´Å²ÄÑõ»¯ÌúµÄ»¯Ñ§Ãû³ÆÊÇÑõ»¯Ìú£¨Fe2O3£©£¬ËüÊǵç×Ó¡¢µçÐŹ¤ÒµµÄ´ÅÐÔ²ÄÁÏ£¬¹¤ÒµÉϲÉÓÃÉú²úîÑ°×·ÛµÄϽÅÁÏ£¨º¬´óÁ¿FeSO4ÈÜÒº£©ÖƱ¸´Å²ÄÑõ»¯ÌúµÄ²½ÖèÈçÏ£º

¢ÙÏòϽÅÁÏ£¨º¬´óÁ¿FeSO4ÈÜÒº£©ÖмÓÈëÉÙÁ¿2 mo1¡¤L-1½ÐH2SO4ºÍÌúƤ£»
¢ÚÏò¢ÙÖÐËùµÃÈÜÒºÖмÓÐõÄý¼ÁºÍË®£¬¾­¹ýÂ˳ýÈ¥Îü¸½ÁËÔÓÖʵÄÐõÄý¼Á£»
¢Û½«¢ÚËùµÃµÄÂËҺŨËõ½á¾§µÃµ½¾§ÌåA£»
¢Ü½«¾§ÌåAÈÜÓÚË®£¬²¢¼ÓÈëNH4HCO3£¬²úÉúCO2ÆøÌåͬʱµÃµ½FeCO3³ÁµíºÍÎÞÉ«ÈÜÒºC£»
¢Ý½«FeCO3³ÁµíÏ´µÓ¡¢ºæ¸É²¢ìÑÉÕ¡££¨ìÑÉÕÖеı仯Ϊ£ºFeCO3="FeO+CO2¡ü; " 4FeO+O2="2" Fe2O3£©
¾ÝÉÏÊöÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÃ18.4mo1¡¤L-1µÄH2SO4ÅäÖÆ500mL 2 mo1¡¤L-1H2SO4£¬ËùÐè²£Á§ÒÇÆ÷³ý
         mLÁ¿Í²¡¢²£Á§°ô¡¢ÉÕ±­¡¢500mI¡£ÈÝÁ¿Æ¿Í⣬»¹ÐèÒª        ¡£
£¨2£©²½Öè¢ÙÖÐ2 mo1¡¤L-1H2SO4ºÍÌúƤµÄ×÷Ó÷ֱðΪ        ¡£
£¨3£©¾§ÌåAµÄ»¯Ñ§Ê½Îª        £¬¼ìÑéÈÜÒºCÖÐÈÜÖÊÑôÀë×ӵķ½·¨ÊÇ        ¡£
£¨4£©ÏòÈÜÒºCÖмÓÈëCaCl2ÈÜÒºÄܵõ½CaSO4³Áµí£¬³£ÎÂÏÂKSP£¨CaSO4£©=9x10-6£¬³£ÎÂÏÂCaSO4ÔÚË®ÖеijÁµíÈܽâƽºâÇúÏßÈçͼËùʾ¡£
¢Ùaµã¶ÔÓ¦µÄKSP        cµã¶ÔÓ¦µÄKSPÌî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©£»
¢ÚÈôÓÉbµã±äµ½aµãÏÂÁдëÊ©¿ÉÐеÄÊÇ         ¡£

A£®¼ÓÈëÊÊÁ¿CaCl2B£®¼ÓÈëÊÊÁ¿BaCl2£®
C£®¼ÓÈËÊÊÁ¿Na2SO4D£®Õô·¢

£¨1£©100£¬£¨2·Ö£©½ºÍ·µÎ¹Ü£¨1·Ö£© £¨2£©¼Ó2mol¡¤·ÀÖ¹Fe·¢ÉúË®½â£¬£¨1·Ö£© ¼ÓÌúƤ·ÀÖ¹Fe±»Ñõ»¯£¨1·Ö£©   £¨3£©£¬£¨1·Ö£©È¡¸ÃÈÜÒºÉÙÐí¼ÓÈëÇâÑõ»¯ÄÆÈÜÒº²¢¼ÓÈÈ£¬Èô²úÉúÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌ壬֤Ã÷¸ÃÈÜÒºµÄÑôÀë×ÓΪ¡££¨3·Ö£© £¨4£©µÈÓÚ£¬£¨2·Ö£©  A¡¢B¡££¨2·Ö£©

