ÌâÄ¿ÄÚÈÝ

ij¿ÆÑÐС×éÓÃMnO2ºÍŨÑÎËáÖÆ±¸Cl2ʱ£¬ÀûÓøÕÎüÊÕ¹ýÉÙÁ¿SO2µÄNaOHÈÜÒº¶ÔÆäÎ²Æø½øÐÐÎüÊÕ´¦Àí¡£
£¨1£©ÇëÍê³ÉSO2ÓëÉÙÁ¿µÄNaOHÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º           £¨2·Ö£©
£¨2£©·´Ó¦Cl2+Na2SO3+2NaOH===2NaCl+Na2SO4+H2OÖÐÃ¿×ªÒÆ2.5molµÄµç×ÓÔò²Î¼Ó·´Ó¦µÄ»¹Ô­¼ÁµÄÎïÖʵÄÁ¿Îª                  mol£®£¨2·Ö£©
£¨3£©ÎüÊÕÎ²ÆøÒ»¶Îʱ¼äºó£¬ÎüÊÕÒº£¨Ç¿¼îÐÔ£©Öп϶¨´æÔÚCl¡¢OHºÍSO£®ÇëÉè¼ÆÊµÑ飬̽¾¿¸ÃÎüÊÕÒºÖпÉÄÜ´æÔ򵀮äËûÒõÀë×Ó£¨²»¿¼ÂÇ¿ÕÆøµÄCO2µÄÓ°Ï죩£®
¢ÙÌá³öºÏÀí¼ÙÉè £®
¼ÙÉè1£ºÖ»´æÔÚSO32-£» ¼ÙÉè2£ºÖ»´æÔÚClO
¼ÙÉè3£º¼È²»´æÔÚSO32-Ò²²»´æÔÚClO£»
¼ÙÉè4£º                                            £®£¨2·Ö£©
¢ÚÉè¼ÆÊµÑé·½°¸½øÐÐʵÑé¡£ÇëÔÚ´ðÌ⿨ÉÏд³öʵÑé²½ÖèÒÔ¼°Ô¤ÆÚÏÖÏóºÍ½áÂÛ
ÏÞѡʵÑéÊÔ¼Á£º3moLL-1H2SO4¡¢0.01molL-1KMnO4¡¢×ÏɫʯÈïÊÔÒº£®£¨Ã¿¿Õ2·Ö£©
ʵÑé²½Öè
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
²½Öè1£ºÈ¡ÉÙÁ¿ÎüÊÕÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó3 moLL-1 H2SO4ÖÁÈÜÒº³ÊËáÐÔ£¬È»ºó½«ËùµÃÈÜÒº·ÖÖÃÓÚA¡¢BÊÔ¹ÜÖУ®
 
²½Öè2£º
 
²½Öè3£º
 
 
(1) SO2+NaOH=NaHSO3(2·Ö);     (2)1.25mol(2·Ö)
(3) ¢ÙSO32-¡¢ClO-¶¼´æÔÚ(2·Ö)
¢Ú(ÿ¿Õ2·Ö)
ʵÑé²½Öè
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
²½Öè1£ºÈ¡ÉÙÁ¿ÎüÊÕÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó3 moLL-1 H2SO4ÖÁÈÜÒº³ÊËáÐÔ£¬È»ºó½«ËùµÃÈÜÒº·ÖÖÃÓÚA¡¢BÊÔ¹ÜÖÐ
 
²½Öè2£ºÔÚAÊÔ¹ÜÖеμÓ×ÏɫʯÈïÊÔÒº
ÈôÏȱäºìºóÍÊÉ«£¬Ö¤Ã÷ÓÐClO-£¬·ñÔòÎÞ
²½Öè3£ºÔÚBÊÔ¹ÜÖеμÓ0.01molL-1KMnO4ÈÜÒº
Èô×ϺìÉ«ÍËÈ¥£¬Ö¤Ã÷ÓÐSO32-£¬·ñÔòÎÞ
 
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨14·Ö£©Ä³ÐËȤС×éµÄѧÉú¸ù¾ÝMgÓëCO2·´Ó¦Ô­Àí£¬ÍÆ²âÄÆÒ²Ó¦ÄÜÔÚCO2ÖÐȼÉÕ£¬ÎªÁËÈ·¶¨ÆäÉú³É²úÎï²¢½øÐÐʵÑéÂÛÖ¤¡£Ä³Í¬Ñ§Éè¼ÆÁËÏÂͼËùʾװÖýøÐÐʵÑ飨ÒÑÖªPdCl2Äܱ»CO»¹Ô­µÃµ½ºÚÉ«µÄPd£©£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÎªÊ¹·´Ó¦Ëæ¿ªËæÓã¬Ëæ¹ØËæÍ££¬·½¿òÄÚӦѡÓÃÏÂͼËùʾ         ×°Öã¨Ìî×Öĸ´úºÅ£©¡£

