Bạn tham khảo:
$a/$
$Fe+2HCl \to FeCl_2+H_2(1)$
$Mg+2HCl \to MgCl_2+H_2(2)$
$b/$
Giả sử:
$n_{Mg}=1(mol)$
$n_{Fe}=a(mol)$
$(1)$
$n_{Fe}=n_{FeCl_2}=a(mol)$
$n_{HCl}=2a(mol)$
$(2)$
$n_{Mg}=n_{MgCl_2}=1(mol)$
$n_{HCl}=2(mol)$
$\to n_{HCl}=2+2a(mol)$
$m_{ddHCl}=\frac{36,5.(2+2a).100}{15}=486,67+486,67a(g)$
$m_{dd}=56a+24+486,67+486,67a-2-2a=508,67+540,67a(g)$
$C\%_{MgCl_2}=\frac{95}{508,67+540,67a}.100\%=5,975\%$
$\to a=2$
$\to C\%_{FeCl_2}=\frac{127.2}{508,67+540,67.2}.100\%=15,97\%$