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

´Å²ÄÑõ»¯ÌúµÄ»¯Ñ§Ãû³ÆÊÇÑõ»¯Ìú£¨Fe2O3£©£¬ËüÊǵç×Ó¡¢µçÐŹ¤ÒµµÄ´ÅÐÔ²ÄÁÏ£¬¹¤ÒµÉϲÉÓÃÉú²úîÑ°×·ÛµÄϽÅÁÏ£¨º¬´óÁ¿FeSO4ÈÜÒº£©ÖƱ¸´Å²ÄÑõ»¯ÌúµÄ²½ÖèÈçÏ£º

    ¢ÙÏòϽÅÁÏ£¨º¬´óÁ¿FeSO4ÈÜÒº£©ÖмÓÈëÉÙÁ¿2 mo1¡¤L-1½ÐH2SO4ºÍÌúƤ£»

¢ÚÏò¢ÙÖÐËùµÃÈÜÒºÖмÓÐõÄý¼ÁºÍË®£¬¾­¹ýÂ˳ýÈ¥Îü¸½ÁËÔÓÖʵÄÐõÄý¼Á£»

    ¢Û½«¢ÚËùµÃµÄÂËҺŨËõ½á¾§µÃµ½¾§ÌåA£»

¢Ü½«¾§ÌåAÈÜÓÚË®£¬²¢¼ÓÈëNH4HCO3£¬²úÉúCO2ÆøÌåͬʱµÃµ½FeCO3³ÁµíºÍÎÞÉ«ÈÜÒºC£»

    ¢Ý½«FeCO3³ÁµíÏ´µÓ¡¢ºæ¸É²¢ìÑÉÕ¡££¨ìÑÉÕÖеı仯Ϊ£ºFeCO3=FeO+CO2¡ü;  4FeO+O2=2 Fe2O3£©

¾ÝÉÏÊöÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÓÃ18.4mo1¡¤L-1µÄH2SO4ÅäÖÆ500mL 2 mo1¡¤L-1H2SO4£¬ËùÐè²£Á§ÒÇÆ÷³ý

          mLÁ¿Í²¡¢²£Á§°ô¡¢ÉÕ±­¡¢500mI¡£ÈÝÁ¿Æ¿Í⣬»¹ÐèÒª         ¡£

£¨2£©²½Öè¢ÙÖÐ2 mo1¡¤L-1H2SO4ºÍÌúƤµÄ×÷Ó÷ֱðΪ         ¡£

£¨3£©¾§ÌåAµÄ»¯Ñ§Ê½Îª         £¬¼ìÑéÈÜÒºCÖÐÈÜÖÊÑôÀë×ӵķ½·¨ÊÇ         ¡£

£¨4£©ÏòÈÜÒºCÖмÓÈëCaCl2ÈÜÒºÄܵõ½CaSO4³Áµí£¬³£ÎÂÏÂKSP£¨CaSO4£©=9x10-6£¬³£ÎÂÏÂCaSO4ÔÚË®ÖеijÁµíÈܽâƽºâÇúÏßÈçͼËùʾ¡£

¢Ùaµã¶ÔÓ¦µÄKSP         cµã¶ÔÓ¦µÄKSPÌî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©£»

