ÌâÄ¿ÄÚÈÝ

ǦÐîµç³ØµÄ»¯Ñ§·´Ó¦Îª£ºPbO2 + 2H2SO4 + Pb 2 PbSO4 + 2H2O¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ

¡¡ A£®³äµçʱתÒÆ1molµç×ÓÔòÉú³É0.5mol H2SO4

¡¡ B£®·ÅµçʱÕý¼«µÄµç¼«·´Ó¦Îª£ºPbO2 + 4H+ + SO42¨D + 2e¨D = PbSO4 + 2H2O

¡¡ C£®³äµçʱ£¬µç³ØÉϱê×¢¡°+¡±µÄµç¼«Ó¦ÓëÍâ½ÓµçÔ´µÄÕý¼«ÏàÁ¬

¡¡D£®·Åµçʱ£¬Pbʧµç×Ó·¢ÉúÑõ»¯·´Ó¦

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÀûÓû¯Ñ§·´Ó¦Ô­ÀíÑо¿Éú²ú¡¢Éú»îÖеÄʵ¼ÊÎÊÌâ¾ßÓÐÊ®·ÖÖØÒªµÄÒâÒ壺
£¨¢ñ£©µªÆøºÍÇâÆøºÏ³É°±ÊÇ»¯Ñ§¹¤ÒµÖм«ÎªÖØÒªµÄ·´Ó¦£¬ÆäÈÈ»¯Ñ§·½³Ìʽ¿É±íʾΪ£ºN2(g)£«3H2(g) 2NH3(g)¡¡¦¤H£½£­92 kJ¡¤mol£­1¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)È¡1 mol N2(g)ºÍ3 mol H2(g)·ÅÔÚÒ»ÃܱÕÈÝÆ÷ÖУ¬ÔÚ´ß»¯¼Á´æÔÚʱ½øÐз´Ó¦£¬²âµÃ·´Ó¦·Å³öµÄÈÈÁ¿£ß£ß£ß£ß£ß92 kJ(Ìî¡°´óÓÚ¡±¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±)£¬Ô­ÒòÊǣߣߣߣߣߣߣߣ»Èô¼ÓÈë´ß»¯¼Á£¬¦¤H¡¡¡¡¡¡(Ìî¡°±ä´ó¡±¡°±äС¡±»ò¡°²»±ä¡±)¡£
(2)ÒÑÖª£º·Ö±ðÆÆ»µ1 mol N¡ÔN¼ü¡¢1 mol H¡ªH¼üÐèÒªÎüÊÕµÄÄÜÁ¿Îª£º946 kJ¡¢436 kJ£¬ÔòÆÆ»µ1 mol N¡ªH¼üÐèÒªÎüÊÕµÄÄÜÁ¿Îª£ß£ß£ß£ß£ß£ßkJ¡£
(3)N2H4¿ÉÊÓΪ£ºNH3·Ö×ÓÖеÄH±»¡ªNH2È¡´úµÄ²úÎï¡£·¢ÉäÎÀÐÇÓÃN2H4(g)ΪȼÁÏ£¬NO2ΪÑõ»¯¼ÁÉú³ÉN2ºÍH2O(g)¡£
ÒÑÖª£ºN2(g)£«2O2(g)===2NO2(g)  ¦¤H1£½£«67£®7 kJ¡¤mol£­1
N2H4(g)£«O2(g)===N2(g)£«2H2O(g)  ¦¤H2£½£­534 kJ¡¤mol£­1¡£
Ôò£º1 mol N2H4ÍêÈ«·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ                                 ¡£
£¨¢ò£©Ä³Ç¦Ðîµç³ØµÄÕý¡¢¸º¼«±ê¼Ç±»Ä¥Ëð¡£ÊÔÓÃÏÂͼװÖÃÉè¼ÆʵÑ飬ʶ±ð³ö´ËǦÐîµç³ØµÄÕý¸º¼«¡£

(1)ÈôA½ÓE£¬B½ÓF£¬¶øBµç¼«³öÏÖ     £¬·´Ó¦Ê½Îª                         £¬Ôò˵Ã÷FΪÕý¼«£»
£¨2£©ÈôǦÐîµç³Ø¹¤×÷ʱ£¨·Åµç£©£¬ÆäEËùÔڵ缫µÄµç¼«·´Ó¦Ê½Îª£º             £¬³äµçʱ¸Ã¼«ÓëÍâ¼ÓµçÔ´µÄ    ¼«ÏàÁ¬¡£
£¨3£©ÈôÓøõç³Øµç½âCu(NO3)2 ÈÜÒº£¬Æäµç½â·½³ÌʽΪ                         
ÈôÓÐ0.2molµç×Ó·¢ÉúתÒÆ£¬ÔòÕý¼«ÏûºÄµÄPbO2µÄÎïÖʵÄÁ¿ÊÇ       £»ÒªÏëCuSO4ÈÜÒº»Ö¸´Ô­Ñù£¬Ðè¼ÓÈëµÄÎïÖÊÊÇ          ,ÖÊÁ¿Îª        

