ÌâÄ¿ÄÚÈÝ

°±ÊǵªÑ­»·¹ý³ÌÖеÄÖØÒªÎïÖÊ,°±µÄºÏ³ÉÊÇÄ¿Ç°ÆÕ±éʹÓõÄÈ˹¤¹Ìµª·½·¨¡£
(1)¸ù¾Ýͼ1ÌṩµÄÐÅÏ¢,д³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ:¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡,ÔÚͼ1ÖÐÇúÏß¡¡¡¡¡¡¡¡¡¡¡¡¡¡(Ìî¡°a¡±»ò¡°b¡±)±íʾ¼ÓÈëÌú´¥Ã½µÄÄÜÁ¿±ä»¯ÇúÏß¡£ 

(2)ÔÚºãÈÝÈÝÆ÷ÖÐ,ÏÂÁÐÃèÊöÖÐÄÜ˵Ã÷ÉÏÊö·´Ó¦ÒÑ´ïƽºâµÄÊÇ¡¡¡¡¡¡¡¡¡£ 
A.3v(H2)Õý=2v(NH3)Äæ
B.µ¥Î»Ê±¼äÄÚÉú³Én mol N2µÄͬʱÉú³É2n mol NH3
C.»ìºÏÆøÌåµÄÃܶȲ»Ôٸıä
D.ÈÝÆ÷ÄÚѹǿ²»Ëæʱ¼äµÄ±ä»¯¶ø±ä»¯
(3)Ò»¶¨Î¶ÈÏÂ,Ïò2 LÃܱÕÈÝÆ÷ÖгäÈë1 mol N2ºÍ3 mol H2,±£³ÖÌå»ý²»±ä,0.5 minºó´ïµ½Æ½ºâ,²âµÃÈÝÆ÷ÖÐÓÐ0.4 mol NH3,Ôòƽ¾ù·´Ó¦ËÙÂÊv(N2)=¡¡¡¡¡¡¡¡¡¡¡¡,¸ÃζÈϵÄƽºâ³£ÊýK=¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ÈôÉý¸ßζÈ,KÖµ±ä»¯¡¡¡¡¡¡¡¡(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)¡£ 
(4)ΪÁËÑ°ÕҺϳÉNH3µÄζȺÍѹǿµÄÊÊÒËÌõ¼þ,ijͬѧÉè¼ÆÁËÈý×éʵÑé,²¿·ÖʵÑéÌõ¼þÒѾ­ÌîÔÚÏÂÃæʵÑéÉè¼Æ±íÖС£

ʵÑé±àºÅ
 
T(¡æ)
 
n(N2)/n(H2)
 
p(MPa)
 
¢¡
 
450
 
1/3
 
1
 
¢¢
 
¡¡ 
 
¡¡ 
 
10
 
¢£
 
480
 
¡¡ 
 
10
 
 
A.ÇëÔÚÉϱí¿Õ¸ñÖÐÌîÈëÊ£ÓàµÄʵÑéÌõ¼þÊý¾Ý¡£
B.¸ù¾Ý·´Ó¦N2(g)+3H2(g)2NH3(g)µÄÌصã,ÔÚ¸ø³öµÄ×ø±êͼ2ÖÐ,»­³öÆäÔÚ1 MPaºÍ10 MPaÌõ¼þÏÂH2µÄת»¯ÂÊËæζȱ仯µÄÇ÷ÊÆÇúÏßʾÒâͼ,²¢±êÃ÷¸÷ÌõÇúÏßµÄѹǿ¡£

(1)N2(g)+3H2(g)2NH3(g)¦¤H="-92" kJ¡¤mol-1¡¡b
(2)BD¡¡(3)0.2 mol¡¤L-1¡¤min-1¡¡0.058¡¡¼õС
(4)A.¢¢.450¡¡1/3¡¡¢£.1/3
B.

