Question

£¨12·Ö£©X¡¢Y¡¢Z¡¢WΪµç×ÓÊýÏàͬµÄ·Ö×Ó»òÀë×Ó¡£XÓÐ5¸öÔ­×Ӻˡ£Í¨³£×´¿öÏ£¬WΪÎÞÉ«ÒºÌå¡£ËüÃÇÖ®¼äת»¯¹ØϵÈçͼËùʾ£¬Çë»Ø´ð£º

£¨1£©¹¤ÒµÉÏÿÖÆÈ¡1molZÒª·Å³ö46.2 kJµÄÈÈÁ¿£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º      ¡£
£¨2£©ÊµÑéÊÒÖÆÈ¡ZµÄ·½·¨²»Ö¹Ò»ÖÖ£¬Ð´³öÆäÖÐÒ»ÖÖ·½·¨µÄ»¯Ñ§·½³Ìʽ£º      ¡£
£¨3£©Ä³»¯Ñ§Ð¡×éͬѧģÄ⹤ҵÉú²úÖÆÈ¡HNO₃£¬Éè¼ÆÏÂͼËùʾװÖã¬ÆäÖÐaΪһ¸ö¿É³ÖÐø¹ÄÈë¿ÕÆøµÄÏðƤÇò¡£

¢ÙAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                                ¡£
¢ÚBÖÐŨH₂SO₄µÄ×÷ÓÃÊÇ      ¡£
£¨4£©Ð´³öD×°ÖÃÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ                          ¡£
£¨5£©Ïòº¬ÓÐag HNO₃µÄÏ¡ÈÜÒºÖУ¬¼ÓÈëbgÌú·Û³ä·Ö·´Ó¦£¬Ìú·ÛÈ«²¿Èܽ⡣ÒÑÖª£¬Ôò±»»¹Ô­µÄHNO₃µÄÖÊÁ¿Îª      g¡£

Answer 1

£¨1£©N₂(g)+3H₂(g)2NH₃(g) ¡÷H="-92.4" kJ£¯mol
£¨2£©2NH₄Cl+Ca(OH)₂2NH₃¡ü+2H₂O+CaCl₂»òNH₃¡¤H₂ONH₃¡ü+H₂O
£¨3£©¢Ù4NH₃+5O₂4NO+6H₂O ¢ÚÎüÊÕH₂O£¨g£©ºÍδ·´Ó¦µÄNH₃
£¨4£©3NO₂+H₂O=2HNO₃+NO   £¨5£©

ÊÔÌâ·ÖÎö£ºWΪÎÞÉ«ÒºÌ壬һ°ãÊÇH₂O£¬¼´ÓÐ10¸öµç×Ó£¬XÓÐ5¸öÔ­×ÓºËÇÒÓÐ10¸öµç×Ó£¬¿ÉÍƶϳöÊÇNH₄^£«£¬ÔòYΪOH^£­£¬ZΪNH₃¡£
£¨1£©¹¤ÒµÉϺϳɰ±£¬ÊÇÓÃÇâÆøºÍµªÆøÖ±½ÓºÏ³É£¬·´Ó¦·½³ÌʽΪN₂(g)+3H₂(g)2NH₃(g) ¡÷H="-92.4" kJ£¯mol
£¨2£©°±µÄʵÑéÊÒÖÆ·¨£¬Ò»°ãÊǼÓÈÈï§ÑκͼîµÄ»ìºÏÎï¡£
£¨3£©£¨4£©°±¾­¹ýһϵÁз´Ó¦¿ÉÒԵõ½ÏõËᣬÆä¹ý³Ì¿É±íʾΪN₂ ¡ªNH₃¡ªNO¡ªNO₂¡ªHNO_3¡£
£¨5£©µ±Ìú²»×ãʱ£¬´Ë·´Ó¦µÄ·½³ÌʽΪ
4HNO₃£«FeFe(NO₃)₃£«NO£«2H₂O
4      1
b/14   b/56
Ôò£¨b/14£©=(a/63) µÃa/b=4.5
µ±Ìú¹ýÁ¿Ê±£¬Ê£ÓàµÄÌúÄܼÌÐøºÍFe(NO₃)₃·´Ó¦£¬¸Ã·´Ó¦·½³ÌʽΪ
8HNO₃£«3Fe3Fe(NO₃)₂£«2NO£«4H₂O
8      3
b/56   b/21
Ôò b/21=(a/63) µÃa/b=3
ÓÖÒòΪ£¨a/b£©´óÓÚµÈÓÚ3£¬Ð¡ÓÚµÈÓÚ4.5£¬ËµÃ÷¸Ã·´Ó¦½éÓÚ¸Õ²ÅËùÁоٵÄÁ½ÕßÖ®¼ä£¬¼´·´Ó¦·½³ÌʽΪ
4Fe£«12HNO₃3Fe(NO₃)₂£«Fe(NO₃)₃£«3NO£«6H₂O
12                       3
a                        a/4
¼´±»»¹Ô­µÄÏõËáµÄÖÊÁ¿Îª(a/4)g¡£
µãÆÀ£º±¾Ìâͨ¹ýÍƶϣ¬ÅжϳöNÔªËصÄÏà¹ØÎïÖÊ£¬Ö÷Òª¿¼²ìÊéÉϹØÓÚNH₃¼°HNO₃µÄÖÆ·¨µÄ»¯Ñ§·½³Ìʽ£¬Ñ§Éú¿ÉÒÔ¸ù¾ÝËùѧ֪ʶ½øÐнâ´ð£¬ÄѶȲ»´ó¡£