ÌâÄ¿ÄÚÈÝ

ÔÚÒ»ºãκãÈÝÃܱÕÈÝÆ÷ÖУ¬A¡¢BÆøÌå¿É½¨Á¢ÈçÏÂÆ½ºâ£º

2A£¨g£©+2B£¨g£©  C£¨g£©+3D£¨g£©ÏÖ·Ö±ð´ÓÁ½Ìõ;¾¶½¨Á¢Æ½ºâ£º

I£®A¡¢BµÄÆðʼÁ¿¾ùΪ2mol£»

¢ò£®C¡¢DµÄÆðʼÁ¿·Ö±ðΪ2molºÍ6mol¡£

ÏÂÁÐÐðÊö²»ÕýÈ·µÄÊÇ£º     £¨   £©

A£®I¡¢¢òÁ½Í¾¾¶×îÖմﵽƽºâʱ£¬ÌåϵÄÚ»ìºÏÆøÌåµÄ°Ù·Ö×é³ÉÏàͬ

B£®I¡¢¢òÁ½Í¾¾¶×îÖմﵽƽºâʱ£¬ÌåϵÄÚ»ìºÏÆøÌåµÄ°Ù·Ö×é³É²»Í¬

C£®´ïµ½Æ½ºâʱ£¬Í¾¾¶IµÄ ºÍ;¾¶¢òÌåϵÄÚ»ìºÏÆøÌ寽¾ùÏà¶Ô·Ö×ÓÖÊÁ¿Ïàͬ

D£®´ïµ½Æ½ºâʱ£¬Í¾¾¶IµÄÆøÌåÃܶÈΪ;¾¶¢òÃܶȵÄ1/2

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
»ðÁ¦·¢µç³§²úÉú´óÁ¿µÄµªÑõ»¯ÎNOx£©¡¢¶þÑõ»¯ÁòºÍ¶þÑõ»¯Ì¼µÈÆøÌå»áÔì³É»·¾³ÎÛȾ£®¶Ôȼú·ÏÆø½øÐÐÍÑÏõ¡¢ÍÑÁòºÍÍÑ̼µÈ´¦Àí£¬¿ÉʵÏÖÂÌÉ«»·±£¡¢½ÚÄܼõÅÅ¡¢·ÏÎïÀûÓõÈÄ¿µÄ£®
£¨1£©ÍÑÏõ£®ÀûÓü×Íé´ß»¯»¹Ô­NOx£º
CH4£¨g£©+4NO2£¨g£©=4NO£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H1=-574kJ?mol-1
CH4£¨g£©+4NO£¨g£©=2N2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H2=-1160kJ?mol-1
¼×ÍéÖ±½Ó½«NO2»¹Ô­ÎªN2µÄÈÈ»¯Ñ§·½³ÌʽΪ
CH4£¨g£©+2NO2£¨g£©=N2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-867kJ/mol
CH4£¨g£©+2NO2£¨g£©=N2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-867kJ/mol
£®
£¨2£©ÍÑ̼£®½«CO2ת»¯Îª¼×´¼µÄ»¯Ñ§·½³ÌʽΪ£º
CO2+3H2CH3OH£¨g£©+H2O£¨g£©
¢ÙÈ¡Îå·ÝµÈÌå»ýCO2ºÍH2µÄ»ìºÏÆøÌ壨ÎïÖʵÄÁ¿Ö®±È¾ùΪ1£º3£©£¬·Ö±ð¼ÓÈëζȲ»Í¬¡¢ÈÝ»ýÏàͬµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÉÏÊö·´Ó¦£¬tminºó£¬²âµÃ¼×´¼µÄÌå»ý·ÖÊý¦Õ£¨CH3OH£© Ó뷴ӦζÈTµÄ¹ØÏµÇúÏßÈçͼËùʾ£¬ÔòÉÏÊöºÏ³É¼×´¼µÄ·´Ó¦µÄÄæ·´Ó¦Îª
ÎüÈÈ
ÎüÈÈ
·´Ó¦£¨Ìî¡°·ÅÈÈ¡±»ò¡°ÎüÈÈ¡±£©£®
¢ÚÔÚÒ»ºãκãÈÝÃܱÕÈÝÆ÷ÖгäÈë1mol CO2ºÍ3mol H2£¬½øÐÐÉÏÊö·´Ó¦£®²âµÃCO2ºÍCH3OH£¨g£©µÄŨ¶ÈËæÊ±¼ä±ä»¯ÈçͼËùʾ£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
CE
CE

