ÌâÄ¿ÄÚÈÝ

ÒÑÖª·´Ó¦£º
¡¢
ÏÖÓÐÎïÖÊA~IµÄת»¯¹ØϵÈçÏÂͼ£º

ÈôBµÄ·Ö×ÓʽΪC8H8O£¬Æä±½»·ÉϵÄһԪȡ´úÎïÖ»ÓÐÁ½ÖÖ£»GΪ¸ß·Ö×Ó»¯ºÏÎï¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)д³öÏÂÁз´Ó¦µÄ·´Ó¦ÀàÐÍ£º·´Ó¦¢Ü               £¬·´Ó¦¢Û               ¡£
(2)д³öÏÂÁÐÎïÖʵĽṹ¼òʽ£ºF               £¬I               £¬A               ¡£
(3)д³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢ÙB¡úC£º                                                            £»
¢ÚC+D¡úH£º                                                         £»
¢ÛF¡úG£º                                                            £»
(4)CµÄͬËØ·ÖÒì¹¹ÌåÇÒÊôÓÚõ¥ÀàµÄ·¼Ïã×廯ºÏÎï¹²ÓР      ÖÖ£¬Çëд³öÆäÖÐÒ»ÖÖͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º                                ¡£
(1)Ëõ¾Û£»õ¥»¯£»(2) HOCH2¡ª¡ªCOOH£»HOOC¡ª¡ªCOOH£»¡ªCH3£»
(3)¢ÙH3C¡ª¡ªCHO + 2Cu(OH)2 H3C¡ª¡ªCOOH + Cu2O¡ý+ 2H2O£»
¢ÚH3C¡ª¡ªCOOH+HOCH2¡ª¡ªCH3            H3C¡ª¡ªCOOCH2¡ª¡ªCH3+H2O

¢ÛnHOCH2¡ª¡ªCOOH                                 + nH2O
(4)6£»

                              ÖÐÈÎдһÖÖ¡£
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
(11·Ö)£¨1£©Ä¿Ç°¹¤ÒµÉÏÓÐÒ»ÖÖ·½·¨ÊÇÓÃCO2À´Éú²úȼÁϼ״¼¡£ÎªÌ½¾¿·´Ó¦Ô­Àí£¬ÏÖ½øÐÐÈçÏÂʵÑ飬ÔÚÌå»ýΪ1 LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë1mol CO2ºÍ3mol H2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º
CO2(g)£«3H2(g)CH3OH(g)£«H2O(g) £¬¡÷H£½£­49.0kJ/mol£»²âµÃCO2ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ¡£

¢Ù´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬Æ½¾ù·´Ó¦ËÙÂÊv(CO2)£½             mol/(L¡¤min)¡£
¢Ú¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ________________________________________¡£
¢ÛÏÂÁдëÊ©ÖÐÄÜʹn(CH3OH)£¯n(CO2)Ôö´óµÄÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
A£®Éý¸ßζÈB£®³äÈëHe(g)£¬Ê¹ÌåϵѹǿÔö´ó
C£®½«H2O(g)´ÓÌåϵÖзÖÀëD£®ÔÙ³äÈë1mol H2
£¨2£©ÔÚÔØÈ˺½ÌìÆ÷µÄÉú̬ϵͳÖУ¬²»½öÒªÇó·ÖÀëÈ¥³ýCO2£¬»¹ÒªÇóÌṩ³ä×ãµÄO2¡£Ä³Öֵ绯ѧװÖÿÉʵÏÖÈçÏÂת»¯£º2CO2£½2CO£«O2¡¡£¬CO¿ÉÓÃ×÷ȼÁÏ¡£
ÒÑÖª¸Ã·´Ó¦µÄÑô¼«·´Ó¦Îª£º4OH¨D¨D4e¨D£½O2¡ü£«2H2O
ÔòÒõ¼«·´Ó¦Ê½Îª£º                                              ¡£
ÓÐÈËÌá³ö£¬¿ÉÒÔÉè¼Æ·´Ó¦2CO£½2C£«O2£¨¡÷H£¾0¡¢¡÷S£¼0£©À´Ïû³ýCOµÄÎÛȾ¡£ÇëÄãÅжÏÊÇ·ñ¿ÉÐв¢Ëµ³öÀíÓÉ£º¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£¬¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
ijÓлúÎïA£¨Ö»º¬C¡¢H¡¢OÈýÖÖÔªËØ£©ÊÇÒ»ÖÖÖØÒª»¯¹¤Éú²úµÄÖмäÌå¡£ÒÑÖª£º
¢ÙAµÄÕôÆøÃܶÈÊÇÏàͬ״¿öÏÂÇâÆøÃܶȵÄ83±¶£¬·Ö×ÓÖÐ̼ԭ×Ó×ÜÊýÊÇÑõÔ­×Ó×ÜÊýµÄ3±¶¡£
¢ÚAÊôÓÚ·¼Ïã×廯ºÏÎÆä±½»·ÉÏÖ»ÓÐÒ»¸öÈ¡´ú»ù£¬ÇÒÈ¡´ú»ù̼Á´ÉÏÎÞÖ§Á´£»
¢ÛA¿ÉÓëNaHCO3ÈÜÒº×÷Ó㬲úÉúÎÞÉ«ÆøÅÝ£»
¢ÜAÔÚÒ»¶¨Ìõ¼þÏ¿ÉÓëÒÒËá·¢Éúõ¥»¯·´Ó¦£»
¢ÝA´æÔÚÈçÏÂת»¯¹Øϵ£º

