a,
$n_{HCl}=0,3.2=0,6(mol)$
$n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)$
$Fe+2HCl\to FeCl_2+H_2$
$2n_{H_2}<n_{HCl}\Rightarrow HCl$ dư
$\Rightarrow Fe$ tan hết
$n_{Fe}=n_{H_2}=0,2(mol)$
$Cu$ không tác dụng với $HCl$ nên $m_{Cu}=1,28g$
$\%m_{Cu}=\dfrac{1,28.100}{1,28+0,2.56}=10,26\%$
$\%m_{Fe}=89,74\%$
b,
$n_{FeCl_2}=n_{Fe}=0,2(mol)$
$n_{HCl\text{pứ}}=2n_{H_2}=0,4(mol)$
$\Rightarrow n_{HCl\text{dư}}=0,6-0,4=0,2(mol)$
$HCl+NaOH\to NaCl+H_2O$
$FeCl_2+2NaOH\to Fe(OH)_2+2NaCl$
$\Rightarrow n_{NaOH}=0,2+0,2.2=0,6(mol)$
$V_{dd}=200ml=0,2l$
$\to C_{M_{NaOH}}=\dfrac{0,6}{0,2}=3M$