¢ÚÈôÓÉbµã±äµ½aµãÏÂÁдëÊ©¿ÉÐеÄÊÇ          ¡£

        A£®¼ÓÈëÊÊÁ¿CaCl2        B£®¼ÓÈëÊÊÁ¿BaCl2£®

        C£®¼ÓÈËÊÊÁ¿Na2SO4       D£®Õô·¢

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

´Å²ÄÑõ»¯ÌúµÄ»¯Ñ§Ãû³ÆÊÇÑõ»¯Ìú£¨Fe2O3£©£¬ËüÊǵç×Ó¡¢µçÐŹ¤ÒµµÄ´ÅÐÔ²ÄÁÏ£¬¹¤ÒµÉϲÉÓÃÉú²úîÑ°×·ÛµÄϽÅÁÏ£¨º¬´óÁ¿FeSO4ÈÜÒº£©ÖƱ¸´Å²ÄÑõ»¯ÌúµÄ²½ÖèÈçÏ£º

¢ÙÏòϽÅÁÏ£¨º¬´óÁ¿FeSO4ÈÜÒº£©ÖмÓÈëÉÙÁ¿2 mo1¡¤L-1½ÐH2SO4ºÍÌúƤ£»

¢ÚÏò¢ÙÖÐËùµÃÈÜÒºÖмÓÐõÄý¼ÁºÍË®£¬¾­¹ýÂ˳ýÈ¥Îü¸½ÁËÔÓÖʵÄÐõÄý¼Á£»

¢Û½«¢ÚËùµÃµÄÂËҺŨËõ½á¾§µÃµ½¾§ÌåA£»

¢Ü½«¾§ÌåAÈÜÓÚË®£¬²¢¼ÓÈëNH4HCO3£¬²úÉúCO2ÆøÌåͬʱµÃµ½FeCO3³ÁµíºÍÎÞÉ«ÈÜÒºC£»

¢Ý½«FeCO3³ÁµíÏ´µÓ¡¢ºæ¸É²¢ìÑÉÕ¡££¨ìÑÉÕÖеı仯Ϊ£ºFeCO3=FeO+CO2¡ü;  4FeO+O2=2 Fe2O3£©

¾ÝÉÏÊöÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÓÃ18.4mo1¡¤L-1µÄH2SO4ÅäÖÆ500mL 2 mo1¡¤L-1H2SO4£¬ËùÐè²£Á§ÒÇÆ÷³ý

          mLÁ¿Í²¡¢²£Á§°ô¡¢ÉÕ±­¡¢500mI¡£ÈÝÁ¿Æ¿Í⣬»¹ÐèÒª        ¡£

£¨2£©²½Öè¢ÙÖÐ2mo1¡¤L-1H2SO4ºÍÌúƤµÄ×÷Ó÷ֱðΪ        ¡£

£¨3£©¾§ÌåAµÄ»¯Ñ§Ê½Îª         £¬¼ìÑéÈÜÒºCÖÐÈÜÖÊÑôÀë×ӵķ½·¨ÊÇ        ¡£

£¨4£©ÏòÈÜÒºCÖмÓÈëCaCl2ÈÜÒºÄܵõ½CaSO4³Áµí£¬³£ÎÂÏÂKSP£¨CaSO4£©=9x10-6£¬³£ÎÂÏÂCaSO4ÔÚË®ÖеijÁµíÈܽâƽºâÇúÏßÈçͼËùʾ¡£

¢Ùaµã¶ÔÓ¦µÄKSP         cµã¶ÔÓ¦µÄKSPÌî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©£»

¢ÚÈôÓÉbµã±äµ½aµãÏÂÁдëÊ©¿ÉÐеÄÊÇ         ¡£

    A£®¼ÓÈëÊÊÁ¿CaCl2                  B£®¼ÓÈëÊÊÁ¿BaCl2£®

    C£®¼ÓÈËÊÊÁ¿Na2SO4                 D£®Õô·¢

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨16·Ö£©´Å²ÄÑõ»¯ÌúµÄ»¯Ñ§Ãû³ÆÊÇÑõ»¯Ìú£¨Fe2O3£©£¬ËüÊǵç×Ó¡¢µçÐŹ¤ÒµµÄ´ÅÐÔ²ÄÁÏ£¬¹¤ÒµÉϲÉÓÃÉú²úîÑ°×·ÛµÄϽÅÁÏ£¨º¬´óÁ¿FeSO4µÄ·ÏÒº£©ÎªÔ­ÁÏÀ´ÖƱ¸´Å²ÄÑõ»¯Ìú¡£