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Ñо¿CO2µÄÀûÓöԴٽøµÍ̼Éç»áµÄ¹¹½¨¾ßÓÐÖØÒªÒâÒå¡£
£¨1£©½«CO2Ó뽹̿×÷ÓÃÉú³ÉCO£¬CO¿ÉÓÃÓÚÁ¶ÌúµÈ¡£
ÒÑÖª£ºFe2O3(s) + 3C(ʯī) =" 2Fe(s)" + 3CO(g) ¡÷H 1 =" +489.0" kJ¡¤mol£­1
C(ʯī) +CO2(g) = 2CO(g)           ¡÷H 2 =" +172.5" kJ¡¤mol£­1
ÔòCO»¹Ô­Fe2O3(s)µÄÈÈ»¯Ñ§·½³ÌʽΪ                  ¡£
£¨2£©¶þÑõ»¯Ì¼ºÏ³É¼×´¼ÊÇ̼¼õÅŵÄз½Ïò£¬½«CO2ת»¯Îª¼×´¼µÄÈÈ»¯Ñ§·½³Ìʽ
CO2(g) +3H2(g)CH3OH(g) +H2O(g) ¡÷H
¢Ù¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪK=                    ¡£
¢ÚÈ¡Ò»¶¨Ìå»ýCO2ºÍH2µÄ»ìºÏÆøÌå(ÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã3)£¬¼ÓÈëºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÉÏÊö·´Ó¦¡£·´Ó¦¹ý³ÌÖвâµÃ¼×´¼µÄÌå»ý·ÖÊý¦Õ(CH3OH)Ó뷴ӦζÈTµÄ¹ØϵÈçͼAËùʾ£¬Ôò¸Ã·´Ó¦µÄ¦¤H      0£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°£½¡±£©¡£

¢ÛÔÚÁ½ÖÖ²»Í¬Ìõ¼þÏ·¢Éú·´Ó¦£¬²âµÃCH3OHµÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯ÈçͼBËùʾ£¬ÇúÏßI¡¢¢ò¶ÔÓ¦µÄƽºâ³£Êý´óС¹ØϵΪK¢ñ     K¢ò£¨Ìî¡°>¡± »ò¡°<¡±£©¡£
£¨3£©ÒÔCO2ΪԭÁÏ»¹¿ÉÒԺϳɶàÖÖÎïÖÊ¡£¢Ù¹¤ÒµÉÏÄòËØ[CO(NH2)2]ÓÉCO2ºÍNH3ÔÚÒ»¶¨Ìõ¼þϺϳɣ¬Æä·´Ó¦·½³ÌʽΪ                                    ¡£µ±°±Ì¼±È£½3£¬´ïƽºâʱCO2µÄת»¯ÂÊΪ60%£¬ÔòNH3µÄƽºâת»¯ÂÊΪ                  ¡£
¢ÚÓÃÁòËáÈÜÒº×÷µç½âÖʽøÐеç½â£¬CO2Ôڵ缫ÉÏ¿Éת»¯Îª¼×Í飬¸Ãµç¼«·´Ó¦µÄ·½³ÌʽΪ                ¡£

(1)ÒÑÖª£º
¢ÙFe(s)£«O2(g)=FeO(s)¡¡¦¤H£½£­272.0 kJ¡¤mol£­1
¢Ú2Al(s)£«O2(g)=Al2O3(s)¡¡¦¤H£½£­1675.7 kJ¡¤mol£­1
AlºÍFeO·¢ÉúÂÁÈÈ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ____________________________________

(2)ij¿ÉÄæ·´Ó¦ÔÚ²»Í¬Ìõ¼þϵķ´Ó¦Àú³Ì·Ö±ðΪA¡¢B(ÈçÉÏͼËùʾ)¡£
¢Ù¸ù¾ÝͼÅжϸ÷´Ó¦´ïµ½Æ½ºâºó£¬ÆäËûÌõ¼þ²»±ä£¬Éý¸ßζȣ¬·´Ó¦ÎïµÄת»¯ÂÊ________(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±)£»
¢ÚÆäÖÐBÀú³Ì±íÃ÷´Ë·´Ó¦²ÉÓõÄÌõ¼þΪ________(Ñ¡ÌîÐòºÅ)¡£
A£®Éý¸ßζȡ¡¡¡¡¡¡¡  B£®Ôö´ó·´Ó¦ÎïµÄŨ¶È   C£®½µµÍζȠ D£®Ê¹Óô߻¯¼Á
(3)1000 ¡æʱ£¬ÁòËáÄÆÓëÇâÆø·¢ÉúÏÂÁз´Ó¦£ºNa2SO4(s)£«4H2(g)Na2S(s)£«4H2O(g)
¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ________________________________£»
ÒÑÖªK1000 ¡æ<K1200 ¡æ£¬Èô½µµÍÌåϵζȣ¬»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿½«»á________(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±)¡£
(4)³£ÎÂÏ£¬Èç¹ûÈ¡0.1 mol¡¤L£­1 HAÈÜÒºÓë0.1 mol¡¤L£­1 NaOHÈÜÒºµÈÌå»ý»ìºÏ(»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ)£¬²âµÃ»ìºÏÒºµÄpH£½8¡£
¢Ù»ìºÏÒºÖÐÓÉË®µçÀë³öµÄOH£­Å¨¶ÈÓë0.1 mol¡¤L£­1 NaOHÈÜÒºÖÐÓÉË®µçÀë³öµÄOH£­Å¨¶ÈÖ®±ÈΪ________£»
¢ÚÒÑÖªNH4AÈÜҺΪÖÐÐÔ£¬ÓÖÖª½«HAÈÜÒº¼Óµ½Na2CO3ÈÜÒºÖÐÓÐÆøÌå·Å³ö£¬ÊÔÍƶÏ(NH4)2CO3ÈÜÒºµÄpH________7(Ìî¡°<¡±¡°>¡±»ò¡°£½¡±)£»ÏàͬζÈÏ£¬µÈÎïÖʵÄÁ¿Å¨¶ÈµÄÏÂÁÐËÄÖÖÑÎÈÜÒº°´pHÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòΪ(ÌîÐòºÅ)________¡£
a£®NH4HCO3  b£®NH4A       c£®(NH4)2CO3  d£®NH4Cl

