\(2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Al_2(SO_4)_3+3BaCl_2\to 2AlCl_3+3BaSO_4\downarrow\\ AlCl_3+3NaOH\to Al(OH)_3\downarrow +3NaCl\\ 2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\\ 2Al_2O_3\xrightarrow{đpnc}4Al+3O_2\)

(7) 

(1) \(CaCO_3\xrightarrow[]{t^o}CaO+CO_2\)

(2) \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

(3) \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

(4) \(CaCl_2+Na_2SO_4\rightarrow CaSO_4+2NaCl\)

\(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

\(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

\(CaCl_2+Na_2SO_4\rightarrow2NaCl+CaSO_4\)

Câu 2:

\(2Fe\left(OH\right)_3\rightarrow\left(t^o\right)Fe_2O_3+3H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ FeCl_3+3AgNO_3\rightarrow Fe\left(NO_3\right)_3+3AgCl\downarrow\\ Fe\left(NO_3\right)_3+3KOH\rightarrow Fe\left(OH\right)_3\downarrow+3KNO_3\\2 Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\)

\((1)2SO_2+O_2\xrightarrow[V_2O_5]{t^o}2SO_3\\ (2)SO_3+H_2O\to H_2SO_4\\ (3)H_2SO_4+MgO\to MgSO_4 +H_2O\\ (4)MgSO_4+2NaOH\to Na_2SO_4+Mg(OH)_2\downarrow\)

\(\left(1\right)2Al_2O_3\xrightarrow[Na_3AlF_6]{đpnc}4Al+3O_2\uparrow\)

\(\left(2\right)2Al+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2\uparrow\)

\(\left(3\right)Al_2\left(SO_4\right)_3+3BaCl_2--->3BaSO_4\downarrow+2AlCl_3\)

\(\left(4\right)AlCl_3+3NaOH--->Al\left(OH\right)_3\downarrow+3NaCl\)

$a)(1)S+O_2\xrightarrow{t^o}SO_2$

$(2)2SO_2+O_2\xrightarrow[V_2O_5]{t^o}2SO_3$

$(3)SO_3+H_2O\to H_2SO_4$

$(4)H_2SO_4+2NaOH\to Na_2SO_4+2H_2O$

$(5)Na_2SO_4+BaCl_2\to BaSO_4\downarrow+2NaCll$

$b)(1)SO_2+2NaOH\to Na_2SO_3+H_2O$

$(2)Na_2SO_3+H_2SO_4\to Na_2SO_4+H_2O+SO_2\uparrow$

$(3)Na_2SO_4+Ba(OH)_2\to BaSO_4\downarrow+2NaOH$

$(4)2NaOH+CO_2\to Na_2CO_3+H_2O$

$c)(1)CaO+CO_2\to CaCO_3$

$(2)CaCO_3\xrightarrow{t^o}CaO+CO_2$

$(3)CaO+H_2O\to Ca(OH)_2$

$(4)Ca(OH)_2+CO_2\to CaCO_3\downarrow+H_2O$

$(5)CaCO_3+H_2SO_4\to CaSO_4+H_2O+CO_2\uparrow$

$d)(1)2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3$

$(2)FeCl_3+3NaOH\to Fe(OH)_3\downarrow+3NaCl$

$(3)2Fe(OH)_3\xrightarrow{t^o}Fe_2O_3+3H_2O$

$(4)Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O$

$(5)Fe_2(SO_4)_3+3BaCl_2\to 2FeCl_3+3BaSO_4\downarrow$

$e)(1)Fe+2HCl\to FeCl_2+H_2$

$(2)FeCl_2+2AgNO_3\to Fe(NO_3)_2+2AgCl\downarrow$

$(3)Fe(NO_3)_2+2NaOH\to Fe(OH)_2\downarrow+2NaNO_3$

$(4)Fe(OH)_2+H_2SO_4\to FeSO_4+2H_2O$

$f)(1)2Cu+O_2\xrightarrow{t^o}2CuO$

$(2)CuO+2HCl\to CuCl_2+H_2O$

$(3)CuCl_2+2NaOH\to Cu(OH)_2\downarrow+2NaCl$

$(4)Cu(OH)_2\xrightarrow{t^o}CuO+H_2O$

$(5)CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$(6)Cu+2H_2SO_{4(đ)}\to CuSO_4+SO_2\uparrow+2H_2O$

