Câu 1:

$n_{Cu}=\dfrac{9,6}{64}=0,15(mol);n_{Cl_2}=\dfrac{2,24}{22,4}=0,1(mol)$

$PTHH:Cu+Cl_2\xrightarrow{t^o}CuCl_2$

Xét tỉ lệ: $n_{Cu}>n_{Cl_2}\Rightarrow Cu$ dư

Theo PT: $n_{CuCl_2}=n_{Cl_2}=0,1(mol)$

$\Rightarrow m_{CuCl_2}=0,1.135=13,5(g)$

$\to m=13,5$

Câu 2:

$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2(1)$

$Al_2O_3+6HCl\to 2AlCl_3+3H_2O(2)$

$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$

$n_{AlCl_3}=\dfrac{53,4}{133,5}=0,4(mol)$

Theo PT(1): $n_{AlCl_3(1)}=n_{Al}=\dfrac{2}{3}n_{H_2}=0,2(mol)$

$\Rightarrow n_{AlCl_3(2)}=0,4-0,2=0,2(mol)$

Theo PT(2): $n_{Al_2O_3}=\dfrac{1}{2}n_{AlCl_3(2)}=0,1(mol)$

$\Rightarrow m_{Al_2O_3}=0,1.102=10,2(g);m_{Al}=0,2.27=5,4(g)$

$\Rightarrow \%m_{Al_2O_3}=\dfrac{10,2}{10,2+5,4}.100\%\approx 65,38\%$

$b)$ Từ cả 2 PT ta có: $\Sigma n_{HCl}=6n_{Al_2O_3}+3n_{Al}$

$\Rightarrow \Sigma n_{HCl}=6.0,1+3.0,2=1,2(mol)$

$\Rightarrow C\%_{HCl}=\dfrac{1,2.36,5}{131,4}.100\%\approx 33,33\%$