A£®10minºó£¬Ïò¸ÃÈÝÆ÷ÖÐÔÙ³äÈë1mol CO2ºÍ3mol H2£¬ÔòÔٴδﵽƽºâʱc£¨CH3OH£©=1.5mol?L-1
B£®0¡«10minÄÚ£¬ÇâÆøµÄƽ¾ù·´Ó¦ËÙÂÊΪ0.075mol?L-1?min-1
C£®´ïµ½Æ½ºâʱ£¬ÇâÆøµÄת»¯ÂÊΪ75%
D£®¸ÃζÈÏ£¬·´Ó¦µÄƽºâ³£ÊýµÄֵΪ3/16
E£®Éý¸ßζȽ«Ê¹n£¨CH3OH£©/n£¨CO2£©¼õС
¢ÛÄÜ˵Ã÷CO2£¨g£©+3H2£¨g£©CH3OH£¨g£©+H2O£¨g£©´ïµ½Æ½ºâ״̬µÄÊÇ
C
C

A£®Éú³É¼×´¼µÄËÙÂÊÓëÉú³ÉË®µÄËÙÂÊÏàµÈ
B£®v£¨H2£©=3v£¨CH3OH£©
C£®ºãÈÝÈÝÆ÷ÖУ¬ÌåϵµÄѹǿ²»Ôٸıä
D£®ºãÈÝÈÝÆ÷ÖУ¬»ìºÏÆøÌåÃܶȲ»ÔÙ·¢Éú¸Ä±ä
¢ÜÓÒͼ±íʾ·´Ó¦CO2£¨g£©+3H2£¨g£©CH3OH£¨g£©+H2O£¨g£©½øÐйý³ÌÖÐÄÜÁ¿£¨µ¥Î»ÎªkJ?mol-1£©µÄ±ä»¯£®ÔÚÌå»ýΪ1LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬³äÈë1mol CO2ºÍ3mol H2£¬´ïµ½Æ½ºâºó£¬²ÉÈ¡ÏÂÁдëÊ©ÖÐÄÜʹc£¨CH3OH£©Ôö´óµÄÊÇ
CD
CD
£®
A£®Ê¹ÓÃиßЧ´ß»¯¼Á              B£®³äÈëHe£¨g£©£¬Ê¹ÌåϵѹǿÔö´ó
C£®½«H2O£¨g£©´ÓÌåϵÖзÖÀë³öÀ´   D£®ÔÙ³äÈë1mol CO2ºÍ3mol H2
¢Ý¼×´¼È¼ÁÏµç³Ø½á¹¹ÈçͼËùʾ£¬Æä¹¤×÷ʱÕý¼«µÄµç¼«·´Ó¦Ê½¿É±íʾΪ
O2+4e-+4H+=2H2O
O2+4e-+4H+=2H2O
£®
£¨3£©ÍÑÁò£®Ä³ÖÖÍÑÁò¹¤ÒÕÖн«·ÏÆø¾­´¦Àíºó£¬ÓëÒ»¶¨Á¿µÄ°±Æø¡¢¿ÕÆø·´Ó¦£¬Éú³ÉÁòËá狀ÍÏõËá淋ĻìºÏÎï×÷Ϊ¸±²úÆ·»¯·Ê£®  
ÁòËáï§ÈÜÒºµÄpH£¼7£¬ÆäÔ­ÒòΪ£º
NH4++H2ONH3?H2O+H+
NH4++H2ONH3?H2O+H+
£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£»ÔÚÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÏõËáï§ÈÜÒºÖеμÓÊÊÁ¿µÄNaOHÈÜÒº£¬Ê¹ÈÜÒºµÄpH=7£¬ÔòÈÜÒºÖÐc£¨Na+£©
СÓÚ
СÓÚ
c£¨NO3-£©£¨Ñ¡Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©
£¨2012?¿ª·â¶þÄ££©ÓÉÓÚÎÂÊÒЧӦºÍ×ÊÔ´¶ÌȱµÈÎÊÌ⣬ÈçºÎ½µµÍ´óÆøÖеÄCO2º¬Á¿²¢¼ÓÒÔ¿ª·¢ÀûÓã¬ÒýÆðÁ˸÷¹úµÄÆÕ±éÖØÊÓ£®
£¨1£©Ä¿Ç°£¬Óó¬ÁÙ½çCO2£¨Æä״̬½éÓÚÆøÌ¬ºÍҺ̬֮¼ä£©´úÌæ·úÀû°º×÷Àä¼ÁÒѳÉΪһÖÖÇ÷ÊÆ£¬ÕâÒ»×ö·¨¶Ô»·¾³µÄ»ý¼«ÒâÒåÔÚÓÚ
±£»¤³ôÑõ²ã
±£»¤³ôÑõ²ã
£®
£¨2£©½«CO2ת»¯³ÉÓлúÎï¿ÉÓÐЧʵÏÖ̼ѭ»·£®CO2ת»¯³ÉÓлúÎïµÄÀý×Ӻܶ࣬È磺
a.6CO2+6H2O
¹âºÏ
×÷ÓÃ
C6H12O6+6O2      b£®CO2+3H2
´ß»¯¼Á