$g)(1)2Al_2O_3\xrightarrow[criolit]{đpnc}4Al+3O_2$

$(2)2Al+6HCl\to 2AlCl_3+3H_2$

$(3)AlCl_3+3NaOH\to Al(OH)_3\downarrow+3NaCl$

$(4)2NaCl+2H_2O\xrightarrow[cmn]{đpdd}2NaOH+Cl_2\uparrow+H_2\uparrow$

$(5)NaOH+CuCl_2\to Cu(OH)_2\downarrow+2NaCl$

Câu 3 : 

\(a) n_{Al} = a(mol) ; n_{Fe} = b(mol) \Rightarrow 27a + 56b =1 1,1(1)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = 1,5a + b = \dfrac{6,72}{22,4} = 0,3(2)\\ (1)(2) \Rightarrow a = 0,1 ; b = 0,15\\ \%m_{Al} = \dfrac{0,1.27}{11,1}.100\% = 24,32\%\\ \%m_{Fe} = 100\% -24,32\% = 75,68\%\)

\(b) n_{HCl} = 2n_{H_2} = 0,3.2 = 0,6(mol)\\ \Rightarrow m_{dd\ HCl} = \dfrac{0,6.36,5}{14,6\%} = 150(gam)\\ m_{dd\ sau\ pư} = m_{hỗn\ hợp} + m_{dd\ HCl} - m_{H_2} = 11,1 + 150 - 0,3.2 = 160,5(gam)\\ n_{AlCl_3} = a = 0,1(mol)\ ;\ n_{FeCl_2} = b = 0,15(mol)\\ C\%_{AlCl_3} = \dfrac{0,1.133,5}{160,5}.100\% =8,32\%\\ C\%_{FeCl_2} = \dfrac{0,15.127}{160,5}.100\% = 11,87\%\)

Câu 2 : 

\(2KMnO_4 + 16HCl \to 2MnCl_2 + 2KCl + 5Cl_2 + 8H_2O\\ Cl_2 + 2Na \xrightarrow{t^o} 2NaCl\\ 2NaCl \xrightarrow{đpnc} 2Na + Cl_2\\ Cl_2 + 2NaBr \to 2NaCl + Br_2\\ Br_2 + 2NaI \to 2NaBr + I_2\)

Câu 4 : 

\(2Na + 2HCl \to 2NaCl + H_2\\ 2K + 2HCl \to 2KCl + H_2\\ 2Na + 2H_2O \to 2NaOH + H_2\\ 2K + 2H_2O \to 2KOH + H_2\\ n_{HCl} = \dfrac{a.C\%}{36,5} = \dfrac{a.C}{3650}(mol)\\ n_{H_2O} = \dfrac{a-a.C\%}{18} (mol)\)

Theo PTHH :

\(2n_{H_2} = n_{HCl} + n_{H_2O}\\ \Leftrightarrow 2.\dfrac{0,05a}{2} = \dfrac{a.C}{3650} + \dfrac{a-a.C\%}{18}\\ \Leftrightarrow 0,05 = \dfrac{C}{3650} + \dfrac{1-C\%}{18}\\ \Leftrightarrow C = 19,73\)

Vậy C% = 19,73%

Câu1:

- Trích mẫu thử

- Cho quỳ tìm vào các mẫu thử

+ mẫu thử nào làm quỳ tím hóa đỏ là 

+ mẫu thử nào làm quỳ tím hóa xanh là 

+ các mẫu thử còn lại không hiện tượng là 

- Cho dd \(AgNO_3\) tới dư vào các mẫu thử còn lại :

+ mẫu thử nào tạo kết tủa trắng là 

\(AgNO_3\)

+ mẫu thử nào tạo kết tủa màu vàng nhạt là 

\(AgNO_3\)​\(NaNO_3\)​

+ mẫu thử nào không có hiện tượng là \(NaNO_3\)

Câu 2:

1. \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2\uparrow+8H_2O\)

2. \(Cl_2+2Na\underrightarrow{t^o}2NaCl\)

3. \(2NaCl\underrightarrow{đpnc}2Na+Cl_2\)

4.\(2HBr+Cl_2\rightarrow2HCl+Br_2\)

\(2NaI_{\left(lạnh\right)}+Br_2\rightarrow2NaBr+I_2\)