Đặt $x$, $2x$, $3x$ là số mol $A$, $B$, $C$
$n_{H_2}=\dfrac{1,68}{22,4}=0,075(mol)$
PTHH:
$A+2HCl\to ACl_2+H_2$
$B+2HCl\to BCl_2+H_2$
$2C+6HCl\to 2CCl_3+3H_2$
$\Rightarrow x+2x+3x.1,5=0,075$
$\Leftrightarrow x=0,01$
$\Rightarrow n_A=0,01(mol); n_B=0,02(mol); n_C=0,03(mol)$
$\Rightarrow 0,01M_A+0,02M_B+0,03M_C= 2,17$ $(1)$
$\dfrac{M_A}{M_B}=\dfrac{3}{7}$
$\Rightarrow M_A=\dfrac{3}{7}M_B=0$ $(2)$
$(1)(2)\Rightarrow \dfrac{17}{700}M_B+0,03M_C=2,17$
$\Rightarrow 17M_B+21M_C=1519$
Mà $M_C<M_B\Rightarrow M_B=56(Fe), M_C=27(Al)$
$\to M_A=24(Mg)$
Vậy $A$, $B$, $C$ là $Mg, Fe, Al$