£¨2£©ÊµÑéÊÒÖÆÈ¡¶þÑõ»¯Ì¼ÆøÌåÒËÓõÄÒ©Æ·ÊÇ               £¨Ìî±êºÅ£©¡£
¢Ùʯ»Òʯ£¬¢Ú´¿¼î£¬¢ÛСËÕ´ò£¬¢Ü18.4 mol¡¤L¡ª1ÁòËᣬ¢Ý11.2mol¡¤L¡ª1ÑÎËᣬ¢ÞÕôÁóË®£¬¢ßľ̿·Û¡£
£¨3£©ÔÚ2×°ÖÃÄÚ½øÐз´Ó¦µÄÀë×Ó·½³ÌʽΪ                                         ¡£¼ì²é×°ÖÃÆøÃÜÐÔ²¢×°ºÃÒ©Æ·ºó£¬µãȼ¾Æ¾«µÆÖ®Ç°´ý×°Öà             £¨ÌîÊý×Ö±àºÅ£©ÖгöÏÖ        ÏÖÏóʱ£¬ÔÙµãȼ¾Æ¾«µÆ¡£
£¨4£©¢ÙÈô×°ÖÃ6ÖÐÓкÚÉ«³Áµí£¬×°ÖÃ4ÖвÐÁô¹ÌÌ壨ֻÓÐÒ»ÖÖÎïÖÊ£©¼ÓÑÎËáºóÓÐÄÜʹ³ÎÇåʯ»ÒË®±ä»ë×ÇµÄÆøÌå·Å³ö£¬ÔòÄÆÓë¶þÑõ»¯Ì¼·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                         £»
¢ÚÈô×°ÖÃ6ÖÐÈÜÒºÎÞÃ÷ÏÔÏÖÏó£¬×°ÖÃ4ÖвÐÁô¹ÌÌ壨ÓÐÁ½ÖÖÎïÖÊ£©¼ÓÑÎËáºóÓÐÄÜʹ³ÎÇåʯ»ÒË®±ä»ë×ÇµÄÆøÌå·Å³ö£¬ÔòÄÆÓë¶þÑõ»¯Ì¼·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                          ¡£
ÓÃÏÂÁÐÒÇÆ÷¡¢Ò©Æ·ÑéÖ¤ÓÉÍ­ºÍÊÊÁ¿ÏõËá·´Ó¦²úÉúµÄÆøÌåÖк¬NO£¨ÒÇÆ÷¿ÉÑ¡ÔñʹÓã¬N2ºÍO2µÄÓÃÁ¿¿É×ÔÓÉ¿ØÖÆ£©¡£

ÒÑÖª£º¢Ù
 ¢Ú ÆøÌåÒº»¯Î¶ȣºNO2 21¡æ£¬ NO -152¡æ
ÊԻشð£º
£¨1£©Ð´³ö×°ÖÃAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ_____                                                              ¡£
£¨2£©ÒÇÆ÷µÄÁ¬½Ó˳Ðò£¨°´×ó¡úÓÒÁ¬½Ó£¬Ìî¸÷½Ó¿ÚµÄ±àºÅ£©Îª____          
                      ¡£ÔÚÌî¼ÓÒ©
ƷǰӦÏȽøÐеIJÙ×÷ÊÇ                                        ¡£
£¨3£©·´Ó¦Ç°ÏÈͨÈëN2£¬Ä¿µÄÊÇ______                                £»
£¨4£©È·ÈÏÆøÌåÖк¬NOµÄÏÖÏóÊÇ________                               ¡£
£¨5£©×°Öã¡¢FµÄ×÷ÓÃÊÇ                                            _______                 ¡¡¡¡¡¡¡¡¡¡¡¡¡¡          ¡¡¡¡                        ¡£
£¨6£©Èô´Ó£Á³öÀ´µÄ»ìºÏÆøÌåÖÐNO2¡¢NOµÄÌå»ý·Ö±ðΪV1mL¡¢V2mL£¬Èô×îÖÕµªµÄÑõ»¯ÎïÍêÈ«±»ÈÜÒºÎüÊÕ£¬×°ÖÃÖÐÖÁÉÙÐèÒª³äÈëÑõÆøµÄÌå»ýΪ£º     £¨Ïàͬ״̬Ï£©¡£

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

¾«Ó¢¼Ò½ÌÍø