Ò»´ÎȤζ»¯Ñ§»î¶¯ÖУ¬ÍõÀÏʦÏòͬѧÃÇչʾÁËһƿ±êÇ©ÊÜËðµÄÎÞÉ«ÈÜÒº£¬ÈçÓÒͼËùʾ¡£ÒªÇóͬѧÃǽøÐÐ̽¾¿£ºÈ·ÈÏÕâÆ¿ÈÜÒº¾¿¾¹ÊÇʲôÈÜÒº£¿
¡¾Ìá³ö²ÂÏë¡¿  ÍõÀÏʦÌáʾ£ºÕâÆ¿ÎÞÉ«ÈÜÒºÖ»ÄÜÊÇÏÂÁÐËÄÖÖÈÜÒºÖеÄÒ»ÖÖ£º
¢ÙÁòËáþÈÜÒº  ¢ÚÁòËáÄÆÈÜÒº  ¢ÛÁòËáÈÜÒº  
¢ÜÁòËáï§ÈÜÒº
¡¾²éÔÄ×ÊÁÏ¡¿ ¢Ù³£ÎÂÏ£¬Ïà¹ØÎïÖʵÄÈܽâ¶ÈÈçÏ£º
¢Ú(NH4)2SO4µÄË®ÈÜÒºÏÔËáÐÔ
£¨1 £©¡¾ÊµÑé̽¾¿¡¿Ð¡Ã÷ÈÏΪÍõÀÏʦµÄÌáʾ²ÂÏë  ¢Ú²»ÕýÈ·£¬Ô­Òò                                         ¡£
£¨2 £©ÎªÈ·¶¨ÆäËü¼¸ÖÖ²ÂÏëÊÇ·ñÕýÈ·£¬Ð¡Ã÷ͬѧ¼ÌÐø½øÐÐ̽¾¿£º
СÑŶÔСÃ÷ʵÑé²Ù×÷¢ÚµÄ½áÂÛÆÀ¼ÛÊǽáÂÛÊDz»ÕýÈ·µÄ£¬ÀíÓÉÊÇ                       £»
£¨3£©ÇëÄãÉè¼ÆʵÑé·½°¸£¬È·ÈϸÃÈÜÒºÊÇÁòËáï§ÈÜÒº²¢Íê³ÉʵÑ鱨¸æ£º
£¨1 £©³£ÎÂÏÂNa2SO4ÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊý²»¿ÉÄÜ´ïµ½20% ¡£
£¨2 £©¢Ù NaOH  £¨»òÇâÑõ»¯ÄÆ£©£¨´ð°¸ºÏÀí¼´¿É£©    (NH4)2SO4ÈÜÒºÒ²³ÊËáÐÔ
£¨3 £©
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¾«Ó¢¼Ò½ÌÍøÒ»´ÎȤζ»¯Ñ§»î¶¯ÖУ¬ÍõÀÏʦÏòͬѧÃÇչʾÁËһƿ±êÇ©ÊÜËðµÄÎÞÉ«ÈÜÒº£¬ÈçͼËùʾ£®ÒªÇóͬѧÃǽøÐÐ̽¾¿£ºÈ·ÈÏÕâÆ¿ÈÜÒºµ¼µçÊÇʲôÈÜÒº£¿
Ìá³ö²ÂÏ룺
ÍõÀÏʦÌáʾ£ºÕâÆ¿ÎÞÉ«ÈÜÒºÖ»ÄÜÊÇÏÂÁËËÄÖÖÈÜÒºÖеÄÒ»ÖÖ£º
¢ÙÁòËáþÈÜÒº¡¢¢ÚÁòËáÄÆÈÜÒº¡¢¢ÛÁòËáÈÜÒº¡¢¢ÜÁòËáï§ÈÜÒº
²éÔÄ×ÊÁÏ£º
£¨1£©³£ÎÂÏ£¬Ïà¹ØÎïÖʵÄÈܽâ¶ÈÈçÏ£º
ÎïÖÊ MgSO4 Na2SO4 £¨NH4£©2SO4 H2SO4
Èܽâ¶È 35.1g 19.5g 75.4g ÓëË®ÈÎÒâ±È»¥ÈÜ
£¨2£©£¨NH4£©2SO4µÄË®ÈÜÒºÏÔËáÐÔ
ʵÑé̽¾¿£º
£¨1£©Í¨¹ý²éÔÄ×ÊÁÏ£¬Ð¡Ã÷ͬѧÈÏΪ²ÂÏë
 
