=100\%-(27,5+27)\%=45,5\%$

$\bullet$ Ta có: $\rm n_{Al_2O_3}=\dfrac{10×45,5\%}{102}=0,0446\ (mol)$

Phản ứng xảy ra:

$\rm\quad\quad\;\;\; Al_2O_3\:+\:6HCl\to 2AlCl_3+3H_2O\\(mol)\;0,0446\hspace{,0cm}\to\hspace{,0cm}0,2676$

$\rm n_{HCl}=0,2676+0,3=0,5676\ (mol)$

$\rm ⇒m_{dd\ HCl}=\dfrac{0,5676×36,5×110\%}{19,25\%}=118,38\ (gam)$

$\bullet$ Từ phản ứng: $\rm n_{AlCl_3}=0,1+0,0446×2=0,1892\ (mol)$

Bảo toàn khối lượng:

$\rm m_{dd\ AlCl_3}=10-2,75+118,38-0,15×2=115,33\ (gam)$

$\rm ⇒C\%_{AlCl_3}=\dfrac{0,1892×133,5}{115,33}×100\%=21,9\%$

$\rm ⇒C\%_{HCl\ (du)}=\dfrac{118,38×10\%}{115,33}×100\%=10,26\%$ 

$\rm m_Y=m_{Cu}=2,75(g)$

$\rm n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)$

$\rm 2Al+6HCl\to 2AlCl_3+3H_2$

$\rm Al_2O_3+6HCl\to 2AlCl_3+3H_2O$

$\rm n_{H_2}=\dfrac{2}{3}n_{H_2}=0,1(mol)$

`%`$\rm m_{Cu}$`={2,75}/{10}.100%=27,5%`

`%`$\rm m_{Al}$`={0,1.27}/{10}.100%=27%`

`%`$\rm m_{Al_2O_3}$`=100%-27,5%-27%=45,5%`

$\rm n_{Al_2O_3}$`={10.45,5%}/{102}~~`$\rm 0,045(mol)$

$\rm n_{HCl}=3n_{Al}+6n_{Al_2O_3}=0,57(mol)$

$\rm m_{dd\ HCl}$`={0,57.36,5}/{19,25%}.110%~~`$\rm 118,9(g)$

$\rm n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=0,19(mol)$

$\rm m_{dd\ spu}=10-2,75+118,9-0,15.2=125,85(g)$

$\rm C$`%`$\rm _{dd\ AlCl_3}$`={0,19.133,5}/{125,85}.100%=20,15%`

$\rm C$`%`$\rm _{dd\ HCl_{(dư)}}$`={118,9.10%}/{125,85}.100%=9,45%`