ÌâÄ¿ÄÚÈÝ

(10·Ö)ij¿ÎÍâ»î¶¯Ð¡×éÓûÖÆÈ¡´ÎÂÈËáÄÆºÍÂÈ»¯ÄƵĻìºÏÈÜÒº£¬ÎªÌá¸ß´ÎÂÈËáÄÆº¬Á¿£¬ÓÃÈçͼËùʾװÖã®Í¼ÖÐÆ¿ÒÒÊ¢±¥ºÍʳÑÎË®£¬Æ¿±ûʢŨÁòËᣬ·ÖҺ©¶·AÖÐʢŨÑÎËᣮ(¾Ý×ÊÁÏÏÔʾ£ºCl2ÓëNaOHÔÚ²»Í¬Î¶ÈÏ£¬²úÎﲻͬ£®ÔڽϸßζÈÏÂÒ×Éú³ÉNaClO3)£®

ÊԻشð£º
(1)ÉÕÆ¿BÖÐÊ¢¡¡¡¡¡¡¡¡£¬ÊÔ¹ÜCÖÐÊ¢¡¡¡¡¡¡¡¡    .
(2)ÓÐͬѧÈÏΪ¿ÉÒÔʡȥijЩװÖã¬ÄãÈÏΪÔõÑù£ºÄÜ·ñʡȥÒÒ×°Öã¿¡¡¡¡¡¡¡¡(Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±)£¬ÀíÓÉÊÇ                                                          
(3)ÓÐͬѧÈÏΪ»¹±ØÐë¼ÓÈëijЩװÖã¬ÄãÈÏΪÔõÑù£¿                     (Ìî¡°ÐèÒª¡±»ò¡°²»ÐèÒª¡±)£¬Èç¹ûÄãÈÏΪÐèÒª£¬ÇëÖ¸³ö¸Ã×°ÖõÄ×÷ÓÃ
                                                                     
                                                                  
(4)¶¡×°ÖÃÖбùË®µÄ×÷ÓÃÊÇ                                         £®

(1)MnO2¡¡£¨1·Ö£© NaOHÈÜÒº£¨1·Ö£©
(2)²»ÄÜ¡¡£¨1·Ö£©HClÆøÌå½øÈëCÖУ¬ÏûºÄNaOH£¬½µµÍNaClOµÄº¬Á¿¡¡£¨2·Ö£©
(3)ÐèÒª¡¡£¨1·Ö£©Ó¦¼ÓÉÏÎ²Æø´¦Àí×°Ö㬷ÀÖ¹Cl2ÎÛȾ¿ÕÆø£¨2·Ö£©¡¡
(4)·ÀÖ¹Cl2ÓëNaOHÈÜÒºÔÚζȽϸßʱ£¬·¢ÉúÆäËû·´Ó¦£¨2·Ö£©

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2009?»ÆÆÖÇø¶þÄ££©ÈçͼÖÐA¡¢B¡¢C·Ö±ðÊÇij¿ÎÍâ»î¶¯Ð¡×éÉè¼ÆµÄÖÆÈ¡°±Æø²¢½øÐÐÅçȪʵÑéµÄÈý×é×°ÖÃʾÒâͼ£¬ÖÆÈ¡NH3Ñ¡ÓÃÊÔ¼ÁÈçͼËùʾ£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÓÃAͼËùʾµÄ×°ÖÿÉÖÆ±¸¸ÉÔïµÄNH3
¢Ù·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
CaCl2+2NH3¡ü+2H2O
2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
CaCl2+2NH3¡ü+2H2O
£®×°ÖÃÖÐÊÕ¼¯NH3µÄÊԹܿڷÅÖÃÃÞ»¨ÍŵÄ×÷ÓÃÊÇ
·ÀÖ¹¿ÕÆø¶ÔÁ÷£¬Ê¹NH3³äÂúÊÔ¹Ü
·ÀÖ¹¿ÕÆø¶ÔÁ÷£¬Ê¹NH3³äÂúÊÔ¹Ü
£®
¢Ú¸ÉÔï¹ÜÖиÉÔï¼ÁÄÜ·ñ¸ÄÓÃÎÞË®CaCl2
²»ÄÜ
²»ÄÜ
£¬ÀíÓÉÊÇ
CaCl2+8NH3=CaCl2?8NH3
CaCl2+8NH3=CaCl2?8NH3
£®
£¨2£©ÓÃBͼËùʾµÄ×°ÖÿɿìËÙÖÆÈ¡½Ï´óÁ¿NH3Äâ×÷ÅçȪʵÑ飮¸ù¾ÝBͼËùʾµÄ×°Öü°ÊÔ¼Á»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÓû¯Ñ§·½³Ìʽ±íʾŨ°±Ë®µÎÈëCaOÖÐÓдóÁ¿NH3ÒݳöµÄ¹ý³Ì£º
CaO+H2O=Ca£¨OH£©2£¬·´Ó¦·ÅÈÈ£¬NH3?H2O
  ¡÷  
.
 