£¨ÌîÐòºÅ£©²»³ÉÁ¢£¬Ô­ÒòÊÇ
 
£®
£¨2£©ÎªÈ·¶¨ÆäËü¼¸ÖÖ²ÂÏëÊÇ·ñÕýÈ·£¬Ð¡Ã÷ͬѧ¼ÌÐø½øÐÐ̽¾¿£º
ʵÑé²Ù×÷ ʵÑéÏÖÏó ʵÑé½áÂÛ
¢ÙÈ¡¸ÃÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬ÏòÆäÖеμӼ¸µÎ
 
ÈÜÒº
ÈÜÒºÖÐÓа×É«³ÁµíÉú³É ²ÂÏë¢Ù³ÉÁ¢
¢ÚÓò£Á§°ôպȡÉÙÐíÔ­ÈÜÒºµÎÔÚpHÊÔֽˮ£¬²¢¸ú±ÈÉ«¿¨¶ÔÕÕ
ÈÜÒºpHСÓÚ7

²ÂÏë¢Û³ÉÁ¢
СÑÅͬѧÈÏΪСÃ÷ʵÑé²Ù×÷¢ÚµÄ½áÂÛ²»ÕýÈ·£¬ËýµÄÀíÓÉÊÇ
 
£»
£¨3£©ÇëÄãÉè¼ÆʵÑé·½°¸£¬È·ÈϸÃÈÜÒºÊÇÁòËáï§ÈÜÒº²¢Íê³ÉʵÑ鱨¸æ£º
ʵÑé²Ù×÷ ʵÑéÏÖÏó ʵÑé½áÂÛ
È¡¸ÃÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬
 

 
²ÂÏë¢Ü³ÉÁ¢£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2013?ÆëºÓÏØһģ£©Ò»´ÎȤζ»¯Ñ§»î¶¯ÖУ¬ÍõÀÏʦÏòͬѧÃÇչʾÁËһƿ±êÇ©ÊÜËðµÄÎÞÉ«ÈÜÒº£¬ÈçͼËùʾ£®ÒªÇóͬѧÃǽøÐÐ̽¾¿£ºÈ·ÈÏÕâÆ¿ÈÜÒº¾¿¾¹ÊÇʲôÈÜÒº£¿
¡¾Ìá³ö²ÂÏë¡¿ÍõÀÏʦÌáʾ£ºÕâÆ¿ÎÞÉ«ÈÜÒºÖ»ÄÜÊÇÏÂÁÐËÄÖÖÈÜÒºÖеÄÒ»ÖÖ£º¢ÙÁòËáþÈÜÒº ¢ÚÁòËáÄÆÈÜÒº ¢ÛÁòËáÈÜÒº
¢ÜÁòËáï§ÈÜÒº
¡¾²éÔÄ×ÊÁÏ¡¿
¢Ù³£ÎÂÏ£¬Ïà¹ØÎïÖʵÄÈܽâ¶ÈÈçÏ£º
ÎïÖÊ MgSO4 Na2SO4 £¨NH4£©2SO4 H2SO4
Èܽâ¶È 35.1g 19.5g 75.4g ÓëË®ÈÎÒâ±È»¥ÈÜ
¢Ú£¨NH4£©2SO4µÄË®ÈÜÒºÏÔËáÐÔ
£¨1£©¡¾ÊµÑé̽¾¿¡¿Ð¡Ã÷¶ÔÍõÀÏʦµÄÌáʾ½øÐÐÁËÆÀ¼Û
²ÂÏë¢Ú²»³ÉÁ¢
²ÂÏë¢Ú²»³ÉÁ¢
£¬Ô­ÒòÊÇ
³£ÎÂÏÂNa2SO4ÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊý²»¿ÉÄÜ´ïµ½20%
³£ÎÂÏÂNa2SO4ÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊý²»¿ÉÄÜ´ïµ½20%
£®
£¨2£©ÎªÈ·¶¨ÆäËü¼¸ÖÖ²ÂÏëÊÇ·ñÕýÈ·£¬Ð¡Ã÷ͬѧ¼ÌÐø½øÐÐ̽¾¿£º
ʵÑé²Ù×÷ ʵÑéÏÖÏó ʵÑé½áÂÛ
¢ÙÈ¡¸ÃÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬ÏòÆäÖеμӼ¸µÎ
NaOH£¨»òÇâÑõ»¯ÄÆ£©£¨´ð°¸ºÏÀí¼´¿É£©
NaOH£¨»òÇâÑõ»¯ÄÆ£©£¨´ð°¸ºÏÀí¼´¿É£©
ÈÜÒº
ÈÜÒºÖÐÓа×É«³ÁµíÉú³É ²ÂÏë¢Ù³ÉÁ¢
¢ÚÓò£Á§°ôպȡÉÙÐíÔ­ÈÜÒºµÎÔÚpHÊÔÖ½ÉÏ£¬²¢¸ú±ê×¼±ÈÉ«¿¨¶ÔÕÕ
ÈÜÒºpHСÓÚ7