¡÷
CH3OH+H2O
c£®CO2+CH4
´ß»¯¼Á
¡÷
CH3COOH          d.2CO2+6H2
´ß»¯¼Á
¡÷
CH2=CH2+4H2O
ÒÔÉÏ·´Ó¦ÖУ¬×î½ÚÄܵÄÊÇ
a
a
£¬Ô­×ÓÀûÓÃÂÊ×î¸ßµÄÊÇ
c
c
£®
£¨3£©ÈôÓÐ4.4kg CO2Óë×ãÁ¿H2Ç¡ºÃÍêÈ«·´Ó¦£¬Éú³ÉÆøÌ¬µÄË®ºÍ¼×´¼£¬¿É·Å³ö4947kJµÄÈÈÁ¿£¬ÊÔд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
CO2£¨g£©+3H2£¨g£©¨TCH3OH£¨g£©+H2O£¨g£©¡÷H=-49.47kJ/mol
CO2£¨g£©+3H2£¨g£©¨TCH3OH£¨g£©+H2O£¨g£©¡÷H=-49.47kJ/mol
£®
£¨4£©ÎªÌ½¾¿ÓÃCO2À´Éú²úȼÁϼ״¼µÄ·´Ó¦Ô­Àí£¬ÏÖ½øÐÐÈçÏÂʵÑ飺ÔÚÒ»ºãκãÈÝÃܱÕÈÝÆ÷£¬³äÈë1mol CO2ºÍ3molH2£¬½øÐз´Ó¦£®²âµÃCO2ºÍCH3OH£¨g£©µÄŨ¶ÈËæÊ±¼ä±ä»¯ÈçͼËùʾ£®´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâv£¨H2£©=
0.225 mol?L-1?min-1
0.225 mol?L-1?min-1
£»CO2µÄת»¯ÂÊ=
75%
75%
£»¸ÃζÈÏÂµÄÆ½ºâ³£ÊýÊýÖµ=
5.33
5.33
£®ÄÜʹƽºâÌåϵÖÐn£¨CH3OH£©/n£¨CO2£©Ôö´óµÄ´ëÊ©ÓÐ
½«H2O£¨g£©´ÓÌåϵÖзÖÀë
½«H2O£¨g£©´ÓÌåϵÖзÖÀë
 £¨ÈÎдһÌõ£©£®
£¨5£©CO2ÔÚ×ÔÈ»½çÑ­»·Ê±¿ÉÓëCaCO3·´Ó¦£¬CaCO3ÊÇÒ»ÖÖÄÑÈÜÎïÖÊ£¬ÆäKsp=2.8¡Á10-9£®CaCl2ÈÜÒºÓëNa2CO3ÈÜÒº»ìºÏ¿ÉÐγÉCaCO3³Áµí£¬ÏÖ½«µÈÌå»ýµÄCaCl2ÈÜÒºÓëNa2CO3ÈÜÒº»ìºÏ£¬ÈôNa2CO3ÈÜÒºµÄŨ¶ÈΪ4¡Á10-4mol/L£¬ÔòÉú³É³ÁµíËùÐèCaCl2ÈÜÒºµÄ×îСŨ¶ÈΪ
2.8¡Á10-5mol/L
2.8¡Á10-5mol/L
£®
»ðÁ¦·¢µç³§Êͷųö´óÁ¿µÄµªÑõ»¯ÎNOx£©¡¢¶þÑõ»¯ÁòºÍ¶þÑõ»¯Ì¼µÈÆøÌå»áÔì³É»·¾³ÎÛȾ£®¶Ôȼú·ÏÆø½øÐÐÍÑ̼ºÍÍÑÁòµÈ´¦Àí£¬¿ÉʵÏÖÂÌÉ«»·±£¡¢½ÚÄܼõÅÅ¡¢·ÏÎïÀûÓõÈÄ¿µÄ£®
£¨1£©ÍÑ̼£º½«CO2ת»¯Îª¼×´¼µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCO2£¨g£©+3H2£¨g£©?CH3OH£¨g£©+H2O£¨g£©¡÷H3
¢ÙÈ¡Îå·ÝµÈÌå»ýµÄCO2ºÍH2µÄ»ìºÏÆøÌ壨ÎïÖʵÄÁ¿Ö®±È¾ùΪ1£º3£©£¬·Ö±ð¼ÓÈëζȲ»Í¬¡¢ÈÝ»ýÏàͬµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÉÏÊö·´Ó¦£¬·´Ó¦Ïàͬʱ¼äºó£¬²âµÃ¼×´¼µÄÌå»ý·ÖÊý¦Õ£¨CH3OH£© Ó뷴ӦζÈTµÄ¹ØÏµÇúÏßÈçͼ¢ñËùʾ£¬ÔòÉÏÊöCO2ת»¯Îª¼×´¼µÄ·´Ó¦µÄ¡÷H3
£¼
£¼
0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©

