ÌâÄ¿ÄÚÈÝ

Q¡¢W¡¢X¡¢Y¡¢ZÊÇ5ÖÖ¶ÌÖÜÆÚÔªËØ£¬Ô­×ÓÐòÊýÖð½¥Ôö´ó£¬QÊÇÖÜÆÚ±íÖÐÔ­×Ӱ뾶×îСµÄÔªËØ£¬QÓëW×é³ÉµÄ»¯ºÏÎïÊÇÒ»ÖÖÎÂÊÒÆøÌ壬WÓëY¡¢XÓëY×é³ÉµÄ»¯ºÏÎïÊÇ»ú¶¯³µÅųöµÄ´óÆøÎÛȾÎYºÍZÄÜÐγÉÔ­×Ó¸öÊý±ÈΪ1£º1ºÍ1£º2µÄÁ½ÖÖÀë×Ó»¯ºÏÎ
£¨1£©WÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÊÇ
µÚ¶þÖÜÆÚIVA×å
µÚ¶þÖÜÆÚIVA×å
£¬Z2YµÄµç×ÓʽÊÇ
£®
£¨2£©¢ÙÒÑÖª¡°·²ÆøÌå·Ö×Ó×ÜÊýÔö¶àµÄ·´Ó¦Ò»¶¨ÊÇìØÔö´óµÄ·´Ó¦¡±£®¸ù¾ÝËùѧ֪ʶÅжÏÈçÏ·´Ó¦2WY£¨g£©=2W£¨s£©+Y2£¨g£©¡÷H£¾0£¬ÔÚ³£ÎÂÏÂ
²»ÄÜ
²»ÄÜ
×Ô·¢½øÐУ¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©£®
¢Ú400¡æʱ£¬ÔÚ0.5LµÄ·´Ó¦ÈÝÆ÷ÖÐÓÃX2¡¢Q2·´Ó¦ºÏ³ÉXQ3·´Ó¦£¬´ïµ½Æ½ºâʱ£¬²âµÃX2¡¢Q2¡¢XQ3µÄÎïÖʵÄÁ¿·Ö±ðΪ2mol¡¢1mol¡¢2mol£¬Ôò´Ë·´Ó¦ÔÚ400¡æʱµÄ»¯Ñ§Æ½ºâ³£ÊýΪ
0.5
0.5

£¨3£©2.24L£¨±ê×¼×´¿ö£©XQ3±»200mL 1mol/L QXY3ÈÜÒºÎüÊÕºó£¬ËùµÃÈÜÒºÖÐÀë×ÓŨ¶È´Ó´óµ½Ð¡µÄ˳ÐòÊÇ
c£¨NO3-£©£¾c£¨H+£©£¾c£¨NH4+£©£¾c£¨OH-£©
c£¨NO3-£©£¾c£¨H+£©£¾c£¨NH4+£©£¾c£¨OH-£©
£®
£¨4£©WQ4YÓëY2µÄ·´Ó¦¿É½«»¯Ñ§ÄÜת»¯ÎªµçÄÜ£¬Æ乤×÷Ô­ÀíÈçͼËùʾ£¬a¼«µÄµç¼«·´Ó¦Ê½ÊÇ
CH3OH+8OH--6e-=CO32-+6H2O
CH3OH+8OH--6e-=CO32-+6H2O