²ÂÏë¢Û³ÉÁ¢
СÑŶÔСÃ÷ʵÑé²Ù×÷¢ÚµÄ½áÂÛÆÀ¼ÛÊÇ
½áÂÛ²»ÕýÈ·
½áÂÛ²»ÕýÈ·
£¬ÀíÓÉÊÇ
£¨NH4£©2SO4ÈÜÒºÒ²³ÊËáÐÔ
£¨NH4£©2SO4ÈÜÒºÒ²³ÊËáÐÔ
£»
£¨3£©ÇëÄãÉè¼ÆʵÑé·½°¸£¬È·ÈϸÃÈÜÒºÊÇÁòËáï§ÈÜÒº²¢Íê³ÉʵÑ鱨¸æ£º
ʵÑé²Ù×÷ ʵÑéÏÖÏó ʵÑé½áÂÛ
È¡¸ÃÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬
ÏòÊÔ¹ÜÖмÓÈëÉÙÁ¿NaOHÈÜÒº²¢¼ÓÈÈ£¬½«ÊªÈóµÄºìɫʯÈïÊÔÖ½·ÅÔÚÊԹܿÚ
ÏòÊÔ¹ÜÖмÓÈëÉÙÁ¿NaOHÈÜÒº²¢¼ÓÈÈ£¬½«ÊªÈóµÄºìɫʯÈïÊÔÖ½·ÅÔÚÊԹܿÚ
Óд̼¤ÐÔÆøζµÄÆøÌå²úÉú£¬
ºìɫʯÈïÊÔÖ½±äÀ¶
Óд̼¤ÐÔÆøζµÄÆøÌå²úÉú£¬
ºìɫʯÈïÊÔÖ½±äÀ¶
²ÂÏë¢Ü³ÉÁ¢£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
£¨NH4£©2SO4+2NaOH
  ¡÷  
.
 
Na2SO4+2NH3¡ü+2H2O
£¨NH4£©2SO4+2NaOH
  ¡÷  
.
 
Na2SO4+2NH3¡ü+2H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

Ò»´ÎȤζ»¯Ñ§»î¶¯ÖУ¬ÍõÀÏʦÏòͬѧÃÇչʾÁËһƿ±êÇ©ÊÜËðµÄÎÞÉ«ÈÜÒº£¬ÈçͼËùʾ£®
ÒªÇóͬѧÃǽøÐÐ̽¾¿£ºÈ·ÈÏÕâÆ¿ÈÜÒº¾¿¾¹ÊÇʲôÈÜÒº£¿
¡¾Ìá³ö²ÂÏë¡¿ÍõÀÏʦÌáʾ£ºÕâÆ¿ÎÞÉ«ÈÜÒºÖ»ÄÜÊÇÏÂÁÐËÄÖÖÈÜÒºÖеÄÒ»ÖÖ£º¢ÙÁòËáþÈÜÒº ¢ÚÁòËáÄÆÈÜÒº ¢ÛÁòËáÈÜÒº ¢ÜÁòËáï§ÈÜÒº
¡¾²éÔÄ×ÊÁÏ¡¿
¢Ù³£ÎÂÏ£¬Ïà¹ØÎïÖʵÄÈܽâ¶ÈÈçÏ£º
ÎïÖÊ MgSO4 Na2SO4 £¨NH4£©2SO4 H2SO4
Èܽâ¶È 35.1g 19.5g 75.4g ÓëË®ÈÎÒâ±È»¥ÈÜ
¢Ú£¨NH4£©2SO4µÄË®ÈÜÒºÏÔËáÐÔ
£¨1£©¡¾ÊµÑé̽¾¿¡¿Ð¡Ã÷ÈÏΪÍõÀÏʦµÄÌáʾ²ÂÏë ¢Ú²»ÕýÈ·£¬Ô­ÒòÊÇ
³£ÎÂÏÂNa2SO4ÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊý²»¿ÉÄÜ´ïµ½20%
³£ÎÂÏÂNa2SO4ÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊý²»¿ÉÄÜ´ïµ½20%
£®
£¨2£©ÎªÈ·¶¨ÆäËü¼¸ÖÖ²ÂÏëÊÇ·ñÕýÈ·£¬Ð¡Ã÷ͬѧ¼ÌÐø½øÐÐ̽¾¿£º
ʵÑé²Ù×÷ ʵÑéÏÖÏó ʵÑé½áÂÛ
¢ÙÈ¡¸ÃÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬ÏòÆäÖеμӼ¸µÎ
NaOH
NaOH
ÈÜÒº
ÈÜÒºÖÐÓа×É«³ÁµíÉú³É ²ÂÏë¢Ù³ÉÁ¢
¢ÚÓò£Á§°ôպȡÉÙÐíÔ­ÈÜÒºµÎÔÚpHÊÔÖ½ÉÏ£¬²¢¸ú±ê×¼±ÈÉ«¿¨¶ÔÕÕ
ÈÜÒºpHСÓÚ7

