Bài 23: Hiđro Clorua - Axit Clohidric và muối Clorua

\(n_{H_2}=\dfrac{0,784}{22,4}=0,035\left(mol\right)\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=1,39\\1,5a+b=0,035\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,01\\b=0,02\end{matrix}\right.\\ \Rightarrow\%m_{Al}=\dfrac{0,01.27}{1,39}.100=37,53\%\\ \Rightarrow\%m_{Fe}=100\%-37,53\%=62,47\%\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\\ FeO+2HCl\rightarrow FeCl_2+H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ Fe_3O_4+8HCl\rightarrow2FeCl_3+FeCl_2+4H_2O\)

2HCl+CuO->CuCl₂+H₂O

FeO+2HCl->FeCl₂+H₂O

Fe₂O₃+6HCl->2FeCl₃+3H₂O

Fe₃O₄+8HCl->2FeCl₃+FeCl₂+4H₂O

\(Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+2H_2O\)

\(Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)

\(Fe\left(OH\right)_2+2HCl\rightarrow FeCl_2+2H_2O\)

\(2HCl+Cu\left(OH\right)_2\rightarrow CuCl_2+2H_2O\\ 3HCl+Fe\left(OH\right)_3\rightarrow FeCl_3+3H_2O\\ 2HCl+Fe\left(OH\right)_2\rightarrow FeCl_2+2H_2O\)

Cu(OH)₂+2HCl->CuCl₂+2H₂O

Fe(OH)₃+3HCl->FeCl₃+3H₂O

Fe(OH)₂+2HCl->FeCl₂+2H₂O

\(2HCl+CaCO_3\rightarrow CaCl_2+CO_2\uparrow+H_2O\\ HCl+AgNO_3\rightarrow AgCl\downarrow+HNO_3\\ HCl+NaHCO_3\rightarrow NaCl+CO_2\uparrow+H_2O\)

CaCO₃+2HCl->CaCl₂+H₂O+CO₂

AgNO₃+HCl->AgCl+HNO₃

NaHCO₃+HCl->NaCl+H₂O+CO₂

- Vì hỗn hợp rắn qua dd HCl có thấy H₂ nên hh rắn chắc chắn có Al dư

\(PTHH:\left(a\right)4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ \left(b\right)Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ \left(c\right)2Al_{dư}+6HCl\rightarrow2AlCl_3+3H_2\\ n_{H_2\left(c\right)}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\\ n_{AlCl_3\left(tổng\right)}=\dfrac{13,35}{133,5}=0,1\left(mol\right)\\ n_{AlCl_3\left(c\right)}=n_{Al\left(dư\right)}=\dfrac{2}{3}.n_{H_2\left(c\right)}=\dfrac{2.0,075}{3}=0,05\left(mol\right)\\ \Rightarrow n_{AlCl_3\left(b\right)}=0,1-0,05=0,05\left(mol\right)\\ \Rightarrow n_{HCl}=\dfrac{6}{2}.n_{AlCl_3\left(tổng\right)}=\dfrac{6}{2}.0,1=0,3\left(mol\right)\\ \Rightarrow V=V_{ddHCl}=\dfrac{0,3}{1}=0,3\left(l\right)\\ n_{Al_2O_3}=\dfrac{n_{AlCl_3\left(b\right)}}{2}=\dfrac{0,05}{2}=0,025\left(mol\right)\)

\(\Rightarrow n_{Al\left(a\right)}=2.n_{Al_2O_3}=2.0,025=0,05\left(mol\right)\\ \Rightarrow n_{Al\left(tổng\right)}=n_{Al\left(a\right)}+n_{Al\left(c\right)}=0,05+0,05=0,1\left(mol\right)\\ \Rightarrow m=m_{Al\left(tổng\right)}=0,1.27=2,7\left(g\right)\\ n_{O_2}=\dfrac{3}{2}.n_{Al_2O_3}=\dfrac{3}{2}.0,025=0,0375\left(mol\right)\\ \Rightarrow V=V_{O_2\left(đktc\right)}=0,0375.22,4=0,84\left(l\right)\)

Ủa em, em cần nhanh và chi tiết mà em cho cái đề còn yêu cầu đề không cho thì ai giúp em được nè?

Fe + 2 HCl -> FeCl2 + H2

2 Al + 6 HCl -> 2 AlCl3 + 3 H2

Mg + 2 HCl -> MgCl2 + H2

K + HCl -> KCl + 1/2 H2

Zn + 2 HCl -> ZnCl2 + H2

Au + HCl -> không xảy ra

Ag+ HCl->không xảy ra

2HCl +Ba ->BaCl2 +H2

\(n_{HCl}=\dfrac{400}{1000}.0,4=0,16\left(mol\right)\)

\(pthh:2A+2nHCl--->2ACl_n+nH_2\)

Theo PT: \(n_A=\dfrac{0,16}{n}\left(mol\right)\)

\(M_A=\dfrac{3,68}{\dfrac{0,16}{n}}=23n\left(\dfrac{g}{mol}\right)\)

Biện luận: 

Do Na cũng trong nhóm IA.

Vậy A là natri (Na)

H₂+Cl₂->2HCl

         0,35---0,7 mol

n _H2=\(\dfrac{8,96}{22,4}\)=0,4 mol

n _Cl2=\(\dfrac{7,84}{22,4}\)=0,35 mol

=>H2 dư :0,05 mol

=>V_HCl=0,7.22,4=15,68l

b) C% _HCl=\(\dfrac{0,7.36,5}{25,55+224,45}\).100=10,22%

a: \(H_2+Cl_2\rightarrow2HCl\)

\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

\(n_{Cl_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)

=>Cl2 thiếu, H2 dư

\(V_{HCl}=2\cdot V_{Cl_2}=0.7\left(mol\right)\)

\(FeS+2HCl\rightarrow FeCl_2+H_2S\)

\(CuS+HCl\rightarrow\) không phản ứng

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)

\(NaCl+AgNO_3\rightarrow AgCl\downarrow+NaNO_3\)

\(HCl+AgNO_3\rightarrow AgCl\downarrow+HNO_3\)