Đáp án:
$11,68\%$
Giải thích các bước giải:
$4Al+3O_2\xrightarrow{t^0} 2Al_2O_3\\ 2Mg+O_2\xrightarrow{t^0} 2MgO\\ 3Fe+2O_2\xrightarrow{t^0} Fe_3O_4\\ Al_2O_3+^HCl\to 2AlCl_3+3H_2O\\ MgO+2HCl\to MgCl_2+H_2O\\ Fe_3O_4+8HCl\to FeCl_2+2FeCl_3+4H_2O\\ n_{O_2}=\dfrac{0,896}{22,4}=0,04(mol)\\ n_{O\ Y}=2.n_{O_2}=0,08(mol)\\ n_{HCl}=2.n_{O\ Y}=0,16(mol)\\ m_{HCl}=0,16.36,5=5,84(g)\\ C\%_{dd\ HCl}=\dfrac{5,84}{50}.100\%=11,68\%$