¿ÎÍ⻯ѧÐËȤС×éµÄͬѧÀûÓÃij»¯¹¤³§µÄ·Ï¼îÒº£¨Ö÷Òª³É·ÖΪNa2CO3£¬»¹º¬ÓÐÉÙÁ¿NaCl£¬ÆäËüÔÓÖʲ»¼Æ£©ºÍʯ»ÒÈéΪԭÁÏÖƱ¸Éռ²¢¶ÔËùµÃµÄÉÕ¼î´Ö²úÆ·µÄ³É·Ö½øÐзÖÎöºÍ²â¶¨£®
£¨Ò»£©´Ö²úÆ·ÖƱ¸£º
£¨1£©½«·Ï¼îÒº¼ÓÈÈÕô·¢Å¨Ëõ£¬ÐγɽÏŨµÄÈÜÒº£¬ÀäÈ´ºóÓëʯ»ÒÈé»ìºÏ£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£®
£¨2£©½«·´Ó¦ºóµÄ»ìºÏÎï¹ýÂË£¬µÃµ½µÄÂËÒº½øÐÐÕô·¢½á¾§£¬ÖƵôֲúÆ·£®
£¨¶þ£©´Ö²úÆ·³É·Ö·ÖÎö£º
£¨1£©È¡ÊÊÁ¿´Ö²úÆ·ÈÜÓÚË®£¬µÎ¼ÓBa£¨NO3£©2ÈÜÒº³öÏÖ°×É«»ë×Ç£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£¬¸Ã´Ö²úÆ·ÖÐÒ»¶¨²»º¬ÓУ¬
ÀíÓÉÊÇ£®
£¨2£©¸ÃС×éͬѧͨ¹ý¶Ô´Ö²úÆ·³É·ÖµÄʵÑé·ÖÎö£¬È·¶¨¸Ã´Ö²úÆ·Öк¬ÓÐÈýÖÖÎïÖÊ£®
£¨Èý£©´Ö²úÆ·º¬Á¿²â¶¨£º
[Na2CO3º¬Á¿µÄ²â¶¨]
£¨1£©¸ÃÐËȤС×éµÄͬѧÉè¼ÆÁËÏÂͼËùʾµÄʵÑé×°Öã®È¡10.0g´Ö²úÆ·£¬½øÐÐʵÑé
˵Ã÷£º¼îʯ»ÒÊÇCaOÓëNaOHµÄ¹ÌÌå»ìºÏÎE×°ÖÃÖеı¥ºÍNaHCO3ÈÜÒºÊÇΪÁ˳ýÈ¥¶þÑõ»¯Ì¼ÆøÌåÖеÄÂÈ»¯Ç⣬·¢ÉúµÄ·´Ó¦Îª NaHCO3Ê®HCl=NaClÊ®CO2¡üÊ®H2O£®
£¨2£©²Ù×÷²½Öè
¢ÙÁ¬½ÓºÃ×°Ö㬼ì²éÆøÃÜÐÔ£»¢Ú´ò¿ªµ¯»É¼ÐC£¬ÔÚA´¦»º»ºÍ¨ÈëÒ»¶Îʱ¼ä¿ÕÆø£»¢Û³ÆÁ¿GµÄÖÊÁ¿£»¢Ü¹Ø±Õµ¯»É¼ÐC£¬ÂýÂýµÎ¼ÓŨÑÎËáÖÁ¹ýÁ¿£¬Ö±ÖÁDÖÐÎÞÆøÅÝð³ö£»¢Ý´ò¿ªµ¯»É¼ÐC£¬Ôٴλº»ºÍ¨ÈëÒ»¶Îʱ¼ä¿ÕÆø£»¢ÞÔٴγÆÁ¿GµÄÖÊÁ¿£¬µÃÇ°ºóÁ½´ÎÖÊÁ¿²îΪ0.48g£®
£¨3£©ÎÊÌâ̽¾¿
FÖеÄÊÔ¼ÁӦΪ£¬B×°ÖõÄ×÷ÓÃÊÇ£¬
H×°ÖõÄ×÷ÓÃÊÇ£®
ÈôûÓÐH×°Öã¬Ôò²â¶¨µÄNa2CO3µÄÖÊÁ¿·ÖÊý»á£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±¡¢¡°²»±ä¡±£©£®
ÊÂʵÉÏ10.0g´Ö²úÆ·Ö»ÄܲúÉú0.44g CO2£®ÇëÄã×Ðϸ·ÖÎöÉÏÊöʵÑ飬½âÊÍʵÑéÖµ0.48g±ÈÕýÈ·Öµ0.44gÆ«´óµÄÔ­Òò£¨¼ÙÉè²Ù×÷¾ùÕýÈ·£©£®
£¨4£©Êý¾Ý¼ÆËã
