Đáp án:
\(1/\\ a,\%m_{Fe}=39,63\%\\ \%m_{Cu}=60,37\%\\ b,m_{HCl}=10,95\ g.\\ m_{dung\ dịch\ HCl}=75\ g.\\ c,C\%_{dung\ dịch\ spư}=22,92\%\\ 2/\\ a,m_{MgCO_3}=21\ g.\\ m_{MgO}=8\ g.\\ b,C\%_{HCl}=21,9\%\)
Giải thích các bước giải:
\(1/\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2↑\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\ mol.\\ Theo\ pt:\ n_{Fe}=n_{H_2}=0,15\ mol.\\ ⇒\%m_{Fe}=\dfrac{0,15.56}{0,15.56+12,8}.100\%=39,63\%\\ ⇒\%m_{Cu}=100\%-39,63\%=60,37\%\\ b,Theo\ pt:\ n_{HCl}=2n_{H_2}=0,3\ mol.\\ ⇒m_{HCl}=0,3.36,5=10,95\ g.\\ ⇒m_{dung\ dịch\ HCl}=\dfrac{10,95}{14,6\%}=75\ g.\\ c,Theo\ pt:\ n_{FeCl_2}=n_{H_2}=0,15\ mol.\\ m_{dung\ dịch\ spư}=m_{Fe}+m_{dung\ dịch\ HCl}-m_{H_2}=(0,15.56)+75-(0,15.2)=83,1\ g.\\ ⇒C\%_{dung\ dịch\ spư}=\dfrac{0,15.127}{83,1}.100\%=22,92\%\\ 2/\\ a,PTHH:\\ MgO+2HCl\to MgCl_2+H_2O\ (1)\\ MgCO_3+2HCl\to MgCl_2+CO_2↑+H_2O\ (2)\\ Theo\ pt\ (2):\ n_{MgCO_3}=n_{MgCl_2}=n_{CO_2}=\dfrac{5,6}{22,4}=0,25\ mol.\\ ⇒m_{MgCO_3}=0,25.84=21\ g.\\ ⇒m_{MgCl_2}(pt2)=0,25.95=23,75\ g.\\ ⇒m_{MgCl_2}(pt1)=m_{muối\ khan}-m_{MgCl_2}(pt2)=42,75-23,75=19\ g.\\ ⇒n_{MgCl_2}(pt1)=\dfrac{19}{95}=0,2\ mol.\\ Theo\ pt\ (1):\ n_{MgO}=n_{MgCl_2}=0,2\ mol.\\ ⇒m_{MgO}=0,2.40=8\ g.\\ b,Theo\ pt\ (1):\ n_{HCl}=2n_{MgCl_2}=0,4\ mol.\\ Theo\ pt\ (2):\ n_{HCl}=2n_{CO_2}=0,5\ mol.\\ ⇒m_{HCl}=(0,4+0,5).36,5=32,85\ g.\\ ⇒C\%_{HCl}=\dfrac{32,85}{150}.100\%=21,9\%.\)
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