ÌâÄ¿ÄÚÈÝ

Ϊ¼õСCO2¶Ô»·¾³µÄÓ°Ï죬ÔÚÏÞÖÆÆäÅÅ·ÅÁ¿µÄͬʱ£¬Ó¦¼ÓÇ¿¶ÔCO2´´ÐÂÀûÓõÄÑо¿¡£
(1)¢Ù °Ñº¬ÓнϸßŨ¶ÈCO2µÄ¿ÕÆøͨÈë±¥ºÍK2CO3ÈÜÒº¡£
¢Ú ÔÚ¢ÙµÄÎüÊÕÒºÖÐͨ¸ßÎÂË®ÕôÆøµÃµ½¸ßŨ¶ÈµÄCO2ÆøÌå¡£
д³ö¢ÚÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ________¡£
(2)È罫CO2ÓëH2ÒÔ1:3µÄÌå»ý±È»ìºÏ¡£
¢ÙÊʵ±Ìõ¼þϺϳÉijÌþºÍË®£¬¸ÃÌþÊÇ_______(ÌîÐòºÅ)¡£

A£®ÍéÌþB£®Ï©ÌþC£®È²ÌþD£®±½µÄͬϵÎï
¢Ú Êʵ±Ìõ¼þϺϳÉȼÁϼ״¼ºÍË®¡£ÔÚÌå»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë2 mol CO2ºÍ6 mol H2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2(g)£«3H2(g)CH3OH(g)£«H2O(g) ¡÷H£½£­49.0 kJ/mol¡£
²âµÃCO2(g)ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ¡£

´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬v(H2)£½______£»ÇâÆøµÄת»¯ÂÊ£½_______£»ÄÜʹƽºâÌåϵÖÐn(CH3OH)Ôö´óµÄ´ëÊ©ÓÐ______¡£
(3)È罫CO2ÓëH2ÒÔ1:4µÄÌå»ý±È»ìºÏ£¬ÔÚÊʵ±µÄÌõ¼þÏ¿ÉÖƵÃCH4¡£
ÒÑÖª£º
CH4 (g) + 2O2(g)  CO2(g)+ 2H2O(l)   ¦¤H1£½¨D 890.3 kJ/mol
H2(g) + 1/2O2(g)  H2O(l)           ¦¤H2£½£­285.8 kJ/mol
ÔòCO2(g)ÓëH2(g)·´Ó¦Éú³ÉCH4(g)ÓëҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽÊÇ________¡£
(4)ijͬѧÓóÁµí·¨²â¶¨º¬ÓнϸßŨ¶ÈCO2µÄ¿ÕÆøÖÐCO2µÄº¬Á¿£¬¾­²éµÃһЩÎïÖÊÔÚ20¡æµÄÊý¾ÝÈçÏÂ±í¡£
Èܽâ¶È(S)/g
ÈܶȻý(Ksp)
Ca(OH)2
Ba(OH)2
CaCO3
BaCO3
0.16
3.89
2.9¡Á10-9
2.6¡Á10-9
 (˵Ã÷£ºKspԽС£¬±íʾ¸ÃÎïÖÊÔÚË®ÈÜÒºÖÐÔ½Ò׳Áµí)
ÎüÊÕCO2×îºÏÊʵÄÊÔ¼ÁÊÇ_________[Ìî¡°Ca(OH)2¡±»ò¡°Ba(OH)2¡±]ÈÜÒº£¬ÊµÑéʱ³ýÐèÒª²â¶¨¹¤Òµ·ÏÆøµÄÌå»ý(ÕÛËã³É±ê×¼×´¿ö)Í⣬»¹ÐèÒª²â¶¨__________¡£

(1)¢Ú 2KHCO3K2CO3£«H2O£«CO2¡ü
(2)¢Ù B
¢Ú v(H2)£½0.24 mol/(L¡¤min)           80%
½µµÍζÈ(»ò¼Óѹ»òÔö´óH2µÄÁ¿µÈ)
(3)CO2(g) + 4H2(g)  CH4 (g)+ 2H2O(l) ¦¤H1£½ --252.9 kJ/mol
(4)Ba(OH)2        BaCO3µÄÖÊÁ¿

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2013?ºìÇÅÇø¶þÄ££©Îª¼õСCO2¶Ô»·¾³µÄÓ°Ï죬ÔÚÏÞÖÆÆäÅÅ·ÅÁ¿µÄͬʱ£¬Ó¦¼ÓÇ¿¶ÔCO2´´ÐÂÀûÓõÄÑо¿£®
£¨1£©¢Ù°Ñº¬ÓнϸßŨ¶ÈCO2µÄ¿ÕÆøͨÈë±¥ºÍK2CO3ÈÜÒº£®
¢ÚÔÚ¢ÙµÄÎüÊÕÒºÖÐͨ¸ßÎÂË®ÕôÆøµÃµ½¸ßŨ¶ÈµÄCO2ÆøÌ壮
д³ö¢ÚÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ
2KHCO3
  ¡÷  
.
 