NH3¡ü+H2O
CaO+H2O=Ca£¨OH£©2£¬·´Ó¦·ÅÈÈ£¬NH3?H2O
  ¡÷  
.
 
NH3¡ü+H2O

¢Ú¼ìÑéNH3ÊÇ·ñÊÕ¼¯ÂúµÄʵÑé·½·¨ÊÇ£º
Óò£Á§°ôպȡÉÙÐíŨÑÎËá¿¿½üÊÕ¼¯NH3µÄÊԹܿڣ¬Èô²úÉú°×ÑÌ£¬ËµÃ÷ÊÔ¹ÜÒÑÊÕ¼¯ÂúNH3£¬·´Ö®£¬ÔòûÓÐÊÕ¼¯Âú£®
»òÓÃʪÈóµÄºìɫʯÈïÊÔÖ½¿¿½üÊÕ¼¯NH3µÄÊԹܿڣ¬ÈôʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬Ôò˵Ã÷NH3ÒÑÊÕ¼¯Âú£¬·´Ö®£¬ÔòûÓÐÊÕ¼¯Âú£®
Óò£Á§°ôպȡÉÙÐíŨÑÎËá¿¿½üÊÕ¼¯NH3µÄÊԹܿڣ¬Èô²úÉú°×ÑÌ£¬ËµÃ÷ÊÔ¹ÜÒÑÊÕ¼¯ÂúNH3£¬·´Ö®£¬ÔòûÓÐÊÕ¼¯Âú£®
»òÓÃʪÈóµÄºìɫʯÈïÊÔÖ½¿¿½üÊÕ¼¯NH3µÄÊԹܿڣ¬ÈôʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬Ôò˵Ã÷NH3ÒÑÊÕ¼¯Âú£¬·´Ö®£¬ÔòûÓÐÊÕ¼¯Âú£®
£®
£¨3£©ÓÃCͼËùʾµÄ×°ÖýøÐÐÅçȪʵÑ飬Éϲ¿ÉÕÆ¿ÒѳäÂú¸ÉÔï°±Æø£¬Òý·¢Ë®ÉÏÅçµÄ²Ù×÷ÊÇ
´ò¿ªÖ¹Ë®¼Ð¼·³ö½ºÍ·µÎ¹ÜÖеÄË®
´ò¿ªÖ¹Ë®¼Ð¼·³ö½ºÍ·µÎ¹ÜÖеÄË®
£¬¸ÃʵÑéµÄÔ­ÀíÊÇ£º
NH3¼«Ò×ÈܽâÓÚË®£¬ÖÂʹÉÕÆ¿ÄÚѹǿѸËÙ¼õС
NH3¼«Ò×ÈܽâÓÚË®£¬ÖÂʹÉÕÆ¿ÄÚѹǿѸËÙ¼õС
£®
Èô²âµÃC×°ÖÃÉÕÆ¿ÖÐNH3µÄÖÊÁ¿ÊÇÏàͬ״¿öÏÂÏàͬÌå»ýH2ÖÊÁ¿µÄ10±¶£¬ÔòÅçȪʵÑéÍê±Ïºó£¬ÉÕÆ¿ÖÐË®¿ÉÉÏÉýÖÁÉÕÆ¿ÈÝ»ýµÄ
3
4
3
4
£¨Ìî¡°¼¸·ÖÖ®¼¸¡±£©£®
ͼÖÐA¡¢B¡¢C·Ö±ðÊÇij¿ÎÍâ»î¶¯Ð¡×éÉè¼ÆµÄÖÆÈ¡°±Æø²¢½øÐÐÅçȪʵÑéµÄÈý×é×°ÖÃʾÒâͼ£¬A¡¢BÊÇÖÆÈ¡NH3µÄʾÒâͼ£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÓÃA×°ÖÿÉÖÆ±¸NH3µÄ»¯Ñ§·½³Ìʽ
2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
CaCl2+2NH3¡ü+2H2O
2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
CaCl2+2NH3¡ü+2H2O
½«²úÉúµÄ°±ÆøÍ¨ÈëNaClOÈÜÒºÖÐN2H4£¬Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ
ClO-+2NH3¨TN2H4+Cl-+H2O
ClO-+2NH3¨TN2H4+Cl-+H2O

