Đáp án + Giải thích các bước giải:

$a)$ Đặt $n_{Al}=x(mol);n_{Fe}=y(mol)$

$\to 27x+56y=11(1)$

$n_{H_2}=\frac{9,6}{24}=0,4(mol)$

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

$Fe+2HCl\to FeCl_2+H_2$

Theo PT: $1,5x+y=0,4(2)$

Từ $(1)(2)\to x=0,2(mol);y=0,1(mol)$

$\%m_{Al}=\frac{0,2.27}{11}.100\%=49,09\%$

$\%m_{Fe}=100-49,09=50,91\%$

$b)$ $n_{AlCl_3}=x=0,2(mol);n_{FeCl_2}=y=0,1(mol)$

$\to m_{muối}=0,2.133,5+0,1.127=39,4(g)$

$c)n_{H_2(pứ)}=0,4.(100\%-20\%)=0,32(mol)$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

Theo PT: $n_{Fe}=\frac{2}{3}n_{H_2}=\frac{16}{75}(mol)$

$\to m_{Fe}=\frac{16}{75}.56\approx 11,95(g)$