¢ÚÈçͼ¢òÊÇÔÚºãÎÂÃܱÕÈÝÆ÷ÖУ¬Ñ¹Ç¿ÎªP1ʱH2µÄÌå»ý·ÖÊýËæÊ±¼ätµÄ±ä»¯ÇúÏߣ¬ÇëÔÚ¸ÃͼÖл­³ö¸Ã·´Ó¦ÔÚP2£¨P2£¾P1£©Ê±µÄH2Ìå»ý·ÖÊýËæÊ±¼ätµÄ±ä»¯ÇúÏß
£®
£¨2£©ÔÚÒ»ºãκãÈÝÃܱÕÈÝÆ÷ÖгäÈë1mol CO2ºÍ3mol H2£¬½øÐÐÉÏÊö·´Ó¦£®²âµÃCO2ºÍCH3OH£¨g£©µÄÎïÖʵÄÁ¿Å¨¶ÈËæÊ±¼ä±ä»¯Èçͼ¢óËùʾ£®
ÊԻشð£º0¡«10minÄÚ£¬ÇâÆøµÄƽ¾ù·´Ó¦ËÙÂÊΪ
0.225mol/£¨L?min£©
0.225mol/£¨L?min£©
£»¸ÃζÈÏ£¬·´Ó¦µÄƽºâ³£ÊýΪ
5.33
5.33
£»£¨±£ÁôÈý¸öÓÐЧÊý×Ö£©µÚ10minºó£¬Ïò¸ÃÈÝÆ÷ÖÐÔÙ³äÈë1mol CO2ºÍ3mol H2£¬ÔòÔٴδﵽƽºâʱCH3OH£¨g£©µÄÌå»ý·ÖÊý
±ä´ó
±ä´ó
£¨Ìî±ä´ó¡¢¼õÉÙ¡¢²»±ä£©£®
£¨3£©ÍÑÁò£ºÄ³ÖÖÍÑÁò¹¤ÒÕÖн«·ÏÆø¾­´¦Àíºó£¬ÓëÒ»¶¨Á¿µÄ°±Æø¡¢¿ÕÆø·´Ó¦£¬Éú³ÉÁòËá狀ÍÏõËá淋ĻìºÏÎï×÷Ϊ¸±²úÆ·»¯·Ê£®ÅäÖÆ100mlÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÁòËáï§ÈÜÒºËùÐè²£Á§ÒÇÆ÷³ýÉÕ±­¡¢½ºÍ·µÎ¹ÜÍ⻹Ð裺
²£Á§°ô¡¢100mLÈÝÁ¿Æ¿
²£Á§°ô¡¢100mLÈÝÁ¿Æ¿
£®
£¨2011?Ìì½òÄ£Ä⣩»ðÁ¦·¢µç³§Êͷųö´óÁ¿µÄµªÑõ»¯ÎNOx£©¡¢¶þÑõ»¯ÁòºÍ¶þÑõ»¯Ì¼µÈÆøÌå»áÔì³É»·¾³ÎÛȾ£®¶Ôȼú·ÏÆø½øÐÐÍÑÏõ¡¢ÍÑÁòºÍÍÑ̼µÈ´¦Àí£¬¿ÉʵÏÖÂÌÉ«»·±£¡¢½ÚÄܼõÅÅ¡¢·ÏÎïÀûÓõÈÄ¿µÄ£®
£¨1£©ÍÑÏõ£®ÀûÓü×Íé´ß»¯»¹Ô­NOx£º
CH4£¨g£©+4NO2£¨g£©=4NO£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H1=-574kJ?mol-1
CH4£¨g£©+4NO£¨g£©=2N2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H2=-1160kJ?mol-1
¼×ÍéÖ±½Ó½«NO2»¹Ô­ÎªN2µÄÈÈ»¯Ñ§·½³ÌʽΪ
CH4£¨g£©+2NO2£¨g£©=N2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-867kJ/mol
CH4£¨g£©+2NO2£¨g£©=N2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-867kJ/mol
£®
£¨2£©ÍÑ̼£®½«CO2ת»¯Îª¼×´¼µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCO2£¨g£©+3H2£¨g£©¨TCH3OH£¨g£©+H2O£¨g£©¡÷H3

