Bạn tham khảo:
$Fe+2HCl \to FeCl_2+H_2$
$Mg+2HCl \to MgCl_2+H_2$
$n_{Fe}=a(mol)$
$n_{Mg}=b(mol)$
$m_{hh}=56a+24b$
$n_{HCl}=2a+2b(mol)$
$m_{ddHCl}=\dfrac{36,5.(2a+2b).100}{20}=365a+365b(g)$
$m_{ddsaupu}=56a+24b+365a+365b-2a-2b=419a+387b(g)$
$C\%_{MgCl_2}=\dfrac{95a}{419a+387b}.100\%=11,787\%$
$\to a=b$
$C\%_{FeCl_2}=\dfrac{127a}{419a+387b}.100\%=15,757\%$