£¨2£©¢ÙÓÃB×°ÖÿɿìËÙÖÆÈ¡½Ï´óÁ¿NH3£®ÊÔ·ÖÎö²úÉú´óÁ¿°±ÆøµÄÔ­Òò£º
NH4++OH-?NH3?H2O
  ¡÷  
.
 
NH3¡ü+H2O£»CaO+H2O¡úCa£¨OH£©2£¬Éú³É¼îͬʱ·Å³ö´óÁ¿ÈÈ£¬¾ùÓÐÀûÓÚ°±ÆøÒݳö
NH4++OH-?NH3?H2O
  ¡÷  
.
 
NH3¡ü+H2O£»CaO+H2O¡úCa£¨OH£©2£¬Éú³É¼îͬʱ·Å³ö´óÁ¿ÈÈ£¬¾ùÓÐÀûÓÚ°±ÆøÒݳö

¢Ú¼ìÑéNH3ÊÇ·ñÊÕ¼¯ÂúµÄʵÑé·½·¨ÊÇ£º
Óò£Á§°ôպȡÉÙÐíŨÑÎËá¿¿½üÊÕ¼¯NH3µÄÊԹܿڣ¬Èô²úÉú°×ÑÌ£¬ËµÃ÷ÊÔ¹ÜÒÑÊÕ¼¯ÂúNH3£¬·´Ö®£¬ÔòûÓÐÊÕ¼¯Âú»òÓÃʪÈóµÄºìɫʯÈïÊÔÖ½¿¿½üÊÕ¼¯NH3µÄÊԹܿڣ¬ÈôʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬Ôò˵Ã÷NH3ÒÑÊÕ¼¯Âú£¬·´Ö®£¬ÔòûÓÐÊÕ¼¯Âú£®
Óò£Á§°ôպȡÉÙÐíŨÑÎËá¿¿½üÊÕ¼¯NH3µÄÊԹܿڣ¬Èô²úÉú°×ÑÌ£¬ËµÃ÷ÊÔ¹ÜÒÑÊÕ¼¯ÂúNH3£¬·´Ö®£¬ÔòûÓÐÊÕ¼¯Âú»òÓÃʪÈóµÄºìɫʯÈïÊÔÖ½¿¿½üÊÕ¼¯NH3µÄÊԹܿڣ¬ÈôʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬Ôò˵Ã÷NH3ÒÑÊÕ¼¯Âú£¬·´Ö®£¬ÔòûÓÐÊÕ¼¯Âú£®
£®
£¨3£©ÓÃC×°ÖýøÐÐÅçȪʵÑ飬Éϲ¿ÉÕÆ¿ÒѳäÂú¸ÉÔï°±Æø£¬Òý·¢Ë®ÉÏÅçµÄ²Ù×÷ÊÇ
´ò¿ªÖ¹Ë®¼Ð¼·³ö½ºÍ·µÎ¹ÜÖеÄË®
´ò¿ªÖ¹Ë®¼Ð¼·³ö½ºÍ·µÎ¹ÜÖеÄË®
£¬¸ÃʵÑéµÄÔ­ÀíÊÇ£º
NH3¼«Ò×ÈܽâÓÚË®£¬ÖÂʹÉÕÆ¿ÄÚѹǿѸËÙ¼õС
NH3¼«Ò×ÈܽâÓÚË®£¬ÖÂʹÉÕÆ¿ÄÚѹǿѸËÙ¼õС
£®
Èô²âµÃC×°ÖÃÉÕÆ¿ÖÐNH3µÄÖÊÁ¿ÊÇÏàͬ״¿öÏÂÏàͬÌå»ýH2ÖÊÁ¿µÄ10±¶£¬ÔòÅçȪʵÑéÍê±Ïºó£¬ÉÕÆ¿ÖÐË®¿ÉÉÏÉýÖÁÉÕÆ¿ÈÝ»ýµÄ
3
4
3
4
£¨Ìî¡°¼¸·ÖÖ®¼¸¡±£©£®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø