Gọi khối lượng mỗi kim loại là a(g)
$\Rightarrow n_{Ca}=\frac{a}{40}(mol)  , n_{Al}=\frac{a}{27}(mol) , n_{Fe}=\frac{a}{56}(mol)$
  $Ca+2HCl\rightarrow CaCl_{2}+H_{2}$           $\Rightarrow n_{H_{2}} =\frac{a}{40}(mol)$      
  $2Al +6HCl\rightarrow 2AlCl_{3}+3H_{2}$   $\Rightarrow n_{H_{2}} = \frac{a}{18}(mol) $
  $Fe+2HCl\rightarrow FeCl_{2}+H_{2}$    $\Rightarrow n_{H_{2}} =\frac{a}{56}(mol)$
$\Rightarrow \frac{a}{18}>\frac{a}{40}>\frac{a}{56}  \Rightarrow $ Al cho nhiều hidro nhất

$H^{+}+OH^{-}\rightarrow H_{2}O$
  Ta có : $n_{H^{+}}=0.02(mol)        , n_{OH^{-}}=0.03(mol)$
$\Rightarrow OH^{- }dư$
$n_{OH^{-}}(dư)=0.01(mol)\rightarrow [OH^{-}]dư=0.1\Rightarrow pOH=1\Rightarrow pH=13$

a) $Mg+2H^{+}\rightarrow Mg^{2+}+H_{2}$
 $n_{Mg}=0,1(mol);          x_{H^{+}}=0.21(mol)$
$\Rightarrow $ Mg hết , HCl dư
$n_{H^{+}}(dư)=0.01(mol)\rightarrow [H^{+}]dư=0.1\Rightarrow pH=1$

$n_{H^{+}}=0.01(mol)$
Vì sau pư , dd có pH=12 nên dư $OH^{-}$
$n_{OH^{-}}(dư)=0.002(mol)$
  PTHH : $H^{+}+OH^{-}\rightarrow H_{2}O$
  $\rightarrow n_{OH^{-}}(pư)=0.01(mol)$
mà $n_{OH^{-}}(bd)=0.1a(mol)$
$\Rightarrow 0.1a=0.01+0.002\Leftrightarrow a=0.12$