£¨5£©XºÍZ×é³ÉµÄÒ»ÖÖÀë×Ó»¯ºÏÎÄÜÓëË®·´Ó¦Éú³ÉÁ½Öּ¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
Na3N+4H2O=3NaOH+NH3?H2O
Na3N+4H2O=3NaOH+NH3?H2O
£®
·ÖÎö£ºQÊÇÖÜÆÚ±íÖÐÔ­×Ӱ뾶×îСµÄÔªËØ£¬Ó¦ÎªH£¬QÓëW×é³ÉµÄ»¯ºÏÎïÊÇÒ»ÖÖÎÂÊÒÆøÌ壬¸ÃÆøÌåΪCH4£¬ÔòWΪC£¬
WÓëY¡¢XÓëY×é³ÉµÄ»¯ºÏÎïÊÇ»ú¶¯³µÅųöµÄ´óÆøÎÛȾÎΪCOºÍNOx£¬ÔòYΪO£¬XΪN£¬YºÍZÄÜÐγÉÔ­×Ó¸öÊý±ÈΪ1£º1ºÍ1£º2µÄÁ½ÖÖÀë×Ó»¯ºÏÎÕâÁ½ÖÖÀë×Ó»¯ºÏÎï·Ö±ðΪNa2O¡¢Na2O2£¬×÷ΪZΪNa£¬
£¨1£©WΪC£¬Î»ÓÚÖÜÆÚ±íµÚ¶þÖÜÆÚIVA×壬Z2YΪNa2O£¬ÎªÀë×Ó»¯ºÏÎ
£¨2£©¢ÙÓÉ×ÔÓÉÄÜÅоݡ÷G=¡÷H-T?¡÷SÅжϷ´Ó¦ÔÚ³£ÎÂÏÂÄÜ·ñ½øÐУ»¸ù¾Ýk=
c2(NH3)
c(N2)¡Ác3(H2 )
¼ÆË㣻
£¨3£©n£¨NH3£©=
2.24L
22.4L/mol
=0.1mol£¬n£¨HNO3£©=0.2L¡Á1mol/L=0.2mol£¬¸ù¾ÝÉú³ÉÎïµÄÐÔÖÊÅжÏÀë×ÓŨ¶È´óС£»
£¨4£©CH3OH¼îÐÔȼÁϵç³ØÖУ¬CH3OHΪµç³ØµÄ¸º¼«·´Ó¦£»
£¨5£©XºÍZ×é³ÉµÄÒ»ÖÖÀë×Ó»¯ºÏÎïΪNa3N£¬ÄÜÓëË®·´Ó¦Éú³ÉÁ½ÖּÉú³ÉÎïΪNaOHºÍNH3?H2O£®
½â´ð£º½â£ºQÊÇÖÜÆÚ±íÖÐÔ­×Ӱ뾶×îСµÄÔªËØ£¬Ó¦ÎªH£¬QÓëW×é³ÉµÄ»¯ºÏÎïÊÇÒ»ÖÖÎÂÊÒÆøÌ壬¸ÃÆøÌåΪCH4£¬ÔòWΪC£¬WÓëY¡¢XÓëY×é³ÉµÄ»¯ºÏÎïÊÇ»ú¶¯³µÅųöµÄ´óÆøÎÛȾÎΪCOºÍNOx£¬ÔòYΪO£¬XΪN£¬YºÍZÄÜÐγÉÔ­×Ó¸öÊý±ÈΪ1£º1ºÍ1£º2µÄÁ½ÖÖÀë×Ó»¯ºÏÎÕâÁ½ÖÖÀë×Ó»¯ºÏÎï·Ö±ðΪNa2O¡¢Na2O2£¬×÷ΪZΪNa£¬Ôò
£¨1£©WΪC£¬Ô­×ÓÐòÊýΪ6£¬Ô­×ÓºËÍâÓÐ2¸öµç×Ӳ㣬×îÍâ²ãµç×ÓÊýΪ6£¬Ó¦Î»ÓÚÖÜÆÚ±íµÚ¶þÖÜÆÚIVA×壬
Z2YΪNa2O£¬ÎªÀë×Ó»¯ºÏÎµç×ÓʽΪ£¬
¹Ê´ð°¸Îª£ºµÚ¶þÖÜÆÚIVA×壻£»                    
£¨2£©¢ÙÓÉ×ÔÓÉÄÜÅоݡ÷G=¡÷H-T?¡÷S£¬¿ÉÖªµ±¡÷H£¼T?¡÷Sʱ·´Ó¦²ÅÄÜ×Ô·¢½øÐУ¬¼´±ØÐëÔڽϸߵÄζÈϲſÉÒÔ½øÐУ¬»¯Ñ§·´Ó¦ÄÜ·ñ×Ô·¢½øÐÐÈ¡¾öÓÚìʱäºÍìرäµÄ×ÛºÏÅоݣ¬²»ÄÜÖ»ÓÉìʱä»òìرäÀ´¾ö¶¨£¬¹Ê´ð°¸Îª£º²»ÄÜ£»
 ¢Ú´ïƽºâʱ£¬¸÷ÎïÖʵÄŨ¶È·Ö±ðΪ£ºc£¨N2£©=4mol/L£¬c£¨H2£©=2mol/L£¬c£¨NH3£©=4mol/L£¬
Ôòk=
c2(NH3)
c(N2)¡Ác3(H2 )
=
42
4¡Á23
=0.5£¬¹Ê´ð°¸Îª£º0.5£»
£¨3£©n£¨NH3£©=
2.24L
22.4L/mol
=0.1mol£¬n£¨HNO3£©=0.2L¡Á1mol/L=0.2mol£¬HNO3¹ýÁ¿£¬Éú³ÉNH4NO3£¬¶þÕßÎïÖʵÄÁ¿ÏàµÈ£¬µ«ÓÉÓÚNH4+Ë®½â£¬Ôòc£¨H+£©£¾c£¨NH4+£©£¬ËùÒÔÈÜÒºÖдæÔÚc£¨NO3-£©£¾c£¨H+£©£¾c£¨NH4+£©£¾c£¨OH-£©£¬
¹Ê´ð°¸Îª£ºc£¨NO3-£©£¾c£¨H+£©£¾c£¨NH4+£©£¾c£¨OH-£©£» 