¢ÙÈ¡Îå·ÝµÈÌå»ýµÄCO2ºÍH2µÄ»ìºÏÆøÌ壨ÎïÖʵÄÁ¿Ö®±È¾ùΪ1£º3£©£¬·Ö±ð¼ÓÈëζȲ»Í¬¡¢ÈÝ»ýÏàͬµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÉÏÊö·´Ó¦£¬·´Ó¦Ïàͬʱ¼äºó£¬²âµÃ¼×´¼µÄÌå»ý·ÖÊý¦Õ£¨CH3OH£© Ó뷴ӦζÈTµÄ¹ØÏµÇúÏßÈçͼ1Ëùʾ£¬ÔòÉÏÊöCO2ת»¯Îª¼×´¼µÄ·´Ó¦µÄ¡÷H3
£¼
£¼
0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®
¢ÚÔÚÒ»ºãκãÈÝÃܱÕÈÝÆ÷ÖгäÈë1mol CO2ºÍ3mol H2£¬½øÐÐÉÏÊö·´Ó¦£®²âµÃCO2ºÍCH3OH£¨g£©µÄŨ¶ÈËæÊ±¼ä±ä»¯Èçͼ2Ëùʾ£®ÊԻشð£º0¡«10minÄÚ£¬ÇâÆøµÄƽ¾ù·´Ó¦ËÙÂÊΪ
0.225
0.225
mol/£¨L?min£©£»¸ÃζÈÏ£¬·´Ó¦µÄƽºâ³£ÊýµÄֵΪ
16
3
16
3
£»µÚ10minºó£¬Ïò¸ÃÈÝÆ÷ÖÐÔÙ³äÈë1mol CO2ºÍ3mol H2£¬ÔòÔٴδﵽƽºâʱCH3OH£¨g£©µÄÌå»ý·ÖÊý
±ä´ó
±ä´ó
£¨Ìî±ä´ó¡¢¼õÉÙ¡¢²»±ä£©£®
£¨3£©ÍÑÁò£®Ä³ÖÖÍÑÁò¹¤ÒÕÖн«·ÏÆø¾­´¦Àíºó£¬ÓëÒ»¶¨Á¿µÄ°±Æø¡¢¿ÕÆø·´Ó¦£¬Éú³ÉÁòËá狀ÍÏõËá淋ĻìºÏÎï×÷Ϊ¸±²úÆ·»¯·Ê£®ÁòËá狀ÍÏõËáï§µÄË®ÈÜÒºµÄpH£¼7£¬ÆäÔ­ÒòÓÃÀë×Ó·½³Ìʽ±íʾΪ
NH4++H2O?NH3?H2O+H+
NH4++H2O?NH3?H2O+H+
£»ÔÚÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÏõËáï§ÈÜÒºÖеμÓÊÊÁ¿µÄNaOHÈÜÒº£¬Ê¹ÈÜÒºµÄpH=7£¬ÔòÈÜÒºÖÐc£¨Na+£©+c£¨H+£©
£¼
£¼
c£¨NO3-£©+c£¨OH-£©£¨Ìîд¡°£¾¡±¡°=¡±»ò¡°£¼¡±£©
»ðÁ¦·¢µç³§Êͷųö´óÁ¿µÄµªÑõ»¯ÎNOx£©¡¢¶þÑõ»¯ÁòºÍ¶þÑõ»¯Ì¼µÈÆøÌå»áÔì³É»·¾³ÎÛȾ£®¶Ôȼú·ÏÆø½øÐÐÍÑÏõ¡¢ÍÑÁòºÍÍÑ̼µÈ´¦Àí£¬¿ÉʵÏÖÂÌÉ«»·±£¡¢½ÚÄܼõÅÅ¡¢·ÏÎïÀûÓõÈÄ¿µÄ£®¾«Ó¢¼Ò½ÌÍø
£¨1£©ÍÑÏõ£®ÀûÓü×Íé´ß»¯»¹Ô­NOx£º
CH4£¨g£©+4NO2£¨g£©=4NO£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H1=-574kJ?mol-1
CH4£¨g£©+4NO£¨g£©=2N2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H2=-1160kJ?mol-1
¼×ÍéÖ±½Ó½«NO2»¹Ô­ÎªN2µÄÈÈ»¯Ñ§·½³ÌʽΪ
 
