ÌâÄ¿ÄÚÈÝ

10£®¹¤Òµ¾Æ¾«ÖгýÁËÒÒ´¼»¹º¬ÓÐÉÙÁ¿¼×´¼£¬¼×´¼Óж¾£¬Òò´ËÒûÓù¤Òµ¾Æ¾«ÅäÖÆµÄ¼Ù¾Æ»áʹÈËÊÓÁ¦Ñ¸ËÙϽµ¡¢Ê§Ã÷£¬ÉõÖÁËÀÍö£®Óë¼×´¼£¨CH3OH£©¡¢ÒÒ´¼£¨C2H5OH£©½á¹¹ÏàËÆµÄ»¯ºÏÎﻹÓбû´¼£¨C3H7OH£©µÈ£¬ÕâÀàÎïÖʳÆÎª´¼À࣮¸ù¾ÝÒÔÉÏ×ÊÁϻشðÏÂÁÐÎÊÌ⣺
£¨1£©´¼ÀàÎïÖʶ¼ÊÇÓÉ3ÖÖÔªËØ×é³É£®
£¨2£©±û´¼£¨C3H7OH£©ÖÐ̼¡¢ÇâÔªËØµÄÖÊÁ¿±ÈΪ9£º2£®
£¨3£©¼×´¼ºÍÒÒ´¼Ò»Ñù¶¼¾ßÓпÉȼÐÔ£¬Ð´³ö¼×´¼ÍêȫȼÉյĻ¯Ñ§·½³Ìʽ2CH3OH+3O2$\frac{\underline{\;µãȼ\;}}{\;}$2CO2+4H2O£®£®

·ÖÎö £¨1£©¸ù¾ÝÎïÖʵĻ¯Ñ§Ê½·ÖÎö½â´ð£»
£¨2£©¸ù¾Ý»¯ºÏÎïÖÐÔªËØÖÊÁ¿±ÈµÄ¼ÆËã·½·¨·ÖÎö£»
£¨3£©¸ù¾Ý»¯Ñ§·½³ÌʽµÄÊéд·½·¨·ÖÎö£»

½â´ð ½â£º£¨1£©Óɼ״¼£¨CH3OH£©¡¢ÒÒ´¼£¨C2H5OH£©¡¢±û´¼£¨C3H7OH£©µÄ»¯Ñ§Ê½¿ÉÒÔ¿´³ö£¬´¼ÀàÎïÖÊÊÇÓÉ̼¡¢Çâ¡¢ÑõÈýÖÖÔªËØ×é³ÉµÄ£»¹Ê´ð°¸Îª£º3£»
£¨2£©¸ù¾Ý»¯ºÏÎïÖÐÔªËØÖÊÁ¿±ÈµÄ¼ÆËã·½·¨£¬±û´¼£¨C3H7OH£©ÖÐ̼¡¢ÇâÔªËØµÄÖÊÁ¿±ÈΪ12¡Á3£º1¡Á8=9£º2£»¹Ê´ð°¸Îª£º9£º2£»
£¨3£©¼×´¼£¨CH3OH£©ÔÚ¿ÕÆøÖÐÍêȫȼÉÕÉú³É¶þÑõ»¯Ì¼ºÍË®£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2CH3OH+3O2$\frac{\underline{\;µãȼ\;}}{\;}$2CO2+4H2O£®¹Ê´ð°¸Îª£º2CH3OH+3O2$\frac{\underline{\;µãȼ\;}}{\;}$2CO2+4H2O£®

