$n_{H_2O}=\dfrac{9}{18}=0,5(mol)$
Bảo toàn H: $n_{OH}=2n_{H_2O}=1(mol)$
$\Rightarrow m_{\text{kim loại}}=50-1.17=33g$
BTKL: $m_{\text{oxit}}=50-9=41g$
$\Rightarrow m_O=41-33=8g$
$\Rightarrow n_O=0,5(mol)$
$2H+O\to H_2O$
$\Rightarrow n_{HCl}=n_H=2n_O=1(mol)$
$V=\dfrac{1}{2}=0,5l=500ml$
$n_{Cl}=1(mol)\Rightarrow m_{\text{muối}}=33+1.35,5=68,5g$