Đáp án:
\(3/\\ a,\ m_{Fe}=11,2\ g.\\ m_{Mg}=4,8\ g.\\ b,\%m_{Fe}=70\%\\ \%m_{Mg}=30\%\\ c,\ m_{\text{hh muối}}=44,4\ g.\\ 4/\\ a,\ \%m_{Zn}=63,52\%\\ \%m_{Fe}=36,48\%\\ b,\ C_{M_{HCl}}=2\ M.\\ c,\ C_{M_{ZnCl_2}}=0,6\ M.\\ C_{M_{FeCl_2}}=0,4\ M.\)
Giải thích các bước giải:
\(3/\\ a,\ PTHH:\\ Fe+2HCl\to FeCl_2+H_2↑\ (1)\\ Mg+2HCl\to MgCl_2+H_2↑\ (2)\\ n_{H_2}=\dfrac{8,96}{22,4}=0,4\ mol.\\ \text{Gọi $n_{Fe}$ là a (mol), $n_{Mg}$ là b (mol).}\\ \text{Theo đề bài ta có hệ pt:}\\ \left\{\begin{matrix} 56a+24b=16 & \\ a+b=0,4 & \end{matrix}\right.\ ⇒\left\{\begin{matrix} a=0,2 & \\ b=0,2 & \end{matrix}\right.\\ ⇒m_{Fe}=0,2\times 56=11,2\ g.\\ ⇒m_{Mg}=0,2\times 24=4,8\ g.\\ b,\ \%m_{Fe}=\dfrac{11,2}{16}\times 100\%=70\%\\ \%m_{Mg}=\dfrac{4,8}{16}\times 100\%=30\%\\ c,\ Theo\ pt\ (1):\ n_{FeCl_2}=n_{Fe}=0,2\ mol.\\ Theo\ pt\ (2):\ n_{MgCl_2}=n_{Mg}=0,2\ mol.\\ ⇒m_{FeCl_2}=0,2\times 127=25,4\ g.\\ ⇒m_{MgCl_2}=0,2\times 95=19\ g.\\ ⇒m_{\text{hh muối}}=25,4+19=44,4\ g.\\ 4/\\ a,\ PTHH:\\ Zn+2HCl\to ZnCl_2+H_2↑\ (1)\\ Fe+2HCl\to FeCl_2+H_2↑\ (2)\\ n_{H_2}=\dfrac{11,2}{22,4}=0,5\ mol.\\ \text{Gọi $n_{Zn}$ là a (mol), $n_{Fe}$ là b (mol).}\\ \text{Theo đề bài ta có hệ pt:}\\ \left\{\begin{matrix} 65a+56b=30,7 & \\ a+b=0,5 & \end{matrix}\right.\ ⇒\left\{\begin{matrix} a=0,3 & \\ b=0,2 & \end{matrix}\right.\\ ⇒\%m_{Zn}=\dfrac{0,3\times 65}{30,7}\times 100\%=63,52\%\\ \%m_{Fe}=\dfrac{0,2\times 56}{30,7}\times 100\%=36,48\%\\ b,\ Theo\ pt\ (1),(2):\ n_{HCl}=2n_{H_2}=1\ mol.\\ \text{Đổi 500 ml = 0,5 lít.}\\ ⇒C_{M_{HCl}}=\dfrac{1}{0,5}=2\ M.\\ c,\ Theo\ pt\ (1):\ n_{ZnCl_2}=n_{Zn}=0,3\ mol.\\ Theo\ pt\ (2):\ n_{FeCl_2}=n_{Fe}=0,2\ mol.\\ ⇒C_{M_{ZnCl_2}}=\dfrac{0,3}{0,5}=0,6\ M.\\ ⇒C_{M_{FeCl_2}}=\dfrac{0,2}{0,5}=0,4\ M.\)
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