ÌâÄ¿ÄÚÈÝ

(12·Ö)ÒÑ֪ij¹¤Òµ·ÏË®Öк¬ÓдóÁ¿FeSO4£¬½Ï¶àµÄCu2+£¬ÉÙÁ¿µÄNa+ ÒÔ¼°²¿·ÖÎÛÄ࣬ͨ¹ýÏÂÁÐÁ÷³Ì¿É´Ó¸Ã·ÏË®ÖлØÊÕFeSO4¡¤7H2O¾§Ìå¼°½ðÊôCu¡£

£¨1£©²½Öè1µÄÖ÷Òª²Ù×÷ÊÇ   £¬ÐèÓõ½µÄ²£Á§ÒÇÆ÷³ýÉÕ±­ÍâÓР      ,         ¡£

£¨2£©²½Öè2Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                   ¡£

£¨3£©²½Öè3Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                   ¡£

£¨4£©²½Öè4ÖÐÉæ¼°µÄ²Ù×÷ÊÇ£ºÕô·¢Å¨Ëõ£®             £®¹ýÂË£®Ï´µÓ£®ºæ¸É¡£

 

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¾«Ó¢¼Ò½ÌÍøÄÜÔ´ÊÇÖÆÔ¼¹ú¼Ò·¢Õ¹½ø³ÌµÄÒòËØÖ®Ò»£®¼×´¼¡¢¶þ¼×Ãѵȱ»³ÆÎª2 1ÊÀ¼ÍµÄÂÌÉ«ÄÜÔ´£¬¹¤ÒµÉÏÀûÓÃÌìÈ»ÆøÎªÖ÷ÒªÔ­ÁÏÓë¶þÑõ»¯Ì¼¡¢Ë®ÕôÆøÔÚÒ»¶¨Ìõ¼þÏÂÖÆ±¸ºÏ³ÉÆø£¨CO¡¢H2£©£¬ÔÙÖÆ³É¼×´¼¡¢¶þ¼×ÃÑ£®
£¨1£©¹¤ÒµÉÏ£¬¿ÉÒÔ·ÖÀëºÏ³ÉÆøÖеÄÇâÆø£¬ÓÃÓںϳɰ±£¬³£Óô×Ëá¶þ°±ºÏÑÇÍ­[Cu£¨NH3£©2]ACÈÜÒº£¨AC=CH3COO-£©À´ÎüÊÕºÏ³ÉÆøÖеÄÒ»Ñõ»¯Ì¼£¬Æä·´Ó¦Ô­ÀíΪ£º[Cu£¨NH3£©2]AC£¨aq£©+CO£¨g£©+NH3£¨g£©?[Cu£¨NH3£©3]AC?CO£¨aq£©¡÷H£¼0³£Ñ¹Ï£¬½«ÎüÊÕÒ»Ñõ»¯Ì¼µÄÈÜÒº´¦ÀíÖØÐ»ñµÃ[Cu£¨NH3£©2]ACÈÜÒºµÄ´ëÊ©ÊÇ
 
£»
£¨2£©¹¤ÒµÉÏÒ»°ã²ÉÓÃÏÂÁÐÁ½ÖÖ·´Ó¦ºÏ³É¼×´¼£º
·´Ó¦a£ºCO2£¨g£©+3H2£¨g£©?CH3OH£¨g£©+H2O£¨g£©¡÷H=-49.0kJ/mol£»·´Ó¦b£ºCO£¨g£©+2H2£¨g£©?CH3OH£¨g£©¡÷H£¼0
¢Ù¶ÔÓÚ·´Ó¦a£¬Ä³Î¶ÈÏ£¬½«4.0mol CO2£¨g£©ºÍ12.0mol H2£¨g£©³äÈëÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬·´Ó¦µ½´ïƽºâʱ£¬²âµÃ¼×´¼ÕôÆøµÄÌå»ý·ÖÊýΪ30%£¬Ôò¸ÃζÈÏ·´Ó¦µÄƽºâ³£ÊýΪ
 