K2CO3+H2O+CO2¡ü
2KHCO3
  ¡÷  
.
 
K2CO3+H2O+CO2¡ü
£®
£¨2£©È罫CO2ÓëH2 ÒÔ1£º3µÄÌå»ý±È»ìºÏ£®
¢ÙÊʵ±Ìõ¼þϺϳÉijÌþºÍË®£¬¸ÃÌþÊÇ
B
B
£¨ÌîÐòºÅ£©£®
A£®ÍéÌþ     B£®Ï©Ìþ     C£®È²Ìþ    D£®±½µÄͬϵÎï
¢ÚÊʵ±Ìõ¼þϺϳÉȼÁϼ״¼ºÍË®£®ÔÚÌå»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë 2mol CO2ºÍ6mol H2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º
CO2£¨g£©+3H2£¨g£©?CH3OH£¨g£©+H2O£¨g£©¡÷H=-49.0kJ/mol£®
²âµÃCO2£¨g£©ºÍCH3OH£¨g£©µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ£®
´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬v£¨H2£©=
0.225mol/£¨L?min£©
0.225mol/£¨L?min£©
£»ÇâÆøµÄת»¯ÂÊ=
75%
75%
£»ÄÜʹƽºâÌåϵÖÐn£¨CH3OH£©Ôö´óµÄ´ëÊ©ÓÐ
½µµÍζȣ¨»ò¼Óѹ»òÔö´óH2µÄÁ¿µÈ£©
½µµÍζȣ¨»ò¼Óѹ»òÔö´óH2µÄÁ¿µÈ£©
£®
£¨3£©È罫CO2ÓëH2 ÒÔ1£º4µÄÌå»ý±È»ìºÏ£¬ÔÚÊʵ±µÄÌõ¼þÏ¿ÉÖƵÃCH4£®ÒÑÖª£º
CH4 £¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H1=-890.3kJ/mol
H2£¨g£©+1/2O2£¨g£©¨TH2O£¨l£©¡÷H2=-285.8kJ/mol
ÔòCO2£¨g£©ÓëH2£¨g£©·´Ó¦Éú³ÉCH4£¨g£©ÓëҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽÊÇ
CO2£¨g£©+4H2£¨g£©¨TCH4£¨g£©+2H2O£¨l£©¡÷H=-252.9kJ/mol
CO2£¨g£©+4H2£¨g£©¨TCH4£¨g£©+2H2O£¨l£©¡÷H=-252.9kJ/mol
£®
£¨4£©Ä³Í¬Ñ§ÓóÁµí·¨²â¶¨º¬ÓнϸßŨ¶ÈCO2µÄ¿ÕÆøÖÐCO2µÄº¬Á¿£¬¾­²éµÃһЩÎïÖÊÔÚ20¡æµÄÊý¾ÝÈçÏÂ±í£®
Èܽâ¶È£¨S£©/g ÈܶȻý£¨Ksp£©
Ca£¨OH£©2 Ba£¨OH£©2 CaCO3 BaCO3
0.16 3.89 2.9¡Á10-9 2.6¡Á10-9
£¨ËµÃ÷£ºKspԽС£¬±íʾ¸ÃÎïÖÊÔÚË®ÈÜÒºÖÐÔ½Ò׳Áµí£©
ÎüÊÕCO2×îºÏÊʵÄÊÔ¼ÁÊÇ
Ba£¨OH£©2
Ba£¨OH£©2
£¨Ìî¡°Ca£¨OH£©2¡±»ò¡°Ba£¨OH£©2¡±£©ÈÜÒº£¬ÊµÑéʱ³ýÐèÒª²â¶¨¹¤Òµ·ÏÆøµÄÌå»ý£¨ÕÛËã³É±ê×¼×´¿ö£©Í⣬»¹ÐèÒª²â¶¨
BaCO3µÄÖÊÁ¿
BaCO3µÄÖÊÁ¿
£®
¾«Ó¢¼Ò½ÌÍøΪ¼õСCO2¶Ô»·¾³µÄÓ°Ï죬ÔÚÏÞÖÆÆäÅÅ·ÅÁ¿µÄͬʱ£¬Ó¦¼ÓÇ¿¶ÔCO2´´ÐÂÀûÓõÄÑо¿£®
£¨1£©¢Ù°Ñº¬ÓнϸßŨ¶ÈCO2µÄ¿ÕÆøͨÈë±¥ºÍK2CO3ÈÜÒº£®¢ÚÔÚ¢ÙµÄÎüÊÕÒºÖÐͨ¸ßÎÂË®ÕôÆøµÃµ½¸ßŨ¶ÈµÄCO2ÆøÌ壮д³ö¢ÚÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ
 