¼×´¼È¼ÁÏ·ÖΪ¼×´¼ÆûÓͺͼ״¼²ñÓÍ¡£¹¤ÒµÉϺϳɼ״¼µÄ·½·¨ºÜ¶à¡£
£¨1£©Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º
¢ÙCO­2£¨g£© +3H2£¨g£© £½CH­­3OH£¨g£©+H2O£¨g£©   ¡÷H1     
¢Ú2CO­£¨g£© +O2£¨g£© £½2CO­2£¨g£©  ¡÷H2
¢Û2H2£¨g£©+O2£¨g£© £½2H2O£¨g£©         ¡÷H3
ÔòCO£¨g£© + 2H2£¨g£© CH3OH£¨g£©¡¡µÄ¡÷H£½               ¡£
£¨2£©ÔÚÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖнøÐз´Ó¦£º CO£¨g£©+2H2£¨g£©CH3OH£¨g£© £¬ÆäËûÌõ¼þ²»±ä£¬ÔÚ300¡æºÍ500¡æʱ£¬ÎïÖʵÄÁ¿n£¨CH3OH£© Ó뷴Ӧʱ¼ätµÄ±ä»¯ÇúÏßÈçͼËùʾ¡£¸Ã·´Ó¦µÄ¡÷H    0 £¨Ìî>¡¢<»ò=£©¡£

£¨3£©ÈôÒªÌá¸ß¼×´¼µÄ²úÂÊ£¬¿É²ÉÈ¡µÄ´ëÊ©ÓÐ____________£¨Ìî×Öĸ£©¡£

A£®ËõСÈÝÆ÷Ìå»ý
B£®½µµÍζÈ
C£®Éý¸ßζÈ
D£®Ê¹ÓúÏÊʵĴ߻¯¼Á
E£®½«¼×´¼´Ó»ìºÏÌåϵÖзÖÀë³öÀ´
£¨4£©CH4ºÍH2OÔÚ´ß»¯¼Á±íÃæ·¢Éú·´Ó¦CH4+H2OCO+3H2£¬T¡æʱ£¬Ïò1 LÃܱÕÈÝÆ÷ÖÐͶÈë1 mol CH4ºÍ1 mol H2O£¨g£©£¬5Сʱºó²âµÃ·´Ó¦Ìåϵ´ïµ½Æ½ºâ״̬£¬´ËʱCH4µÄת»¯ÂÊΪ50% £¬¼ÆËã¸ÃζÈϵÄƽºâ³£Êý     £¨½á¹û±£ÁôСÊýµãºóÁ½Î»Êý×Ö£©¡£
£¨5£©ÒÔ¼×´¼ÎªÈ¼ÁϵÄÐÂÐ͵ç³Ø£¬Æä³É±¾´ó´óµÍÓÚÒÔÇâΪȼÁϵĴ«Í³È¼Áϵç³Ø£¬Ä¿Ç°µÃµ½¹ã·ºµÄÑо¿£¬ÈçͼÊÇÄ¿Ç°Ñо¿½Ï¶àµÄÒ»Àà¹ÌÌåÑõ»¯ÎïȼÁϵç³Ø¹¤×÷Ô­ÀíʾÒâͼ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙB¼«µÄµç¼«·´Ó¦Ê½Îª                     ¡£
¢ÚÈôÓøÃȼÁϵç³Ø×öµçÔ´£¬ÓÃʯī×öµç¼«µç½âÁòËáÍ­ÈÜÒº£¬µ±µç·ÖÐתÒÆ1mole- ʱ£¬Êµ¼ÊÉÏÏûºÄµÄ¼×´¼µÄÖÊÁ¿±ÈÀíÂÛÉϴ󣬿ÉÄÜÔ­ÒòÊÇ                     ¡£
£¨6£©25¡æʱ£¬²ÝËá¸ÆµÄKsp=4.0¡Á10-8,̼Ëá¸ÆµÄKsp=2.5¡Á10-9¡£Ïò20ml̼Ëá¸ÆµÄ±¥ºÍÈÜÒºÖÐÖðµÎ¼ÓÈë8.0¡Á10-4 mol¡¤L-1µÄ²ÝËá¼ØÈÜÒº20ml£¬ÄÜ·ñ²úÉú³Áµí       £¨Ìî¡°ÄÜ¡±»ò¡°·ñ¡±£©¡£