£»
¢Ú¶ÔÓÚ·´Ó¦b£¬Ä³Î¶ÈÏ£¬½«1.0mol CO£¨g£©ºÍ2.0mol H2£¨g£©³äÈë¹Ì¶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷ÖУ¬·´Ó¦µ½´ïƽºâʱ£¬¸Ä±äζȺÍѹǿ£¬Æ½ºâÌåϵÖÐCH3OHµÄÎïÖʵÄÁ¿·ÖÊý±ä»¯Çé¿öÈçͼËùʾ£¬ÓÚκÍѹǿµÄ¹ØÏµÅжÏÕýÈ·µÄÊÇ
 
£»£¨Ìî×Öĸ´úºÅ£©
 A£®p3£¾p2£¬T3£¾T2  B£®p2£¾p4£¬T4£¾T2   C£®p1£¾p3£¬T1£¾T3    D£®p1£¾p4£¬T2£¾T3
£¨3£©CO¿ÉÒԺϳɶþ¼×ÃÑ£¬¶þ¼×ÃÑ¿ÉÒÔ×÷ΪȼÁÏµç³ØµÄÔ­ÁÏ£¬»¯Ñ§·´Ó¦Ô­ÀíΪ£ºCO£¨g£©+4H2£¨g£©?CH3OCH3£¨g£©+H2O£¨g£©¡÷H£¼0
¢ÙÔÚºãÈÝÃܱÕÈÝÆ÷Àï°´Ìå»ý±ÈΪ1£º4³äÈëÒ»Ñõ»¯Ì¼ºÍÇâÆø£¬Ò»¶¨Ìõ¼þÏ·´Ó¦´ïµ½Æ½ºâ״̬£®µ±¸Ä±ä·´Ó¦µÄijһ¸öÌõ¼þºó£¬ÏÂÁб仯ÄÜ˵Ã÷ƽºâÒ»¶¨ÏòÕý·´Ó¦·½ÏòÒÆ¶¯µÄÊÇ
 
£»
A£®Äæ·´Ó¦ËÙÂÊÏÈÔö´óºó¼õС
B£®·´Ó¦ÎïµÄÌå»ý°Ù·Öº¬Á¿¼õС
C£®Õý·´Ó¦ËÙÂÊÏÈÔö´óºó¼õС
D£®»¯Ñ§Æ½ºâ³£ÊýKÖµÔö´ó
¢Úд³ö¶þ¼×ÃѼîÐÔȼÁÏµç³ØµÄ¸º¼«µç¼«·´Ó¦Ê½
 
£»
¢Û¼ºÖª²ÎÓëµç¼«·´Ó¦µÄµç¼«²ÄÁϵ¥Î»ÖÊÁ¿·Å³öµçÄܵĴóС³ÆÎª¸Ãµç³ØµÄ±ÈÄÜÁ¿£®¹ØÓÚ¶þ¼×ÃѼîÐÔȼÁÏµç³ØÓëÒÒ´¼¼îÐÔȼÁÏµç³Ø£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
 
£¨Ìî×Öĸ£©
A£®Á½ÖÖȼÁÏ»¥ÎªÍ¬·ÖÒì¹¹Ì壬·Ö×ÓʽºÍĦ¶ûÖÊÁ¿Ïàͬ£¬±ÈÄÜÁ¿Ïàͬ
B£®Á½ÖÖȼÁÏËùº¬¹²¼Û¼üÊýÄ¿Ïàͬ£¬¶Ï¼üʱËùÐèÄÜÁ¿Ïàͬ£¬±ÈÄÜÁ¿Ïàͬ
C£®Á½ÖÖȼÁÏËùº¬¹²¼Û¼üÀàÐͲ»Í¬£¬¶Ï¼üʱËùÐèÄÜÁ¿²»Í¬£¬±ÈÄÜÁ¿²»Í¬
£¨4£©ÒÑÖªlg¶þ¼×ÃÑÆøÌåÍêȫȼÉÕÉú³ÉÎȶ¨µÄÑõ»¯Îï·Å³öµÄÈÈÁ¿Îª31.63kJ£¬Çëд³ö±íʾ¶þ¼×ÃÑȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ
 
