ÌâÄ¿ÄÚÈÝ
ÔÚËáÐÔFe(NO3)3ÈÜÒºÖÐÖð½¥Í¨ÈëH2SÆøÌ壬²»¿ÉÄÜ·¢ÉúµÄÀë×Ó·´Ó¦ÊÇ
A£®3H2S£«2NO3£ £«2H£« =2NO¡ü£«3S¡ý£« 4H2O
B£®3Fe3£«£« 3NO3£ £«6H2S =3NO¡ü£« 6S¡ý£«3Fe2£«£«6H2O
C£®Fe3£«£«3NO3£ £«5H2S£«2H£«=3NO¡ü£«5S¡ý£«Fe2£«£«6H2O
D£®Fe3£«£«9NO3£ £«14H2S£« 8H£«=9NO¡ü£«14S¡ý£«Fe2£«£«18H2O
ºÏ³É°±ÊÇÈËÀà¿ÆÑ§¼¼ÊõÉϵÄÒ»ÏîÖØ´óÍ»ÆÆ£¬¹¤ÒµÉÏÒÔÌìÈ»ÆøÎªÔÁϺϳɰ±¡£ÆäÉú²ú
¹¤ÒÕÈçÏ£ºÔìÆø½×¶Î¡úת»¯½×¶Î¡ú·ÖÀë¾»»¯¡úºÏ³É½×¶Î
£¨1£©ÔìÆø½×¶ÎµÄ·´Ó¦Îª£ºCH4£¨g£©£«H2O£¨g£©
CO£¨g£©£«3H2£¨g£© ¦¤H£½£«206.1 kJ/mol
¢ÙÔÚÒ»¸öÃܱÕÈÝÆ÷ÖнøÐÐÉÏÊö·´Ó¦£¬ ²âµÃCH4µÄÎïÖʵÄÁ¿Å¨¶ÈËæ·´Ó¦Ê±¼äµÄ±ä»¯ÈçÏÂͼ1Ëùʾ£¬10minʱ£¬¸Ä±äµÄÍâ½çÌõ¼þ¿ÉÄÜÊÇ ¡£
¢ÚÈçͼ2Ëùʾ£¬ÔÚ³õʼÈÝ»ýÏàµÈµÄ¼×¡¢ÒÒÁ½ÈÝÆ÷Öзֱð³äÈëµÈÎïÖʵÄÁ¿µÄCH4ºÍH2O¡£ÔÚÏàͬζÈÏ·¢Éú·´Ó¦£¬²¢Î¬³Ö·´Ó¦¹ý³ÌÖÐζȲ»±ä¡£Ôò´ïµ½Æ½ºâʱ£¬Á½ÈÝÆ÷ÖÐCH4µÄת»¯ÂÊ´óС¹ØÏµÎª£º¦Á¼×£¨CH4£©
¦ÁÒÒ£¨CH4£©
![]()
£¨2£©×ª»¯½×¶Î·¢ÉúµÄ¿ÉÄæ·´Ó¦Îª£ºCO£¨g£©£«H2O£¨g£©
CO2£¨g£©£«H2£¨g£©£¬ÔÚÒ»¶¨Î¶ÈÏ£¬·´Ó¦µÄƽºâ³£ÊýΪK£½1¡£Ä³Ê±¿Ì²âµÃ¸ÃζÈϵÄÃܱÕÈÝÆ÷Öи÷ÎïÖʵÄÎïÖʵÄÁ¿¼ûÏÂ±í£º
CO | H2O | CO2 | H2 |
0.5 mol | 8.5 mol | 2.0 mol | 2.0 mol |
´Ëʱ·´Ó¦ÖÐÕý¡¢Äæ·´Ó¦ËÙÂʵĹØÏµÊ½ÊÇ £¨ÌîÐòºÅ£©¡£
a£®v£¨Õý£©£¾v£¨Ä棩 b£®v£¨Õý£©£¼v£¨Ä棩 c£®v£¨Õý£©£½v£¨Ä棩 d£®ÎÞ·¨ÅжÏ
£¨3£©ºÏ³É°±·´Ó¦Îª£ºN2£¨g£©+3H2£¨g£©
2NH3£¨g£© ∆H=£92.4kJ/mol£¬¸ù¾ÝÀÕÏÄÌØÁÐÔÀí£¬¼òÊöÌá¸ßºÏ³É°±ÔÁÏת»¯ÂʵÄÒ»ÖÖ·½·¨ ¡£
£¨4£©¹¤ÒµºÏ³É°±µÄÈÈ»¯Ñ§·½³ÌʽΪN2£¨g£©£«3H2£¨g£©
2NH3£¨g£© ¦¤H£½£92.4 kJ¡¤mol£1¡£ÔÚijѹǿºã¶¨µÄÃܱÕÈÝÆ÷ÖмÓÈë2mol N2ºÍ4mol H2£¬´ïµ½Æ½ºâʱ£¬N2µÄת»¯ÂÊΪ50%£¬Ìå»ý±äΪ10 L¡£Çó£º
¢Ù¸ÃÌõ¼þÏÂµÄÆ½ºâ³£ÊýΪ_________£»
¢ÚÈôÏò¸ÃÈÝÆ÷ÖмÓÈëa mol N2¡¢b mol H2¡¢c mol NH3£¬ÇÒa¡¢b¡¢c ¾ù´óÓÚ0£¬ÔÚÏàͬÌõ¼þÏ´ﵽƽºâʱ£¬»ìºÏÎïÖи÷×é·ÖµÄÎïÖʵÄÁ¿ÓëÉÏÊöƽºâÏàͬ¡£·´Ó¦·Å³öµÄÈÈÁ¿__________£¨Ìî¡°>¡±¡°<¡±»ò¡°£½¡±£©92.4 kJ¡£
¢ÛÔÚÒ»¶¨Î¶ÈÏ£¬ÔÚÒ»¸öÈÝ»ý²»±äµÄÃܱÕÈÝÆ÷ÖУ¬·¢ÉúºÏ³É°±·´Ó¦£¬ÏÂÁÐÄÜÅжϸ÷´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ____________
a£®v£¨N2£©=3v£¨NH3£©
b£®»ìºÏÆøÌåµÄÃܶȲ»ËæÊ±¼ä¸Ä±ä
c£®»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»ËæÊ±¼ä¸Ä±ä
d£®ÈÝÆ÷ÖеÄѹǿ²»ËæÊ±¼ä¸Ä±ä
e£®c£¨N2£©=c£¨NH3£©