a)
$2Al + 6HCl → 2AlCl_3 + 3H_2↑$
$a\hspace{5cm}1,5a$
$Mg + 2HCl → MgCl_2 + H_2↑$
$b\hspace{5cm}b$
$m_{Al, Mg}=12(g)$
$⇒27a+24b=12\quad (1)$
$n_{H_2}=\frac{14,56}{22,4}=0,65(mol)$
$⇒1,5a+b=0,65\quad(2)$
Từ (1), (2) $⇒\left\{\begin{array}{}27a+24b=12\\1,5a+b=0,65\end{array}\right.$
$⇔\left\{\begin{array}{}a=0,4\\b=0,05\end{array}\right.$
$⇒m_{Al}=0,4.27=10,8 (g)$
$⇒m_{Mg}=12-10,8=1,2 (g)$
$⇒m_{HCl\text{ pư}}=0,8.36,5+0,1.36,5=32,85 (g)$
b)
$HCl + NaOH → NaCl + H_2O$
$n_{HCl}=n_{NaOH}=0,8.0,25=0,2 (mol)$
$⇒m_{HCl\text{ dư}}=0,2.36,5=7,3 (g)$
$⇒∑m_{HCl}=7,3+32,85=40,15 (g)$
$⇒m_{\text{dd }HCl}=\frac{40,15}{36,5}.100=110 (g)$