£®

£¨15·Ö£©
£¨1£©ÏÂÁÐÊÂʵÖУ¬ÄÜÖ¤Ã÷ÑÇÁòËáµÄËáÐÔÇ¿ÓÚÇâÁòËáµÄÊÇ       £¨Ìî×Öĸ£©¡£
a£®ÑÇÁòËáÊÜÈÈʱÒ×·Ö½â
b£®ÏàͬÌõ¼þÏ£¬µÈŨ¶ÈµÄÑÇÁòËáÈÜÒºµ¼µçÄÜÁ¦Ç¿ÓÚÇâÁòËá
c£®ÑÇÁòËáÈÜÒº¿ÉʹƷºìÈÜÒºÍÊÉ«£¬¶øÇâÁòËá²»ÄÜ
d£®³£ÎÂÏ£¬Å¨¶È¾ùΪ0.k^s*5#u1mol£¯LµÄH2SO3ÈÜÒººÍH2SÈÜÒºµÄpH·Ö±ðÊÇ2.k^s*5#u1ºÍ4.k^s*5#u5
£¨2£©¹¤ÒµÉϳýÈ¥¸ßѹ¹øÓÃË®ÖÐÈܽâµÄÑõÆø³£ÓõÄÊÔ¼ÁÓÐNa2SO3ºÍN2H4£¨ë£©¡£
¢ÙÒÑÖª16gҺ̬µÄëÂÓëÑõÆø·´Ó¦µÃµ½µªÆøºÍҺ̬ˮʱ£¬·ÅÈÈ354.k^s*5#u87kJ£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ                                ¡£
¢Ú³ýÈ¥µÈÖÊÁ¿µÄO2£¬ËùºÄNa2SO3ºÍN2H4µÄÖÊÁ¿±ÈÊÇ          £¨Ìî×î¼òÕûÊý±È£©¡£
£¨3£©ÏòNa2SO3ºÍNa2SµÄ»ìºÏÈÜÒºÖмÓÈëÏ¡ÑÎËᣬÈÜÒºÖлá²úÉú´óÁ¿µ­»ÆÉ«³Áµí¡£Ôò¸Ã·´Ó¦ÖÐÑõ»¯¼ÁºÍ»¹Ô­¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈÊÇ                         ¡£
£¨4£©ÒÑÖªNa2SO3ÔÚ¸ßÎÂÏ·¢Éú·Ö½â£¬µÃµ½Á½ÖÖ²úÎijͬѧ³ÆÈ¡25.k^s*5#u2g´¿¾»µÄNa2SO3¡¤7H2O¾§ÌåÔÚ¸ßÎÂϸô¾ø¿ÕÆø¼ÓÈÈÖÁºãÖØ£¬ÀäÈ´ºó³ÆµÃ¹ÌÌåΪ12.k^s*5#u6g£¬½«ÆäÍêÈ«ÈÜÓÚË®Åä³É1LÈÜÒº£¬²¢²âÈÜÒºµÄpH¡£
¢ÙNa2SO3¸ßηֽâµÄ»¯Ñ§·½³ÌʽÊÇ                                   ¡£
¢Ú²âµÃÈÜÒºµÄpH´óÓÚ0.k^s*5#u025mol£¯LNa2SO3ÈÜÒºµÄpH£¬ÊÔ½âÊÍÔ­Òò£¨½áºÏÀë×Ó·½³Ìʽ˵Ã÷£©                                                            ¡£

(12·Ö)ÒÑ֪ij¹¤Òµ·ÏË®Öк¬ÓдóÁ¿FeSO4£¬½Ï¶àµÄCu2+£¬ÉÙÁ¿µÄNa+ ÒÔ¼°²¿·ÖÎÛÄ࣬ͨ¹ýÏÂÁÐÁ÷³Ì¿É´Ó¸Ã·ÏË®ÖлØÊÕFeSO4¡¤7H2O¾§Ìå¼°½ðÊôCu¡£

£¨1£©²½Öè1µÄÖ÷Òª²Ù×÷ÊÇ   £¬ÐèÓõ½µÄ²£Á§ÒÇÆ÷³ýÉÕ±­ÍâÓР       ,         ¡£

£¨2£©²½Öè2Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                   ¡£

£¨3£©²½Öè3Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                   ¡£

£¨4£©²½Öè4ÖÐÉæ¼°µÄ²Ù×÷ÊÇ£ºÕô·¢Å¨Ëõ£®              £®¹ýÂË£®Ï´µÓ£®ºæ¸É¡£

 

 

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

¾«Ó¢¼Ò½ÌÍø