$n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)$
$\Rightarrow n_{OH^-}=2n_{H_2}=0,3(mol)$
$n_{HCl}=n_{H_2SO_4}=x (mol)$
$\Rightarrow n_{H^+}=n_{HCl}+2n_{H_2SO_4}=3x (mol)$
$n_{H^+}=n_{OH^-}\Rightarrow x=0,1$
$\Rightarrow n_{Cl^-}=n_{SO_4^{2-}}=0,1(mol)$
BTKL: $m=m_X+m_{Cl^-}+m_{SO_4^{2-}}=16+0,1.35,5+0,1.96=29,15g$