Đáp án:
\(a,\ m_{Mg}=7,2\ g.\\ m-{Fe}=5,6\ g.\\ b,\ \%m_{Fe}=43,75\%\\ \%m_{Mg}=56,25\%\\ c,\ V_{HCl}=0,4\ lít.\\ d,\ m_{\text{kết tủa}}=26,4\ g.\)
Giải thích các bước giải:
\(a,\ PTHH:\\ Mg+2HCl\to MgCl_2+H_2↑\ (1)\\ Fe+2HCl\to FeCl_2+H_2↑\ (2)\\ n_{H_2}=\dfrac{8,96}{22,4}=0,4\ mol.\\ \text{Gọi $n_{Mg}$ là a (mol), $n_{Fe}$ là b (mol).}\\ \text{Theo đề bài ta có hệ pt:}\ \left\{\begin{matrix} 24a+56b=12,8 & \\ a+b=0,4 & \end{matrix}\right.\ ⇔\left\{\begin{matrix} a=0,3 & \\ b=0,1 & \end{matrix}\right.\\ ⇒m_{Mg}=0,3\times 24=7,2\ g.\\ m_{Fe}=0,1\times 56=5,6\ g.\\ b,\ \%m_{Mg}=\dfrac{7,2}{12,8}\times 100\%=56,25\%\\ \%m_{Fe}=\dfrac{5,6}{12,8}\times 100\%=43,75\%\\ c,\ Theo\ pt\ (1),(2):\ n_{HCl}=2n_{H_2}=0,8\ mol.\\ ⇒V_{HCl}=\dfrac{0,8}{2}=0,4\ lít.\\ d,\ PTHH:\\ FeCl_2+2NaOH\to Fe(OH)_2↓+2NaCl\ (3)\\ MgCl_2+2NaOH\to Mg(OH)_2↓+2NaCl\ (4)\\ Theo\ pt\ (1):\ n_{MgCl_2}=n_{Mg}=0,3\ mol.\\ Theo\ pt\ (2):\ n_{FeCl_2}=n_{Fe}=0,1\ mol.\\ Theo\ pt\ (3):\ n_{Fe(OH)_2}=n_{FeCl_2}=0,1\ mol.\\ Theo\ pt\ (4):\ n_{Mg(OH)_2}=n_{MgCl_2}=0,3\ mol.\\ ⇒m_{Fe(OH)_2}=0,1\times 90=9\ g.\\ m_{Mg(OH)_2}=0,3\times 58=17,4\ g.\\ ⇒m_{\text{kết tủa}}=9+17,4=26,4\ g.\)
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