ÊÔ¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ðÎÊÌ⣺
£¨1£©AµÄ·Ö×Óʽ            £»AÖк¬Ñõ¹ÙÄÜÍŵÄÃû³Æ               ¡£
£¨2£©DµÄ½á¹¹¼òʽΪ              ¡£
£¨3£©Ð´³öA¡úCºÍA¡úBµÄ»¯Ñ§·´Ó¦·½³Ìʽ£¨×¢Ã÷·´Ó¦Ìõ¼þ£©£¬²¢×¢Ã÷·´Ó¦ÀàÐÍ£º
A¡úC£º                         _____________________________________£¬
·´Ó¦ÀàÐÍ£º               £»
A¡úB£º                         ______________________________________£¬
·´Ó¦ÀàÐÍ£º               ¡£
£¨4£©·ûºÏÏÂÁÐÈý¸öÌõ¼þµÄAµÄͬ·ÖÒì¹¹ÌåµÄÊýÄ¿¹²ÓР       ¸ö¡£
¢Ù±½»·ÉÏÖ»ÓÐÁ½¸öÁÚλȡ´ú»ù
¢ÚÄÜÓëÈýÂÈ»¯ÌúÈÜÒº·¢ÉúÏÔÉ«·´Ó¦
¢ÛÒ»¶¨Ìõ¼þÏ¿ÉÒÔË®½âÉú³ÉÁ½ÖÖÎïÖÊ
д³öÆäÖÐÈÎÒâÒ»¸öͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ                          ¡£
£¨15·Ö£©
ÓÉ̼¡¢Çâ¡¢ÑõÈýÖÖÔªËØ×é³ÉµÄijÓлúÎïXµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª136£¬ÆäÖÐ̼¡¢ÇâÔªËصÄÖÊÁ¿·ÖÊýÖ®ºÍΪ76.5©‡.ÓÖÖªXÖк¬Óм׻ù£¬±½»·ÉϵÄÒ»ÂÈ´úÎïÓÐÁ½ÖÖ¡£X¾­ÈȵÄËáÐÔKMnO4ÈÜÒº´¦Àíºóת»¯ÎªY,1mol YÓë×ãÁ¿NaHCO3ÈÜÒº·´Ó¦Éú³É±ê׼״̬ÏÂ44.8L CO2ÆøÌå¡£
¾Ý´ËÍê³ÉÏÂÁÐÒªÇó¡£
£¨1£©XÖк¬Ñõ¹ÙÄÜÍŵÄÃû³ÆΪ                 £¬1mol Y×î¶à¿ÉÓë                mol Na·´Ó¦¡£
£¨2£©ÒÑÖªA¡¢B¡¢C¡¢D¾ùÓëX»¥ÎªÍ¬·ÖÒì¹¹Ì壬ÇÒ¾ùÊÇÓÐÒ»¸öÈ¡´ú»ùµÄ·¼Ïã×廯ºÏÎÆäÖÐB¿ÉÒÔ·¢ÉúÒø¾µ·´Ó¦¡£Ïà¹Øת»¯¹ØϵÈçÏÂͼ¡£

¢ÙAµÄ½á¹¹¼òʽΪ                 ¡£A¡«KµÄ¸÷ÓлúÎïÖУ¬ÊôÓÚËáµÄÊÇ                 £¨Ìî×Öĸ£©¡£
¢ÚCÓë×ãÁ¿µÄNaOHÈÜÒº¼ÓÈÈʱ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                 £¬¸Ã·´Ó¦ÖÐÉæ¼°µ½µÄÓлú·´Ó¦ÀàÐÍΪ                 ¡£
¢ÛÊôÓÚ·¼Ïã×廯ºÏÎïµÄEµÄͬ·ÖÒì¹¹ÌåÓР               ÖÖ£¨²»º¬E£©¡£
¢Ü×ãÁ¿µÄGÓëYÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                        ¡£

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

¾«Ó¢¼Ò½ÌÍø