£®
£¨2£©ÍÑ̼£®½«CO2ת»¯Îª¼×´¼µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCO2£¨g£©+3H2£¨g£©?CH3OH£¨g£©+H2O£¨g£©£»¡÷H3£¼0
¢ÙÔÚÒ»ºãκãÈÝÃܱÕÈÝÆ÷ÖгäÈë1mol CO2ºÍ3mol H2£¬½øÐÐÉÏÊö·´Ó¦£®²âµÃCO2ºÍCH3OH£¨g£©µÄŨ¶ÈËæÊ±¼ä±ä»¯Èçͼ1Ëùʾ£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
 
£¨Ìî×Öĸ´úºÅ£©£®
A£®µÚ10minºó£¬Ïò¸ÃÈÝÆ÷ÖÐÔÙ³äÈë1mol CO2ºÍ3mol H2£¬ÔòÔٴδﵽƽºâʱc£¨CH3OH£©=1.5mol?L-1
B£®0¡«10minÄÚ£¬ÇâÆøµÄƽ¾ù·´Ó¦ËÙÂÊΪ0.075mol/£¨L?min£©
C£®´ïµ½Æ½ºâʱ£¬ÇâÆøµÄת»¯ÂÊΪ75%
D£®¸ÃζÈÏ£¬·´Ó¦µÄƽºâ³£ÊýµÄֵΪ
3
16

E£®Éý¸ßζȽ«Ê¹
n(CH3OH)
n(CO2)
¼õС
¢Ú¼×´¼È¼ÁÏµç³Ø½á¹¹Èçͼ2Ëùʾ£®Æä¹¤×÷ʱÕý¼«µÄµç¼«·´Ó¦Ê½¿É±íʾΪ£º
 
£®
£¨3£©ÍÑÁò£®Ä³ÖÖÍÑÁò¹¤ÒÕÖн«·ÏÆø¾­´¦Àíºó£¬ÓëÒ»¶¨Á¿µÄ°±Æø¡¢¿ÕÆø·´Ó¦£¬Éú³ÉÁòËá狀ÍÏõËá淋ĻìºÏÎï×÷Ϊ¸±²úÆ·»¯·Ê£®ÁòËá狀ÍÏõËáï§µÄË®ÈÜÒºµÄpH£¼7£¬ÆäÖÐÔ­Òò¿ÉÓÃÒ»¸öÀë×Ó·½³Ìʽ±íΪ£º
 
£»ÔÚÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÏõËáï§ÈÜÒºÖеμÓÊÊÁ¿µÄNaOHÈÜÒº£¬Ê¹ÈÜÒºµÄpH=7£¬ÔòÈÜÒºÖУºc£¨Na+£©+c£¨H+£©
 
c£¨NO3-£©+c£¨OH-£©£¨Ìîд¡°£¾¡±¡°=¡±»ò¡°£¼¡±£©£®

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

¾«Ó¢¼Ò½ÌÍø