Bạn tham khảo:
$Fe+2HCl \to FeCl_2+H_2$
$n_{Fe}=0,1(mol)$
$a/$
$n_{H_2}=0,1(mol)$
$V_{H_2}=0,1.22,4=2,24(l)$
$b/$
$n_{FeCl_2}=0,1(mol)$
$m_{FeCl_2}=0,1.127=12,7(g)$
$c/$
$n_{HCl}=0,2(mol)$
$C\%_{HCl}=\frac{0,2.36,5}{200}.100\%=3,65\$%
$d/$
$C\%_{FeCl_2}=\frac{12,7}{5,6+200-0,1.2}.100\%=6,18\%$
$e/$
$m_{dd}=5,6+200-0,1.2=205,4(g)$
$V_{dd}=171(ml)$
$CM=\frac{0,1}{0,171}=0,58M$
