a)
$m_{O\ trong\ oxit} = 40,6 - 26,2 = 14,4(gam)$
$n_O = \dfrac{14,4}{16} =0,9(mol)$
$2H^+ + O^{2-} \to H_2O$
$n_{HCl} = n_{H^+} = 2n_O = 1,8(mol)$
$\Rightarrow V = \dfrac{1,8}{0,5} = 3,6(lít)$
b) $n_{Cl} = n_{HCl} = 1,8(mol)$
$m_{muối} = m_{kim\ loại} + m_{Cl} = 26,2 + 1,8.35,5 = 90,1(gam)$