Ñо¿»¯Ñ§·´Ó¦Ô­Àí¶ÔÉú²úÉú»îºÜÓÐÒâÒ壬ÇëÓû¯Ñ§·´Ó¦Ô­ÀíµÄÏà¹Ø֪ʶ»Ø´ðÏÂÁÐÎÊÌ⣺

   £¨1£©ÓÃǦÐîµç³Øµç½â¼×¡¢ÒÒÁ½µç½â³ØÖеÄÈÜÒº¡£ÒÑ֪ǦÐîµç³ØµÄ×Ü·´Ó¦Îª£ºPb£¨s£©£«PbO2£¨s£©£«2H2SO4£¨aq£©2PbSO4£¨s£©£«2H2O£¨1£©¡£µç½âÒ»¶Îʱ¼äºó£¬Ïòc¼«ºÍd¼«¸½½ü·Ö±ðµÎ¼Ó·Ó̪ÊÔ¼Á£¬c¼«¸½½üÈÜÒº±äºì£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ____________

£¨ÌîдÐòºÅ£©

    A£®d¼«ÎªÒõ¼«

    B£®ÈôÀûÓü׳ؾ«Á¶Í­£¬b¼«Ó¦Îª´ÖÍ­                              

    C£®·ÅµçʱǦÐîµç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½Îª£º

       PbO2£¨s£©£«4H£«£¨aq£©£«£¨aq£©£«4e£­PbSO4£¨s£©£«2H2O£¨1£©

D£®ÈôËĸöµç¼«²ÄÁϾùΪʯī£¬µ±Îö³ö6£®4g Cuʱ£¬Á½³ØÖй²²úÉúÆøÌå3£®36L£¨±ê×¼

×´¿öÏ£©

   £¨2£©Ä³¶þÔªËáH2AÔÚË®ÖеĵçÀë·½³ÌʽÊÇ£ºH2A£½H£«£«HA£­£»HA£­H£«£«A2£­£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

    ¢ÙNa2AÈÜÒºÏÔ_________£¨Ìî¡°ËáÐÔ¡±¡¢¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£©£¬ÀíÓÉÊÇ________________

£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£

    ¢ÚÒÑÖª0£®1mol¡¤L£­1µÄNaHAÈÜÒºµÄpH£½2£¬Ôò0£®1mol¡¤L£­1µÄH2AÈÜÒºÖÐÇâÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶È¿ÉÄÜÊÇ__________0£®1lmol¡¤L£­1£¨Ìî¡°<¡±¡¢¡°>¡±»ò¡°£½¡±£©£¬ÀíÓÉÊÇ£º___________________________.

   £¨3£©¶þ¼×ÃÑÊÇÒ»ÖÖÖØÒªµÄÇå½àȼÁÏ£¬ÀûÓÃˮúÆøºÏ³É¶þ¼×ÃѵÄÈý²½·´Ó¦ÈçÏ£º

    ¢Ù2H2£¨g£©£«CO£¨g£©CH3OH£¨g£©£» ¡÷H£½£­90£®8kJ¡¤mol£­1

    ¢Ú2CH3OH£¨g£©CH3OCH3£¨g£©£«H2O£¨g£©£» ¡÷H£½£­23£®5kJ¡¤mol£­1

    ¢ÛCO£¨g£©£«H2O£¨g£©CO2£¨g£©£«H2£¨g£©£»¡÷£½£­41£®3kJ¡¤mol£­1

    д³öˮúÆøÖ±½ÓºÏ³É¶þ¼×ÃÑͬʱÉú³ÉCO2µÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ___________________.