¸ù¾ÝÕýÈ·Öµ0.44g¿ÉÇóµÃ´Ö²úÆ·ÖÐNa2CO3µÄÖÊÁ¿·ÖÊýΪ%£®
[NaOHº¬Á¿µÄ²â¶¨]
¸ÃС×éͬѧÓÖÁíÈ¡10.0g´Ö²úÆ·£¬ÖðµÎ¼ÓÈë20%µÄÑÎËáÖÁÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ÏûºÄÑÎËáµÄÖÊÁ¿Îª36.5g£¬·Å³öCO20.44g£¨²»¿¼ÂǶþÑõ»¯Ì¼ÆøÌåµÄÈܽ⣩£®ÇóÔ­´Ö²úÆ·ÖÐNaOHµÄÖÊÁ¿·ÖÊý£®£¨Ð´³ö¼ÆËã¹ý³Ì£©
¡¾´ð°¸¡¿·ÖÎö£ºÌ¼ËáÄÆÓëÇâÑõ»¯¸Æ·´Ó¦Éú³É̼Ëá¸Æ³ÁµíºÍÇâÑõ»¯ÄÆ£®ÎÒÃÇ¿ÉÀûÓÃ̼ËáÄƵÄÐÔÖÊÀ´È·¶¨Ì¼ËáÄÆÊÇ·ñÍêÈ«·´Ó¦£®
½â´ð£º½â£º£¨Ò»£©´Ö²úÆ·ÖƱ¸£º
£¨1£©Òò·Ï¼îÒºÖк¬ÓÐ̼ËáÄÆËùÒÔ£¬¼ÓÇâÑõ»¯¸Æʱ£¬Ì¼ËáÄÆ»áÓëÇâÑõ»¯¸Æ·´Ó¦Éú³É³Áµí£®
£¨¶þ£©´Ö²úÆ·³É·Ö·ÖÎö£º
£¨1£©µ±¼ÓÏõËá±µ³öÏÖ³Áµí˵Ã÷£¬´Ö²úÆ·Öк¬ÓÐ̼ËáÄÆ£¬Òòº¬ÓÐ̼ËáÄÆÄÜÓëÇâÑõ»¯¸Æ·´Ó¦£¬ËùÒÔÒ»¶¨²»º¬ÇâÑõ»¯¸Æ£®
£¨Èý£©´Ö²úÆ·º¬Á¿²â¶¨£º
[Na2CO3º¬Á¿µÄ²â¶¨]
£¨3£©ÀûÓóýÈ¥¶þÑõ»¯Ì¼ÇÒ¸ÉÔïµÄ¿ÕÆø½«×°ÖÃÄڵĿÕËùÅž»£®È»ºóÀûÓÃŨÑÎËáÓë´Ö²úÆ··´Ó¦²úÉú¶þÑõ»¯Ì¼£¬ÀûÓüîʯ»ÒÎüÊÕ¶þÑõ»¯Ì¼£¬ÔöÖصÄÖÊÁ¿¾ÍÊǶþÑõ»¯Ì¼µÄÖÊÁ¿£®È»ºóÇó³ö´Ö²úÆ·ÖÐ̼ËáÄƵĺ¬Á¿£®ÔÚ×°ÖÃÖÐÓõÄÊÇŨÑÎËᣬÇÒ×°ÖÃÖл¹ÓÐ̼ËáÇâÄÆÓëŨÑÎËá·´Ó¦²úÉú¶þÑõ»¯Ì¼£¬ËùÒÔÔö¼ÓµÄÖÊÁ¿±Èʵ¼Ê´ó£®
£¨4£©ÀûÓöþÑõ»¯Ì¼µÄÖÊÁ¿¿ÉÇó³ö̼ËáÄƵÄÖÊÁ¿·ÖÊý£®Ò²¿ÉÇó³öÑÎËáµÄŨ¶È£¬¸ù¾ÝÑÎËáµÄ¿ÉÇó³ö´Ö²úÆ·ÖÐÇâÑõ»¯ÄƵÄÖÊÁ¿·ÖÊý£®
¹Ê´ð°¸Îª£º
£¨Ò»£©´Ö²úÆ·ÖƱ¸£º
£¨1£©Na2CO3+Ca£¨OH£©2=CaCO3¡ý+2NaOH
£¨2£©NaOH
£¨¶þ£©´Ö²úÆ·³É·Ö·ÖÎö£º
£¨1£©Na2CO3+Ba£¨NO3£©2=BaCO3¡ý+2NaNO3£»Ca£¨OH£©2£»Ca£¨OH£©2ºÍNa2CO3ÄÜ·¢Éú»¯Ñ§·´Ó¦£¬ÔÚÈÜÒºÖв»Äܹ²´æ£®
£¨Èý£©´Ö²úÆ·º¬Á¿²â¶¨£º
£¨3£©Å¨ÁòË᣻³ýÈ¥¿ÕÆøÖеĶþÑõ»¯Ì¼£»·ÀÖ¹¿ÕÆøÖеĶþÑõ»¯Ì¼ºÍË®ÕôÆø±»GÖмîʯ»ÒÎüÊÕ£»Æ«´ó£»×°ÖÃDÖÐŨÑÎËá»Ó·¢³öµÄÂÈ»¯ÇâÓë×°ÖÃEÖÐNaHCO3·´Ó¦²úÉú¶þÑõ»¯Ì¼£®