еġ¶»·¾³¿ÕÆøÖÊÁ¿±ê×¼¡·½«ÓÚ2016Äê1ÔÂ1ÈÕÔÚÎÒ¹úÈ«Ãæʵʩ¡£¾Ý´Ë£¬»·¾³¿ÕÆøÖÊÁ¿Ö¸Êý(AQI)ÈÕ±¨ºÍʵʱ±¨¸æ°üÀ¨ÁËSO2¡¢NO2¡¢CO¡¢O3¡¢PM10¡¢PM2.5µÈÖ¸±ê£¬Îª¹«ÖÚÌṩ½¡¿µÖ¸Òý£¬Òýµ¼µ±µØ¾ÓÃñºÏÀí°²ÅųöÐкÍÉú»î¡£
(1)Æû³µÅųöµÄβÆøÖк¬ÓÐCOºÍNOµÈÆøÌ壬Óû¯Ñ§·½³Ìʽ½âÊͲúÉúNOµÄÔ­Òò________________________________________
(2)Æû³µÅÅÆø¹ÜÄÚ°²×°µÄ´ß»¯×ª»¯Æ÷£¬¿ÉʹÆû³µÎ²ÆøÖеÄÖ÷ÒªÎÛȾÎïת»¯ÎªÎÞ¶¾µÄ´óÆøÑ­»·ÎïÖÊ¡£ÒÑÖª£º
N2(g)£«O2(g)===2NO(g)¡¡¦¤H£½£«180.5 kJ¡¤mol£­1
2C(s)£«O2(g)===2CO(g)¡¡¦¤H£½£­221.0 kJ¡¤mol£­1
C(s)£«O2(g)===CO2(g)¡¡¦¤H£½£­393.5 kJ¡¤mol£­1
Ôò·´Ó¦2NO(g)£«2CO(g)??N2(g)£«2CO2(g)µÄ¦¤H£½________kJ¡¤mol£­1£»¸Ã·´Ó¦µÄ¦¤S________0(Ìî¡°>¡±¡°<¡±»ò¡°£½¡±)¡£
(3)½«0.20 mol NOºÍ0.10 mol CO³äÈëÒ»¸öÈÝ»ýºã¶¨Îª1 LµÄÃܱÕÈÝÆ÷ÖУ¬ÔÚ²»Í¬Ìõ¼þÏ·´Ó¦¹ý³ÌÖв¿·ÖÎïÖʵÄŨ¶È±ä»¯×´¿öÈçͼËùʾ¡£

¢Ù¼ÆËã²úÎïN2ÔÚ6¡«9 minʱµÄƽ¾ù·´Ó¦ËÙÂÊv(N2)£½________mol¡¤L£­1¡¤min£­1£»
¢ÚµÚ12 minʱ¸Ä±äµÄ·´Ó¦Ìõ¼þΪ________(Ìî¡°ÉýΡ±»ò¡°½µÎ¡±)£»
¢Û¼ÆËã·´Ó¦ÔÚµÚ24 minʱµÄƽºâ³£ÊýK£½________¡£Èô±£³ÖζȲ»±ä£¬ÔÙÏòÈÝÆ÷ÖгäÈëCO¡¢N2¸÷0.060 mol£¬Æ½ºâ½«________Òƶ¯(Ìî¡°ÕýÏò¡±¡°ÄæÏò¡±»ò¡°²»¡±)¡£
(4)»·¾³¼à²âÖл¹¿ÉÓóÁµí·¨²â¶¨¿ÕÆøÖк¬ÓнϸßŨ¶ÈSO2µÄº¬Á¿£¬¾­²éµÃһЩÎïÖÊÔÚ20 ¡æµÄÊý¾ÝÈçÏÂ±í£º

