Sơ lược về hợp chất có oxi của clo

$\begin{array}{l} NaClO+2HCl\to NaCl+C{{l}_{2}}+{{H}_{2}}O \\ 3C{{l}_{2}}+6KOH\xrightarrow{{{t}^{o}}}5KCl+KCl{{O}_{3}}+3{{H}_{2}}O \end{array}$

$\begin{array}{l} & {{n}_{C{{l}_{2}}}}=x(mol) \\ & 3C{{l}_{2}}+6KOH\xrightarrow{{{100}^{o}}C}5KCl+KCl{{O}_{3}}+3{{H}_{2}}O \\ & x\to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{5x}{3}\,\,\,\,\,\,\,\,\,\,\dfrac{\,x}{3} \\ & \to {{m}_{muoi}}={{m}_{KCl}}+{{m}_{KCl{{O}_{3}}}}=\dfrac{5x}{3}.74,5+122,5.\dfrac{x}{3}=16,5 \\ & \to x=0,1\to {{V}_{C{{l}_{2}}}}=2,24(lit) \end{array}$

Clorua vôi: $CaOC{{l}_{2}}$ , trong phân tử clorua vôi có $C{{l}^{+1}}$ nên có tính oxi hóa mạnh

Các phản ứng sinh ra khí là:

$\begin{array}{l} (1)2HCl+CaOC{{l}_{2}}\to CaC{{l}_{2}}+C{{l}_{2}}\uparrow +{{H}_{2}}O \\ (3)NaClO+2HCl\to NaCl+C{{l}_{2}}\uparrow +{{H}_{2}}O \\ (4)6HCl+KCl{{O}_{3}}\to KCl+3C{{l}_{2}}\uparrow +3{{H}_{2}}O \end{array}$

Trong phản ứng $C{{l}^{+1}}$ của clorua vôi giảm xuống $C{{l}^{-1}}$ nên clorua vôi thể hiện tính oxi hóa

$\begin{array}{l} & {{n}_{C{{l}_{2}}}}=0,25\to {{n}_{C{{l}_{2}}\,\,p/u}}=0,25.80\%=0,2 \\ & C{{l}_{2}}+Ca{{(OH)}_{2}}\xrightarrow{{{30}^{0}}C}CaOC{{l}_{2}}+{{H}_{2}}O \\ & 0,2\to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,2 \\ & \to {{m}_{Clorua\,\,voi}}=25,4(gam) \end{array}$

$\begin{array}{l} & {{n}_{C{{l}_{2}}}}=x\to {{n}_{C{{l}_{2}}\,\,p/u}}=0,8x \\ & C{{l}_{2}}+Ca{{(OH)}_{2}}\to CaOC{{l}_{2}}+{{H}_{2}}O \\ & 0,8x\to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,8x \\ & \to 0,8x={{n}_{CaOC{{l}_{2}}}}=0,16\to x=0,2 \\ & \to {{V}_{C{{l}_{2}}}}=4,48(l) \end{array}$

$\begin{array}{l} NaCl+{{H}_{2}}S{{O}_{4}}\to NaHS{{O}_{4}}+HCl \\ 4HCl+Mn{{O}_{2}}\to MnC{{l}_{2}}+C{{l}_{2}}+2{{H}_{2}}O \\ C{{l}_{2}}+2NaOH\to NaCl+NaClO+{{H}_{2}}O \end{array}$

$\begin{array}{l} C{{l}_{2}}+2NaOH\to NaCl+NaClO+{{H}_{2}}O \\ NaClO+{{H}_{2}}O+C{{O}_{2}}\to NaHC{{O}_{3}}+HClO \\ NaHC{{O}_{3}}+HCl\to NaCl+{{H}_{2}}O+C{{O}_{2}} \end{array}$

$\begin{array}{l} CaO+{{H}_{2}}O\to Ca{{(OH)}_{2}} \\ 2KMn{{O}_{4}}+16HCl\to 2KCl+2MnC{{l}_{2}}+5C{{l}_{2}}+8{{H}_{2}}O \\ C{{l}_{2}}+Ca{{(OH)}_{2}}\to CaOC{{l}_{2}}+{{H}_{2}}O \end{array}$

Do nước Gia ven có tính tẩy màu nên cánh hoa hồng sẽ bị mất màu

$\begin{array}{l} & Goi\,\,\,{{n}_{C{{l}_{2}}}}=x\to {{n}_{C{{l}_{2}}\,\,p/u}}=0,8x \\ & {{m}_{NaClO}}=50.29,8\%=14,9(gam) \\ & \to {{n}_{NaClO}}=0,2(mol) \\ & C{{l}_{2}}+2NaOH\to NaCl+NaClO+{{H}_{2}}O \\ & 0,8x\to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,8x \\ & \to 0,8x=0,2\to x=0,25 \\ & \to {{V}_{C{{l}_{2}}}}=5,6(lit) \end{array}$

Tính sát trùng và tẩy màu của nước Gia-ven là do trong $NaClO$ , Cl có số oxi hóa +1, thể hiện tính oxi hóa mạnh.

