a,
$n_{H_2}=\dfrac{1,68}{22,4}=0,075(mol)$
$2A+2HCl\to 2ACl_2+H_2$
$\Rightarrow n_A=2n_{H_2}=0,15(mol)$
$M_A=\dfrac{5,85}{0,15}=39(K)$
Vậy A là kali.
b,
$n_{HCl}=n_A=0,15(mol)$
$\Rightarrow C_{M_{HCl}}=\dfrac{0,15}{0,2}=0,75M$
c,
$n_{KCl}=n_{HCl}=0,15(mol)$
$\Rightarrow C_{M_{KCl}}=0,75M$