Èܽâ¶È(S)/g
ÈܶȻý(Ksp)
Ca(OH)2
Ba(OH)2
CaSO3
BaSO3
 0.160
3.89
6.76¡Á10£­3
5.48¡Á10£­9
 
¢ÙÎüÊÕSO2×îºÏÊʵÄÊÔ¼ÁÊÇ________[Ìî¡°Ca(OH)2¡±»ò¡°Ba(OH)2¡±]ÈÜÒº£»
¢ÚÔÚ20 ¡æʱ£¬ÏòCaSO3Ðü×ÇÒºÖеμÓÊÊÁ¿µÄBaCl2ÈÜÒº£¬µ±CaSO3ÏòBaSO3µÄת»¯´ïµ½Æ½ºâʱ£¬ÈÜÒºÖеģ½____________(д³ö±í´ïʽ¼´¿É)¡£

¸ß¯Á¶ÌúÊÇÒ±Á¶ÌúµÄÖ÷Òª·½·¨£¬·¢ÉúµÄÖ÷Òª·´Ó¦Îª£ºFe2O3£¨s£©£«3CO£¨g£©??2Fe£¨s£©£«3CO2£¨g£©
¦¤H£½a kJ¡¤mol£­1¡£
£¨1£©ÒÑÖª£º¢ÙFe2O3£¨s£©£«3C£¨s£¬Ê¯Ä«£©=2Fe£¨s£©£«3CO£¨g£©
¦¤H1£½£«489.0 kJ¡¤mol£­1£»
¢ÚC£¨s£¬Ê¯Ä«£©£«CO2£¨g£©=2CO£¨g£©¡¡¦¤H2£½£«172.5 kJ¡¤mol£­1¡£Ôòa£½________¡£
£¨2£©Ò±Á¶Ìú·´Ó¦µÄƽºâ³£Êý±í´ïʽK£½________£¬Î¶ÈÉý¸ßºó£¬KÖµ________£¨Ìî¡°Ôö´ó¡±¡¢¡°²»±ä¡±»ò¡°¼õС¡±£©¡£
£¨3£©ÔÚT ¡æʱ£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK£½64£¬ÔÚ2 LºãÈÝÃܱÕÈÝÆ÷¼×ºÍÒÒÖУ¬·Ö±ð°´Ï±íËùʾ¼ÓÈëÎïÖÊ£¬·´Ó¦¾­¹ýÒ»¶Îʱ¼äºó´ïµ½Æ½ºâ¡£

 
Fe2O3
CO
Fe
CO2
¼×/mol
1.0
1.0
1.0
1.0
ÒÒ/mol
1.0
2.0
1.0
1.0
 
¢Ù¼×ÈÝÆ÷ÖÐCOµÄƽºâת»¯ÂÊΪ________¡£
¢ÚÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ________£¨Ìî×Öĸ£©¡£
a£®ÈôÈÝÆ÷ÄÚÆøÌåÃܶȺ㶨ʱ£¬±êÖ¾·´Ó¦´ïµ½Æ½ºâ״̬
b£®Ôö¼ÓFe2O3µÄÁ¿£¬¿ÉÒÔÌá¸ßCOµÄת»¯ÂÊ
c£®¼×ÈÝÆ÷ÖÐCOµÄƽºâת»¯ÂÊ´óÓÚÒÒµÄƽºâת»¯ÂÊ
d£®¼×¡¢ÒÒÈÝÆ÷ÖУ¬COµÄƽºâŨ¶ÈÖ®±ÈΪ2¡Ã3
£¨4£©²ÉÈ¡Ò»¶¨´ëÊ©¿É·ÀÖ¹¸ÖÌú¸¯Ê´¡£ÏÂÁÐ×°ÖÃÖеÄÉÕ±­Àï¾ùÊ¢ÓеÈŨ¶È¡¢µÈÌå»ýµÄNaClÈÜÒº¡£
¢ÙÔÚa¡¢b¡¢c×°ÖÃÖÐÄܱ£»¤ÌúµÄÊÇ________£¨Ìî×Öĸ£©¡£
¢ÚÈôÓÃd×°Öñ£»¤Ìú£¬X¼«µÄµç¼«²ÄÁÏÓ¦ÊÇ________£¨ÌîÃû³Æ£©¡£