²ÂÏë¢Û³ÉÁ¢
СÑŶÔСÃ÷ʵÑé²Ù×÷¢ÚµÄ½áÂÛÆÀ¼ÛÊǽáÂÛ²»ÕýÈ·£¬ÀíÓÉÊÇ
£¨NH4£©2SO4ÈÜÒºÒ²³ÊËáÐÔ
£¨NH4£©2SO4ÈÜÒºÒ²³ÊËáÐÔ
£»
£¨3£©ÇëÄãÉè¼ÆʵÑé·½°¸£¬È·ÈϸÃÈÜÒºÊÇÁòËáï§ÈÜÒº²¢Íê³ÉʵÑ鱨¸æ£º
ʵÑé²Ù×÷ ʵÑéÏÖÏó ʵÑé½áÂÛ
È¡¸ÃÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬
ÏòÊÔ¹ÜÖмÓÈëÉÙÁ¿NaOHÈÜÒº²¢¼ÓÈÈ£¬
ÏòÊÔ¹ÜÖмÓÈëÉÙÁ¿NaOHÈÜÒº²¢¼ÓÈÈ£¬


½«ÊªÈóµÄºìɫʯÈïÊÔÖ½·ÅÔÚÊԹܿÚ
½«ÊªÈóµÄºìɫʯÈïÊÔÖ½·ÅÔÚÊԹܿÚ
Óд̼¤ÐÔÆøζµÄÆøÌå²úÉú£¬
Óд̼¤ÐÔÆøζµÄÆøÌå²úÉú£¬


ºìɫʯÈïÊÔÖ½±äÀ¶
ºìɫʯÈïÊÔÖ½±äÀ¶
²ÂÏë¢Ü³ÉÁ¢£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
£¨NH4£©2SO4+2NaOH
£¨NH4£©2SO4+2NaOH


  ¡÷  
.
 
=Na2SO4+2NH3¡ü+2H2O
  ¡÷  
.
 