Khi cho dung dịch axit clohidric vào ống nghiệm đựng clorua vôi thấy có khí màu vàng lục $2HCl+CaOC{{l}_{2}}\to CaC{{l}_{2}}+C{{l}_{2}}+{{H}_{2}}O$

$\begin{array}{l} KCl+{{H}_{2}}S{{O}_{4}}\to KHS{{O}_{4}}+HCl\,(1) \\ 4HCl+Mn{{O}_{2}}\to MnC{{l}_{2}}+C{{l}_{2}}+2{{H}_{2}}O\,\,(2) \\ CaO+{{H}_{2}}O\to Ca{{(OH)}_{2}}\,\,(3) \\ C{{l}_{2}}+Ca{{(OH)}_{2}}\to CaOC{{l}_{2}}+{{H}_{2}}O\,\,(4) \end{array}$

Phản ứng oxi hóa – khử là (2), (4)

$\begin{array}{l} CaOC{{l}_{2}}+2HCl\to CaC{{l}_{2}}+C{{l}_{2}}+{{H}_{2}}O \\ C{{l}_{2}}+2NaOH\to NaCl+NaClO+{{H}_{2}}O \end{array}$

$\begin{array}{l} NaClO+2HCl\to NaCl+C{{l}_{2}}\uparrow +{{H}_{2}}O \\ 3C{{l}_{2}}+2Fe\xrightarrow{{{t}^{o}}}2FeC{{l}_{3}} \\ FeC{{l}_{3}}+3NaOH\to Fe{{(OH)}_{3}}+3NaCl \end{array}$

$\begin{array}{l} & KCl{{O}_{3}}+6HCl\to KCl+3C{{l}_{2}}+3{{H}_{2}}O \\ & 0,2\to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,6\, \\ & CaOC{{l}_{2}}+2HCl\to CaC{{l}_{2}}+C{{l}_{2}}+{{H}_{2}}O \\ & 0,1\to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1 \\ & \to {{n}_{C{{l}_{2}}}}=0,7\to {{V}_{C{{l}_{2}}}}=15,68(lit) \end{array}$

$\begin{array}{l} & 2NaCl+2{{H}_{2}}O\xrightarrow{dien\,\,phan\,\,dung\,\,dich,co\,\,mang\,\,ngan}2NaOH+C{{l}_{2}}+{{H}_{2}} \\ & 2KCl+2{{H}_{2}}O\xrightarrow{dien\,\,phan\,\,dung\,\,dich,co\,\,mang\,\,ngan}2KOH+C{{l}_{2}}+{{H}_{2}} \\ & \to {{n}_{NaOH}}+{{n}_{KOH}}=2{{n}_{C{{l}_{2}}}}=0,1(mol) \\ & \to {{C}_{M}}(NaOH+KOH)=\dfrac{0,1}{2}=0,05M \end{array}$

$\begin{array}{l} N{{a}_{2}}O+{{H}_{2}}O\to 2NaOH \\ NaCl+{{H}_{2}}S{{O}_{4}}\to NaHS{{O}_{4}}+HCl \\ 6HCl+2HN{{O}_{3}}\to 3C{{l}_{2}}+2NO+4{{H}_{2}}O \\ C{{l}_{2}}+2NaOH\to NaCl+NaClO+{{H}_{2}}O \end{array}$

Phản ứng clo đóng vai trò vừa là chất khử, vừa là chất oxi hóa là:

$\begin{array}{l} (1)\,C{{l}_{2}}+{{H}_{2}}O\to \\ (2) C{{l}_{2}}+Ca{{(OH)}_{2}} \\ (4)C{{l}_{2}}+NaOH\to \end{array}$

$Ca{{(OH)}_{2}}\xrightarrow{C{{l}_{2}}}CaOC{{l}_{2}}\xrightarrow{HCl}CaC{{l}_{2}}\xrightarrow{AgN{{O}_{3}}}AgCl$

Ban đầu, quỳ tím trong dung dịch NaOH có màu xanh, khi sục khí clo vào sinh ra NaClO có tính tẩy màu làm quỳ tím mất màu

$C{{l}_{2}}+2NaOH\to NaCl+NaClO+{{H}_{2}}O$

$KCl{{O}_{3}}$ vừa có tính oxi hóa vừa có tính khử

$\left( 1 \right)\,\,2K\overset{+5}{\mathop{Cl}}\,{{O}_{3}}_{\left( r \right)}\to 2K{{\overset{-1}{\mathop{Cl}}\,}_{\left( r \right)}}+3{{O}_{2}}_{\left( k \right)}$ ® tính oxi hóa

$\left( 2 \right)\,\,4K\overset{+5}{\mathop{Cl}}\,{{O}_{3}}_{\left( r \right)}\to 3K\overset{+7}{\mathop{Cl}}\,{{O}_{4}}_{\left( r \right)}+K{{\overset{-1}{\mathop{Cl}}\,}_{\left( r \right)}}.$ ®tính khử