µãÆÀ ±¾ÌâÄѶȲ»´ó£¬¿¼²éѧÉúÔªËØÖÊÁ¿±ÈµÄ¼ÆËã¼°Êéд»¯Ñ§·½³ÌʽµÄÄÜÁ¦£¬»¯Ñ§·½³ÌʽÊéд¾­³£³öÏֵĴíÎóÓв»·ûºÏ¿Í¹ÛÊÂʵ¡¢²»×ñÊØÖÊÁ¿Êغ㶨ÂÉ¡¢²»Ð´Ìõ¼þ¡¢²»±ê·ûºÅµÈ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
2£®ÅòËɼÁÊÇÒ»ÖÖʳƷÌí¼Ó¼Á£®ËüÔÚÃæ°üÖÆ×÷¹ý³ÌÖз¢Éú·´Ó¦²úÉúÆøÌå£¬Ê¹Ãæ°ü±äµÃËÉÈí£®ÒÑ֪ijÅòËɼÁº¬Ì¼ËáÇâÄÆ¡¢Ì¼ËáÇâï§£¨NH4HCO3£©£¬¼×¡¢ÒÒ»¯Ñ§ÐËȤС×é¶Ô¸ÃÅòËɼÁ¸÷ÎïÖʵĺ¬Á¿½øÐÐÈçÏÂ̽¾¿£®
ÒÑÖª£º¢Ù2NaHCO3$\frac{\underline{\;\;¡÷\;\;}}{\;}$ Na2CO3+H2O+CO2¡ü£» NH4HCO3$\frac{\underline{\;\;¡÷\;\;}}{\;}$ NH3¡ü+H2O+CO2¡ü£®
¢ÚŨÁòËá³£ÓÃÓÚÎüÊÕË®ÕôÆøºÍ°±Æø£®
¢Û¼îʯ»ÒÊǹÌÌåÇâÑõ»¯ÄƺÍÑõ»¯¸ÆµÄ»ìºÏÎ²»Óë°±Æø·´Ó¦£®
¼××éͬѧ°´ÈçͼËùʾװÖã¨ÆøÃÜÐÔÁ¼ºÃ£¬¸÷×°ÖÃÖеÄÊÔ¼ÁΪ×ãÁ¿£©Ì½¾¿Éú³É¶þÑõ»¯Ì¼µÈÎïÖʵÄÖÊÁ¿£®

¢ñ£®½«ÖÊÁ¿ÎªmÅòËɼÁ×°Èë×°ÖÃBÖУ¬ÏȶÔÒÑÁ¬½ÓºÃµÄ×°ÖÃA¡¢BºÍCͨһ¶Îʱ¼äµÄ¿ÕÆø£¬ÔÙÁ¬½ÓÉÏ×°ÖÃD¡¢E£®¹Ø±Õµ¯»É¼Ð£®´ËʱÒÑ×°ÈëÒ©Æ·µÄB¡¢C¡¢DµÄÖÊÁ¿£¬·Ö±ðΪ£ºb1¡¢c1¡¢d1
¢ò£®¼ÓÈȲ£Á§¹Ü£¬´ýÑùÆ·ÍêÈ«·´Ó¦£¬´ò¿ªµ¯»É¼ÐͨÈë¿ÕÆø£¬Ï¨Ãð¾Æ¾«µÆ£¬Ö±µ½²£Á§¹ÜÀäÈ´£®
¢ó£®³ÆÁ¿B¡¢C¡¢DµÄÖÊÁ¿£¬·Ö±ðΪ£ºb2¡¢c2¡¢d2
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¼ÓÈȹý³ÌÖУ¬×°ÖÃCµÄ×÷ÓÃÊÇÎüÊÕË®ÕôÆøºÍ°±Æø£®
£¨2£©Í¨¹ý²â¶¨ÊµÑéǰºó×°ÖÃD£¨Ìî×ÖĸÐòºÅ£©µÄÖÊÁ¿£¬ÆäÖÊÁ¿²î¼´ÎªÑùÆ·²úÉúµÄ¶þÑõ»¯Ì¼ÆøÌåµÄÖÊÁ¿£®Í¬Ñ§ÃÇÈÏΪ²»ÄÜ£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©¾Ý´ËÀ´¼ÆËã̼ËáÇâÄÆµÄº¬Á¿£¬ÀíÓÉÊÇ̼ËáÇâï§¡¢Ì¼ËáÇâÄÆ·Ö½â¶¼»áÉú³É¶þÑõ»¯Ì¼£®
£¨3£©Í¬Ñ§ÃÇÒÔ×°ÖÃBµÄÖÊÁ¿±ä»¯Çó³ö̼ËáÇâÄÆµÄÖÊÁ¿Îª$\frac{84}{53}$£¨ b2-b1+m£©£®
£¨4£©ÒÒ×éͬѧ»¹²â¶¨Á˰±ÆøµÄÖÊÁ¿£ºµ÷ÕûÉÏͼװÖõÄÁ¬½Ó˳ÐòΪABDCE£¬ÁíÈ¡ÖÊÁ¿ÎªmµÄÅòËɼÁ×°Èë×°ÖÃBÖУ¬Í¨¹ý²â¶¨ÊµÑéǰºó×°ÖÃCµÄÖÊÁ¿£¬ÆäÖÊÁ¿²î¼´ÎªÑùÆ·²úÉúµÄ°±ÆøµÄÖÊÁ¿£¬´Ó¶ø²â³ö̼ËáÇâ淋ĺ¬Á¿£®

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

¾«Ó¢¼Ò½ÌÍø