=Na2SO4+2NH3¡ü+2H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ò»´ÎȤζ»¯Ñ§»î¶¯ÖУ¬ÍõÀÏʦÏòͬѧÃÇչʾÁËһƿ±êÇ©ÊÜËðµÄÎÞÉ«ÈÜÒº£¬ÈçͼËùʾ£®ÒªÇóͬѧÃǽøÐÐ̽¾¿£ºÈ·ÈÏÕâÆ¿ÈÜÒºÊÇʲôÈÜÒº£¿
Ìá³ö²ÂÏ룺ÍõÀÏʦÌáʾ£ºÕâÆ¿ÎÞÉ«ÈÜÒºÖ»ÄÜÊÇÒÔÏÂËÄÖÖÈÜÒºÖеÄÒ»ÖÖ£º¢ÙÁòËáþÈÜÒº¢ÚÁòËáÄÆÈÜÒº¢ÛÁòËáÈÜÒº¢ÜÁòËáï§ÈÜÒº
²éÔÄ×ÊÁÏ£º
£¨1£©³£ÎÂÏ£¬Ïà¹ØÎïÖʵÄÈܽâ¶ÈÈçÏ£º
 ÎïÖÊ  MgSO4  Na2SO4  £¨NH4£©2SO4  H2SO4
 Èܽâ¶È  35.1g  19.5g  75.4g  ÓëË®ÈÎÒâ±È»¥ÈÜ
£¨2£©£¨NH4£©2SO4µÄË®ÈÜÒºÏÔËáÐÔ£®
ʵÑé̽¾¿£º
£¨1£©Í¨¹ý²éÔÄ×ÊÁÏ£¬Ð¡Ã÷ͬѧÈÏΪ²ÂÏë
¢Ú
¢Ú
£¨ÌîÐòºÅ£©²»³ÉÁ¢£¬Ô­ÒòÊÇ
³£ÎÂÏÂÁòËáÄƵÄÈÜÖÊÖÊÁ¿·ÖÊý²»»á´ïµ½20%
³£ÎÂÏÂÁòËáÄƵÄÈÜÖÊÖÊÁ¿·ÖÊý²»»á´ïµ½20%
£®
£¨2£©ÎªÈ·¶¨ÆäËü¼¸ÖÖ²ÂÏëÊÇ·ñÕýÈ·£¬Ð¡Ã÷ͬѧ¼ÌÐø½øÐÐ̽¾¿£º
ʵÑé²Ù×÷ ʵÑéÏÖÏó ʵÑé½áÂÛ
¢ÙÈ¡¸ÃÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬ÏòÆäÖеμӼ¸µÎ
NaOH
NaOH
ÈÜÒº
ÈÜÒºÖÐÓа×É«³ÁµíÉú³É ²ÂÏë¢Ù³ÉÁ¢
¢ÚÓò£Á§°ôպȡÉÙÐíÔ­ÈÜÒºµÎÔÚpHÊÔÖ½ÉÏ£¬²¢¸ú±ÈÉ«¿¨¶ÔÕÕ ÈÜÒºpHСÓÚ7 ²ÂÏë¢Û³ÉÁ¢
СÑÅͬѧÈÏΪСÃ÷ʵÑé²Ù×÷¢ÚµÄ½áÂÛ²»ÕýÈ·£¬ËýµÄÀíÓÉÊÇ
£¨NH4£©2SO4ÈÜÒºÒ²³ÊËáÐÔ
£¨NH4£©2SO4ÈÜÒºÒ²³ÊËáÐÔ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

Ò»´ÎȤζ»¯Ñ§»î¶¯ÖУ¬ÍõÀÏʦÏòͬѧÃÇչʾÁËһƿ±êÇ©ÊÜËðµÄÎÞÉ«ÈÜÒº£¬ÈçͼËùʾ£®ÒªÇóͬѧÃǽøÐÐ̽¾¿£ºÈ·ÈÏÕâÆ¿ÈÜÒº¾¿¾¹ÊÇʲôÈÜÒº£¿
¡¾Ìá³ö²ÂÏë¡¿ÍõÀÏʦÌáʾ£ºÕâÆ¿ÎÞÉ«ÈÜÒºÖ»ÄÜÊÇÏÂÃæËÄÖÖÈÜÒºÖеÄÒ»ÖÖ£º
¢ÙÁòËáþÈÜÒº¢ÚÁòËáÄÆÈÜÒº¢ÛÁòËáÈÜÒº¢ÜÁòËáï§ÈÜÒº
¡¾²éÔÄ×ÊÁÏ¡¿£¨1£©³£ÎÂÏ£¬Ïà¹ØÎïÖʵÄÈܽâ¶ÈÈçÏ£º
ÎïÖÊ MgSO4 Na2SO4 £¨NH4£©2SO4 H2SO4
Èܽâ¶È 35.1g 19.5g 75.4g ÓëË®ÒÔÈÎÒâ±È»¥ÈÜ
£¨2£©£¨NH4£©2SO4µÄË®ÈÜÒºÏÔËáÐÔ
£¨3£©º¬NH4+µÄÑÎÓë¼î»ìºÏÊÜÈÈ£¬»á·Å³öNH3
¡¾ÊµÑé̽¾¿¡¿£¨1£©Í¨¹ý²éÔÄ×ÊÁÏ£¬Ð¡Ã÷ͬѧÈÏΪ²ÂÏë¢Ú²»³ÉÁ¢£¬Ô­ÒòÊÇ
³£ÎÂÏÂNa2SO4ÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊý²»¿ÉÄÜ´ïµ½20%
³£ÎÂÏÂNa2SO4ÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊý²»¿ÉÄÜ´ïµ½20%
£®
£¨2£©ÎªÈ·¶¨ÆäËû¼¸ÖÖ²ÂÏëÊÇ·ñÕýÈ·£¬Ð¡Ã÷ͬѧ¼ÌÐø½øÐÐ̽¾¿£º
ʵÑé²Ù×÷ ʵÑéÏÖÏó ʵÑé½áÂÛ
¢ÙÈ¡¸ÃÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬ÏòÆäÖеμӼ¸µÎ
NaOH
NaOH
ÈÜÒº
ÈÜÒºÖÐÓа×É«³ÁµíÉú³É ²ÂÏë¢Ù³ÉÁ¢
¢ÚÓò£Á§°ôպȡÉÙÐíÔ­ÈÜÒºµÎÔÚpHÊÔÖ½ÉÏ£¬ºÍ±ÈÉ«¿¨¶ÔÕÕ ÈÜÒºµÄpHСÓÚ7 ²ÂÏë¢Û³ÉÁ¢
СÑÅͬѧÈÏΪСÃ÷ʵÑé²Ù×÷¢ÚµÄ½áÂÛ²»ÕýÈ·£¬ËýµÄÀíÓÉÊÇ
£¨NH4£©2SO4ÈÜÒºÒ²³ÊËáÐÔ
£¨NH4£©2SO4ÈÜÒºÒ²³ÊËáÐÔ
£®
£¨3£©ÇëÄãÉè¼ÆʵÑé·½°¸£¬È·ÈϸÃÈÜÒºÊÇÁòËáï§ÈÜÒº²¢Íê³ÉʵÑ鱨¸æ£º
ʵÑé²Ù×÷ ʵÑéÏÖÏó ʵÑé½áÂÛ
È¡¸ÃÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬
¼ÓÈëÉÙÁ¿µÄNaOHÈÜÒº£¬È»ºó½«ÊªÈóµÄºìɫʯÈïÊÔÖ½·ÅÔÚÊԹܿÚ
¼ÓÈëÉÙÁ¿µÄNaOHÈÜÒº£¬È»ºó½«ÊªÈóµÄºìɫʯÈïÊÔÖ½·ÅÔÚÊԹܿÚ