ÒÑÖªìÑÉÕÖеĻ¯Ñ§·´Ó¦·½³ÌʽΪ£ºFeCO3 ¡ú FeO + CO2¡ü£¬4FeO + O2 ¡ú 2Fe2O3

£¨1£©ÓÃ98%µÄH2SO4µÄÅäÖÆ500mLµÄ20%µÄH2SO4£¬ËùÐè²£Á§ÒÇÆ÷ÊÇ         

A¡¢²£Á§°ô          B¡¢ÉÕ±­         C¡¢Â©¶·          D¡¢250mLÈÝÁ¿Æ¿

E¡¢500mLÈÝÁ¿Æ¿         F¡¢½ºÍ·µÎ¹Ü

£¨2£©Å¨Ëõ½á¾§ºóµÃµ½µÄ¾§ÌåÊÇ               £¨Ìѧʽ£©£¬AÖÐÈÜÖÊÊÇ           £»

´ËÈÜÒºÖи÷Àë×ÓµÄŨ¶È±È½Ï´óСΪ£º                                     ¡£

£¨3£©20%H2SO4ºÍÌúƤµÄ×÷Ó÷ֱðÊÇ                                           ¡£

£¨4£©¼ìÑéÂËÒºÖк¬ÓÐNH4+µÄ·½·¨ÊÇ                                                

                                                                            

£¨5£©Ð´³ö²½Öè¡°ºÏ³É¡±Öз¢ÉúµÄ»¯Ñ§±ä»¯£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£º

                                                              ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011½ì¹ã¶«Ê¡ÁùУ¸ßÈýÉÏѧÆÚµÚ¶þ´ÎÁª¿¼£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÊµÑéÌâ

£¨16·Ö£©´Å²ÄÑõ»¯ÌúµÄ»¯Ñ§Ãû³ÆÊÇÑõ»¯Ìú£¨Fe2O3£©£¬ËüÊǵç×Ó¡¢µçÐŹ¤ÒµµÄ´ÅÐÔ²ÄÁÏ£¬¹¤ÒµÉϲÉÓÃÉú²úîÑ°×·ÛµÄϽÅÁÏ£¨º¬´óÁ¿FeSO4µÄ·ÏÒº£©ÎªÔ­ÁÏÀ´ÖƱ¸´Å²ÄÑõ»¯Ìú¡£

ÒÑÖªìÑÉÕÖеĻ¯Ñ§·´Ó¦·½³ÌʽΪ£ºFeCO3 ¡ú FeO + CO2¡ü£¬4FeO + O2 ¡ú 2Fe2O3
£¨1£©ÓÃ98%µÄH2SO4µÄÅäÖÆ500mLµÄ20%µÄH2SO4£¬ËùÐè²£Á§ÒÇÆ÷ÊÇ        

A£®²£Á§°ôB£®ÉÕ±­C£®Â©¶·D£®250mLÈÝÁ¿Æ¿
E¡¢500mLÈÝÁ¿Æ¿         F¡¢½ºÍ·µÎ¹Ü
£¨2£©Å¨Ëõ½á¾§ºóµÃµ½µÄ¾§ÌåÊÇ               £¨Ìѧʽ£©£¬AÖÐÈÜÖÊÊÇ           £»
´ËÈÜÒºÖи÷Àë×ÓµÄŨ¶È±È½Ï´óСΪ£º                                     ¡£
£¨3£©20%H2SO4ºÍÌúƤµÄ×÷Ó÷ֱðÊÇ                                           ¡£
£¨4£©¼ìÑéÂËÒºÖк¬ÓÐNH4+µÄ·½·¨ÊÇ                                                
                                                                            
£¨5£©Ð´³ö²½Öè¡°ºÏ³É¡±Öз¢ÉúµÄ»¯Ñ§±ä»¯£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£º
                                                              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