£¨5£©25 ¡æʱÓйØÎïÖʵÄÈܶȻýÈçÏ£ºKsp[Mg£¨OH£©2]£½5.61¡Á10£­12£¬Ksp[Fe£¨OH£©3]£½2.64¡Á10£­39¡£25 ¡æʱ£¬Ïòº¬ÓÐMg2£«¡¢Fe3£«µÄÈÜÒºÖеμÓNaOHÈÜÒº£¬µ±Á½ÖÖ³Áµí¹²´æÇÒÈÜÒºµÄpH£½8ʱ£¬c£¨Mg2£«£©¡Ãc£¨Fe3£«£©£½________¡£

×ÊÔ´»¯ÀûÓöþÑõ»¯Ì¼²»½ö¿É¼õÉÙÎÂÊÒÆøÌåµÄÅÅ·Å£¬»¹¿ÉÖØлñµÃȼÁÏ»òÖØÒª¹¤Òµ²úÆ·¡£
£¨1£©ÓпÆѧ¼ÒÌá³ö¿ÉÀûÓÃFeOÎüÊÕºÍÀûÓÃCO2£¬Ïà¹ØÈÈ»¯Ñ§·½³ÌʽÈçÏ£º6FeO(s)+CO2(g)=2Fe3O4(s)+C(s) ¡÷H="-76.0" kJ¡¤molÒ»1
¢ÙÉÏÊö·´Ó¦ÖÐÿÉú³É1 mol Fe3O4£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿Îª_______mol¡£
¢ÚÒÑÖª£ºC(s)+2H2O(g)=CO2(g)+2H2(g) ¡÷H="+113.4" kJ¡¤molÒ»1£¬Ôò·´Ó¦£º3FeO(s)+ H2O(g)= Fe3O4(s)+ H2(g)µÄ¡÷H=__________¡£
£¨2£©ÔÚÒ»¶¨Ìõ¼þÏ£¬¶þÑõ»¯Ì¼×ª»¯Îª¼×ÍéµÄ·´Ó¦ÈçÏ£ºCO2(g)+4 H2 (g)  C H4 (g)+2 H2O(g)£¬ÏòÒ»ÈÝ»ýΪ2 LµÄºãÈÝÃܱÕÈÝÆ÷ÖгäÈËÒ»¶¨Á¿µÄCO2ºÍH2£¬ÔÚ300¡æʱ·¢ÉúÉÏÊö·´Ó¦£¬´ïµ½Æ½ºâʱ¸÷ÎïÖʵÄŨ¶È·Ö±ðΪCO2 0.2 mol¡¤LÒ»1£¬H2 0.8 mol¡¤LÒ»1£¬CH40.8 mol¡¤LÒ»1£¬H2O1.6 mol¡¤LÒ»1¡£ÔòCO2µÄƽºâת»¯ÂÊΪ________¡£300 ¡æʱÉÏÊö·´Ó¦µÄƽºâ³£ÊýK=____________________¡£200¡æʱ¸Ã·´Ó¦µÄƽºâ³£ÊýK=64.8£¬Ôò¸Ã·´Ó¦µÄ¡÷H_____(Ìî¡°£¾¡¯¡¯»ò¡°<¡±)O¡£
£¨3£©»ªÊ¢¶Ù´óѧµÄÑо¿ÈËÔ±Ñо¿³öÒ»ÖÖ·½·¨£¬¿ÉʵÏÖË®ÄàÉú²úʱCO2ÁãÅÅ·Å£¬Æä»ù±¾Ô­ÀíÈç
ͼËùʾ:

¢ÙÉÏÊöÉú²ú¹ý³ÌµÄÄÜÁ¿×ª»¯·½Ê½ÊÇ____________________¡£
¢ÚÉÏÊöµç½â·´Ó¦ÔÚζÈСÓÚ900¡æʱ½øÐУ¬Ì¼Ëá¸ÆÏÈ·Ö½âΪCaOºÍCO2£¬µç½âÖÊΪÈÛÈÚ̼ËáÄÆ£¬ÔòÑô¼«µÄµç¼«·´Ó¦Ê½Îª___________________¡£

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

¾«Ó¢¼Ò½ÌÍø