   £¨4£©SO2ÊÇÁòËáÉú²úµÄÖØÒªÖмä²úÎҲÊÇ¿ÕÆøÎÛȾµÄÖ÷ÒªÔ­ÒòÖ®Ò»£¬ÆäÑõ»¯Éú³ÉSO3µÄ·´Ó¦Îª£º2SO2£¨g£©£«O2£¨g£©2SO3£¨g£©¡£ÔÚÒ»¶¨Î¶ÈÏ£¬½«0£®23 mol SO2ºÍ0£®11 molÑõÆø·ÅÈëÈÝ»ýΪl LµÄÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦£¬´ïµ½Æ½ºâºóµÃµ½0£®12 mol SO3£¬Ôò·´Ó¦µÄƽºâ³£ÊýK£½________¡£ÈôζȲ»±ä£¬ÔÙ¼ÓÈë0£®50 molÑõÆøºóÖØдﵽƽºâ£¬ÔòSO3µÄÌå»ý·ÖÊý½«___________£¨Ìî¡°Ôö´ó¡±¡¢¡°²»±ä¡±»ò¡°¼õС¡±£©¡£

 

ÀûÓû¯Ñ§·´Ó¦Ô­ÀíÑо¿Éú²ú¡¢Éú»îÖеÄʵ¼ÊÎÊÌâ¾ßÓÐÊ®·ÖÖØÒªµÄÒâÒ壺

£¨¢ñ£©µªÆøºÍÇâÆøºÏ³É°±ÊÇ»¯Ñ§¹¤ÒµÖм«ÎªÖØÒªµÄ·´Ó¦£¬ÆäÈÈ»¯Ñ§·½³Ìʽ¿É±íʾΪ£ºN2(g)£«3H2(g) 2NH3(g)¡¡¦¤H£½£­92 kJ¡¤mol£­1¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)È¡1 mol N2(g)ºÍ3 mol H2(g)·ÅÔÚÒ»ÃܱÕÈÝÆ÷ÖУ¬ÔÚ´ß»¯¼Á´æÔÚʱ½øÐз´Ó¦£¬²âµÃ·´Ó¦·Å³öµÄÈÈÁ¿£ß£ß£ß£ß£ß92 kJ(Ìî¡°´óÓÚ¡±¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±)£¬Ô­ÒòÊǣߣߣߣߣߣߣߣ»Èô¼ÓÈë´ß»¯¼Á£¬¦¤H¡¡¡¡¡¡(Ìî¡°±ä´ó¡±¡°±äС¡±»ò¡°²»±ä¡±)¡£

(2)ÒÑÖª£º·Ö±ðÆÆ»µ1 mol N¡ÔN¼ü¡¢1 mol H¡ªH¼üÐèÒªÎüÊÕµÄÄÜÁ¿Îª£º946 kJ¡¢436 kJ£¬ÔòÆÆ»µ1 mol N¡ªH¼üÐèÒªÎüÊÕµÄÄÜÁ¿Îª£ß£ß£ß£ß£ß£ßkJ¡£

(3)N2H4¿ÉÊÓΪ£ºNH3·Ö×ÓÖеÄH±»¡ªNH2È¡´úµÄ²úÎï¡£·¢ÉäÎÀÐÇÓÃN2H4(g)ΪȼÁÏ£¬NO2ΪÑõ»¯¼ÁÉú³ÉN2ºÍH2O(g)¡£

ÒÑÖª£ºN2(g)£«2O2(g)===2NO2(g)   ¦¤H1£½£«67£®7 kJ¡¤mol£­1

N2H4(g)£«O2(g)===N2(g)£«2H2O(g)   ¦¤H2£½£­534 kJ¡¤mol£­1¡£

Ôò£º1 mol N2H4ÍêÈ«·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ                                  ¡£

£¨¢ò£©Ä³Ç¦Ðîµç³ØµÄÕý¡¢¸º¼«±ê¼Ç±»Ä¥Ëð¡£ÊÔÓÃÏÂͼװÖÃÉè¼ÆʵÑ飬ʶ±ð³ö´ËǦÐîµç³ØµÄÕý¸º¼«¡£

 (1)ÈôA½ÓE£¬B½ÓF£¬¶øBµç¼«³öÏÖ      £¬·´Ó¦Ê½Îª                          £¬Ôò˵Ã÷FΪÕý¼«£»

£¨2£©ÈôǦÐîµç³Ø¹¤×÷ʱ£¨·Åµç£©£¬ÆäEËùÔڵ缫µÄµç¼«·´Ó¦Ê½Îª£º              £¬³äµçʱ¸Ã¼«ÓëÍâ¼ÓµçÔ´µÄ     ¼«ÏàÁ¬¡£

£¨3£©ÈôÓøõç³Øµç½âCu(NO3)2 ÈÜÒº£¬Æäµç½â·½³ÌʽΪ                          

ÈôÓÐ0.2molµç×Ó·¢ÉúתÒÆ£¬ÔòÕý¼«ÏûºÄµÄPbO2µÄÎïÖʵÄÁ¿ÊÇ        £»ÒªÏëCuSO4ÈÜÒº»Ö¸´Ô­Ñù£¬Ðè¼ÓÈëµÄÎïÖÊÊÇ           ,ÖÊÁ¿Îª        

 

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

¾«Ó¢¼Ò½ÌÍø