$n_{HCl}=\dfrac{350.14,6\%}{36,5}=1,4(mol)$
$n_{KOH}=0,14.2=0,28(mol)$
$KOH+HCl\to KCl+H_2O$
$\to n_{HCl\text{dư}}=0,28(mol)$
$\to n_{HCl\text{pứ}}=1,4-0,28=1,12(mol)$
$\to n_H=n_{HCl}=1,12(mol)$
Dư $HCl$ nên oxit tan hết
Phản ứng oxit $+$ axit:
$2H+O\to H_2O$
Phản ứng khử:
$H_2+O\to H_2O$
Ta có: $n_O=n_{H_2}=\dfrac{n_H}{2}=0,56(mol)$
$\to V=0,56.22,4=12,544l$