Óд̼¤ÐÔÆøζµÄÆøÌå²úÉú£¬ºìɫʯÈïÊÔÖ½±äÀ¶
Óд̼¤ÐÔÆøζµÄÆøÌå²úÉú£¬ºìɫʯÈïÊÔÖ½±äÀ¶
²ÂÏë¢Ü³ÉÁ¢£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ
£¨NH4£©2SO4+2NaOH=
Na2SO4+2NH3¡ü+2H2O
£¨NH4£©2SO4+2NaOH=
Na2SO4+2NH3¡ü+2H2O
£¨4£©¶¨Á¿Ì½¾¿
ÁòËá淋ÄÖƱ¸£º¹¤ÒµÉÏÓÃÁòÌú¿ó£¨Ö÷Òª³É·ÖÊÇFeS2£©£¬¾­¹ý·ÐÌÚ±ºÉÕ¯ÄÚͨ¿ÕÆøȼÉÕÉú³ÉSO2ºÍFe2O3£¬SO2¾­´ß»¯¼ÁºÍÑõÆøµÄ¹²Í¬×÷ÓÃÔÚ½Ó´¥ÊÒÄÚת»¯ÎªSO3£¬×îºóSO3ÔÚÎüÊÕËþÄÚºÍË®µÄ¹²Í¬×÷ÓÃת»¯ÎªH2SO4£®ÊµÑéÊÒ¾ÍÊÇÓÃÖƵÃH2SO4ÈÜÒºÖÐͨNH3Ï໥»¯ºÏÖƵÃһƿÉÏÊö132g20%£¨NH4£©2SO4ÈÜÒº£®£¨·´Ó¦Öв»¿¼ÂÇÎïÖʵÄËðºÄ£©Ë¼¿¼£º¢ÙÔÚ¹¤ÒµÉÏÒª°ÑÁòÌú¿óÏÈÄë³É·ÛÄ©ÔÙ·´Ó¦µÄÔ­ÒòÊÇ
Ôö´óÁòÌú¿óÓë¿ÕÆøµÄ½Ó´¥Ãæ»ý
Ôö´óÁòÌú¿óÓë¿ÕÆøµÄ½Ó´¥Ãæ»ý
£®¢ÚÒªÖƵÃ132g20%£¨NH4£©2SO4ÈÜÒº£¬Ð躬40% FeS2µÄÁòÌú¿óµÄÖÊÁ¿ÊÇ
30
30
 g£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