£®
£¨2£©È罫CO2ÓëH2 ÒÔ1£º3µÄÌå»ý±È»ìºÏ£®
¢ÙÊʵ±Ìõ¼þϺϳÉijÌþºÍË®£¬¸ÃÌþ¿ÉÄÜÊÇ
 
£¨ÌîÐòºÅ£©£®
A£®ÍéÌþ      B£®Ï©Ìþ     C£®È²Ìþ    D£®±½µÄͬϵÎï
¢ÚÊʵ±Ìõ¼þϺϳÉȼÁϼ״¼ºÍË®£®ÔÚÌå»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë2mol CO2ºÍ6mol H2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2£¨g£©+3H2£¨g£©¨TCH3OH£¨g£©+H2O£¨g£©¡÷H=-49.0kJ/mol£®²âµÃCO2£¨g£©ºÍCH3OH£¨g£©µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ£®´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬ÇâÆøµÄ·´Ó¦ËÙÂÊv£¨H2£©=
 
£»ÇâÆøµÄת»¯ÂÊ=
 
£»ÄÜʹƽºâÌåϵÖÐn£¨CH3OH£©Ôö´óµÄ´ëÊ©ÓÐ
 
£®£¨ÈÎдÁ½ÖÖ¼´¿É£©
£¨3£©È罫CO2ÓëH2 ÒÔ1£º4µÄÌå»ý±È»ìºÏ£¬ÔÚÊʵ±µÄÌõ¼þÏ¿ÉÖƵÃCH4£®
ÒÑÖª£ºCH4£¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H1=-890.3kJ/mol
H2£¨g£©+
1
2
O2£¨g£©¨TH2O£¨l£©¡÷H2=-285.8kJ/molÔòCO2£¨g£©ÓëH2£¨g£©·´Ó¦Éú³ÉCH4£¨g£©ÓëҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽÊÇ
 
£®
£¨4£©×î½ü¿Æѧ¼ÒÌá³ö¡°ÂÌÉ«×ÔÓÉ¡±¹¹Ï룺°Ñ¿ÕÆø´µÈë̼Ëá¼ØÈÜÒº£¬È»ºóÔÙ°ÑCO2´ÓÈÜÒºÖÐÌáÈ¡³öÀ´£¬¾­»¯Ñ§·´Ó¦ºóʹ¿ÕÆøÖеÄCO2ת±äΪ¿ÉÔÙÉúȼÁϼ״¼£®¼×´¼¿ÉÖÆ×÷ȼÁϵç³Ø£®Ð´³öÒÔÇâÑõ»¯¼ØΪµç½âÖʵļ״¼È¼Áϵç³Ø¸º¼«·´Ó¦Ê½
 
£®µ±µç×ÓתÒƵÄÎïÖʵÄÁ¿Îª
 
ʱ£¬²Î¼Ó·´Ó¦µÄÑõÆøµÄÌå»ýÊÇ6.72L£¨±ê×¼×´¿öÏ£©£®
£¨5£©Ä³Í¬Ñ§ÓóÁµí·¨²â¶¨º¬ÓнϸßŨ¶ÈCO2µÄ¿ÕÆøÖÐCO2µÄº¬Á¿£¬¾­²éµÃһЩÎïÖÊÔÚ20¡æµÄÊý¾ÝÈçÏÂ±í£®
Èܽâ¶È£¨S£©/g ÈܶȻý£¨Ksp£©
Ca£¨OH£©2 Ba£¨OH£©2 CaCO3 BaCO3
0.16 3.89 2.9¡Á10-9 2.6¡Á10-9
ÎüÊÕCO2×îºÏÊʵÄÊÔ¼ÁÊÇ
 
