1.
Thu được 3 khí: $Cl_2$, $H_2$, $CO_2$
PTHH:
$NaClO+2HCl\to NaCl+Cl_2+H_2O$
$2KMnO_4+16HCl\to 2KCl+2MnCl_2+5Cl_2+8H_2O$
$Mg+2HCl\to MgCl_2+H_2$
$CaCO_3+2HCl\to CaCl_2+CO_2+H_2O$
2.
$n_{H_2}=\dfrac{13,44}{22,4}=0,6(mol)$
Bảo toàn e: $2n_{Fe}=2n_{H_2}$
$\to n_{Fe}=0,6(mol)$
$\to n_{Cu}=\dfrac{52,8-0,6.56}{64}=0,3(mol)$
$n_{Cu}: n_{Fe}=0.3:0,6=1:2$
Đặt $x$, $2x$ là số mol $Cu$, $Fe$ trong m.
$n_{SO_2}=\dfrac{4,48}{22,4}=0,2(mol)$
Bảo toàn e: $2n_{Cu}+3n_{Fe}=2n_{SO_2}$
$\to 2x+3.2x=0,2.2$
$\to x=0,05$
$\to m=64x+56.2x=8,8g$
3.
Gọi $n$ là hoá trị $A$
$n_{SO_2}=\dfrac{21,504}{22,4}=0,96(mol)$
Bảo toàn e: $n.n_A=2n_{SO_2}$
$\to n_A=\dfrac{1,92}{n}(mol)$
$\to M_A=\dfrac{61,44n}{1,92}=32n$
$\to n=2; M_A=64(Cu)$