£¨4£©CH3OH¼îÐÔȼÁϵç³ØÖУ¬CH3OHΪµç³ØµÄ¸º¼«·´Ó¦£¬±»Ñõ»¯£¬µç¼«·´Ó¦Ê½ÎªCH3OH+8OH--6e-=CO32-+6H2O£¬
¹Ê´ð°¸Îª£ºCH3OH+8OH--6e-=CO32-+6H2O£»
£¨5£©XºÍZ×é³ÉµÄÒ»ÖÖÀë×Ó»¯ºÏÎïΪNa3N£¬ÄÜÓëË®·´Ó¦Éú³ÉÁ½ÖּÉú³ÉÎïΪNaOHºÍNH3?H2O£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪNa3N+4H2O=3NaOH+NH3?H2O£¬
¹Ê´ð°¸Îª£ºNa3N+4H2O=3NaOH+NH3?H2O£®
µãÆÀ£º±¾Ì⿼²éÔªËصÄÍƶϣ¬ÌâÄ¿ÄѶÈÖеȣ¬±¾Ìâ²àÖØÓڵ绯ѧºÍÈÜÒºÖеÄË®½âºÍÈõµç½âÖʵĵçÀëµÈ֪ʶµÄ¿¼²é£¬ÌâÄ¿×ۺ϶Ƚϸߣ¬´ðÌâʱעÒâ°ÑÎÕÏà¹Ø֪ʶºÍ·½·¨µÄÓ¦Óã¬Îª¸ß¿¼³£¿¼²é·½Ê½£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2011?Î÷³ÇÇøһģ£©Q¡¢W¡¢X¡¢Y¡¢ZÊÇ5ÖÖ¶ÌÖÜÆÚÔªËØ£¬Ô­×ÓÐòÊýÖð½¥Ôö´ó£¬QÓëW×é³ÉµÄ»¯ºÏÎïÊÇÒ»ÖÖÎÂÊÒÆøÌ壬WÓëY¡¢XÓëY×é³ÉµÄ»¯ºÏÎïÊÇ»ú¶¯³µÅųöµÄ´óÆøÎÛȾÎYºÍZÄÜÐγÉÔ­×Ó¸öÊý±ÈΪ1£º1ºÍ1£º2µÄÁ½ÖÖÀë×Ó»¯ºÏÎ
£¨1£©WÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÊÇ
µÚ¶þÖÜÆÚIVA×å
µÚ¶þÖÜÆÚIVA×å
£¬Z2YµÄµç×ÓʽÊÇ
£®
£¨2£©¹¤ÒµºÏ³ÉXQ3ÊÇ·ÅÈÈ·´Ó¦£®ÏÂÁдëÊ©ÖУ¬¼ÈÄܼӿ췴ӦËÙÂÊ£¬ÓÖÄÜÌá¸ßÔ­ÁÏת»¯ÂʵÄÊÇ
d
d
£®
a£®Éý¸ßζȠ                  b£®¼ÓÈë´ß»¯¼Á
c£®½«XQ3¼°Ê±·ÖÀë³öÈ¥        d£®Ôö´ó·´Ó¦ÌåϵµÄѹǿ
£¨3£©2.24L£¨±ê×¼×´¿ö£©XQ3±»200mL 1mol/L QXY3ÈÜÒºÎüÊÕºó£¬ËùµÃÈÜÒºÖÐÀë×ÓŨ¶È´Ó´óµ½Ð¡µÄ˳ÐòÊÇ
c£¨NO3-£©£¾c£¨H+£©£¾c£¨NH4+£©£¾c£¨OH-£©
c£¨NO3-£©£¾c£¨H+£©£¾c£¨NH4+£©£¾c£¨OH-£©
£®
£¨4£©WQ4YÓëY2µÄ·´Ó¦¿É½«»¯Ñ§ÄÜת»¯ÎªµçÄÜ£¬Æ乤×÷Ô­ÀíÈçͼËùʾ£¬a¼«µÄµç¼«·´Ó¦Ê½ÊÇ
CH3OH-6e-+8OH-=CO32-+6H2O
CH3OH-6e-+8OH-=CO32-+6H2O
£®

£¨5£©ÒÑÖª£ºW£¨s£©+Y2 £¨g£©=WY2£¨g£©¡÷H=-393.5kJ/mol      WY£¨g£©+
12
Y2 £¨g£©=WY2£¨g£©¡÷H=-283.0kJ/mol
24g WÓëÒ»¶¨Á¿µÄY2·´Ó¦£¬·Å³öÈÈÁ¿362.5kJ£¬ËùµÃ²úÎïµÄÎïÖʵÄÁ¿Ö®±ÈÊÇ
n£¨CO2£©£ºn£¨CO£©=1£º3
n£¨CO2£©£ºn£¨CO£©=1£º3
£®
£¨6£©XºÍZ×é³ÉµÄÒ»ÖÖÀë×Ó»¯ºÏÎÄÜÓëË®·´Ó¦Éú³ÉÁ½Öּ¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
Na3N+4H2O=3NaOH+NH3?H2O
Na3N+4H2O=3NaOH+NH3?H2O
£®

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

¾«Ó¢¼Ò½ÌÍø