[Ìî¡°Ca£¨OH£©2¡±»ò¡°Ba£¨OH£©2¡±]ÈÜÒº£¬ÊµÑéʱ³ýÐèÒª²â¶¨¹¤Òµ·ÏÆøµÄÌå»ý£¨ÕÛËã³É±ê×¼×´¿ö£©Í⣬»¹ÐèÒª²â¶¨
 
£®

Ϊ¼õСCO2¶Ô»·¾³µÄÓ°Ï죬ÔÚÏÞÖÆÆäÅÅ·ÅÁ¿µÄͬʱ£¬Ó¦¼ÓÇ¿¶ÔCO2´´ÐÂÀûÓõÄÑо¿¡£

(1)¢Ù °Ñº¬ÓнϸßŨ¶ÈCO2µÄ¿ÕÆøͨÈë±¥ºÍK2CO3ÈÜÒº¡£

¢Ú ÔÚ¢ÙµÄÎüÊÕÒºÖÐͨ¸ßÎÂË®ÕôÆøµÃµ½¸ßŨ¶ÈµÄCO2ÆøÌå¡£

д³ö¢ÚÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ________¡£

(2)È罫CO2ÓëH2 ÒÔ1:3µÄÌå»ý±È»ìºÏ¡£

¢ÙÊʵ±Ìõ¼þϺϳÉijÌþºÍË®£¬¸ÃÌþÊÇ_______(ÌîÐòºÅ)¡£

A£®ÍéÌþ     B£®Ï©Ìþ      C£®È²Ìþ      D£®±½µÄͬϵÎï

¢Ú Êʵ±Ìõ¼þϺϳÉȼÁϼ״¼ºÍË®¡£ÔÚÌå»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë2 mol CO2ºÍ6 mol H2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2(g)£«3H2(g)CH3OH(g)£«H2O(g) ¡÷H£½£­49.0 kJ/mol¡£

²âµÃCO2(g)ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ¡£

´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬v(H2)£½______£»ÇâÆøµÄת»¯ÂÊ£½_______£»ÄÜʹƽºâÌåϵÖÐn(CH3OH)Ôö´óµÄ´ëÊ©ÓÐ______¡£

(3)È罫CO2ÓëH2 ÒÔ1:4µÄÌå»ý±È»ìºÏ£¬ÔÚÊʵ±µÄÌõ¼þÏ¿ÉÖƵÃCH4¡£

ÒÑÖª£º

CH4 (g) + 2O2(g)  CO2(g)+ 2H2O(l)   ¦¤H1£½¨D 890.3 kJ/mol

H2(g) + 1/2O2(g)  H2O(l)            ¦¤H2£½£­285.8 kJ/mol

ÔòCO2(g)ÓëH2(g)·´Ó¦Éú³ÉCH4(g)ÓëҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽÊÇ________¡£

(4)ijͬѧÓóÁµí·¨²â¶¨º¬ÓнϸßŨ¶ÈCO2µÄ¿ÕÆøÖÐCO2µÄº¬Á¿£¬¾­²éµÃһЩÎïÖÊÔÚ20¡æµÄÊý¾ÝÈçÏÂ±í¡£

Èܽâ¶È(S)/g

ÈܶȻý(Ksp)

Ca(OH)2

Ba(OH)2

CaCO3

BaCO3

0.16

3.89

2.9¡Á10-9

2.6¡Á10-9

 (˵Ã÷£ºKspԽС£¬±íʾ¸ÃÎïÖÊÔÚË®ÈÜÒºÖÐÔ½Ò׳Áµí)

ÎüÊÕCO2×îºÏÊʵÄÊÔ¼ÁÊÇ_________[Ìî¡°Ca(OH)2¡±»ò¡°Ba(OH)2¡±]ÈÜÒº£¬ÊµÑéʱ³ýÐèÒª²â¶¨¹¤Òµ·ÏÆøµÄÌå»ý(ÕÛËã³É±ê×¼×´¿ö)Í⣬»¹ÐèÒª²â¶¨__________¡£

 

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

¾«Ó¢¼Ò½ÌÍø