£¨4£©½â£ºµ±Éú³É0.44g¶þÑõ»¯Ì¼Ê±£¬¶ÔÓ¦µÄ̼ËáÄƵÄÖÊÁ¿·ÖÊýΪa
Na2CO3+2HCl=2NaCl+H2O+CO2¡ü
106                   44
10g×a               0.44g
=£¬a=10.6%
ÓÉÓÚ»ìºÏÎïÖл¹º¬ÓÐÆäËûÔÓÖÊ£¬ËùÒÔÇâÑõ»¯ÄƵÄÖÊÁ¿·ÖÊý²»ÄÜÓÃ1-10.6%À´¼ÆË㣮
ÉèÓë´Ö²úÆ·ÖÐ̼ËáÄÆ·´Ó¦ÏûºÄµÄÂÈ»¯ÇâµÄÖÊÁ¿Îªx£®
Na2CO3+2HCl=2NaCl+H2O+CO2¡ü
       73             44
        x            0.44g
=£¬x=0.73g
36.5g 20%ÑÎËáÈÜÒºÖк¬ÂÈ»¯ÇâµÄÖÊÁ¿Îª£º36.5g×20%=7.3g
Óë´Ö²úÆ·ÖÐÇâÑõ»¯ÄÆ·´Ó¦ÏûºÄÂÈ»¯ÇâµÄÖÊÁ¿Îª£º
7.3g-0.73g=6.57g
Éè´Ö²úÆ·Öк¬ÇâÑõ»¯ÄƵÄÖÊÁ¿Îªy£®
NaOH+HCl=NaCl+H2O
40  36.5
y   6.57g
=
y=7.2g
´Ö²úÆ·Öк¬ÇâÑõ»¯ÄƵÄÖÊÁ¿·ÖÊýΪ£º7.2g÷10g×100%=72.0%
´ð£º´Ö²úÆ·Öк¬ÇâÑõ»¯ÄƵÄÖÊÁ¿·ÖÊýΪ72.0%£®
µãÆÀ£ºÔÚ¹¤ÒµÉÏÈËÃÇÀûÓÃʯ»ÒË®À´ÖÆÈ¡ÇâÑõ»¯ÄÆ£¬Òòʯ»ÒË®µÄ¼Û¸ñ±ÈÇâÑõ»¯ÄƵļ۸ñµÍ£¬ËùÒÔÎÒÃÇ¿ÉÒÔÓñãÒ˵ÄÀ´ÖÆÈ¡¹óµÄÒ©Æ·£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¾«Ó¢¼Ò½ÌÍøijУ»¯Ñ§¿ÎÍâÐËȤС×éµÄͬѧÔÚÑо¿ÐÔѧϰ¿ÎÉÏչʾÁËÒ»Ì×ÈçÏÂͼËùʾʵÑé×°Ö㨼ÙÉèÿ²½»¯Ñ§·´Ó¦¶¼ÍêÈ«£¬Ñõ»¯ÌúÑùÆ·ÖеÄÔÓÖʲ»²Î¼Ó·´Ó¦£©£®²éÔÄ×ÊÁÏ£º²ÝËáÔÚŨÁòËá´æÔÚʱ¼ÓÈÈ·¢ÉúÈçÏ·´Ó¦£ºH2C2O4
ŨÁòËá
¡÷
CO¡ü+CO2¡ü+H2O
ͨ¹ýÌÖÂÛ£¬Í¬Ñ§ÃǶÔÕâÌ××°ÖÃÓÐÁ˶àÖÖÈÏʶ£®

£¨1£©µÚһС×éͬѧ˵£º´ÓʵÑ鰲ȫºÍʵÑé²Ù×÷³ÌÐò¿´£¬Ê×ÏÈÒª¼ì²é×°ÖõÄÆøÃÜÐÔ£¬ÊµÑ鿪ʼÏȼÓÈÈ
 
£¨ÌîA»ò
D£©´¦£¬ÊµÑé½áÊøʱ£¬Ó¦
 
£¨ÌîÏÈ»òºó£©Í£Ö¹¼ÓÈÈ£»´Ó»·±£µÄ½Ç¶È½²£¬ÔÚE×°Öúó»¹Ó¦¶ÔβÆø½øÐд¦Àí£¬Æä·½·¨ÊÇ£º
 
£®
£¨2£©µÚ¶þС×éµÄͬѧ˵£ºÓÃʵÑé×°ÖÿÉÒÔ¼ìÑéÒ»Ñõ»¯Ì¼ÓëÑõ»¯Ìú·´Ó¦µÄ²úÉú£¬ÆäÖÐB×°ÖõÄ×÷ÓÃ
 
£»E×°ÖõÄ×÷ÓÃ
 
£»Ò»Ñõ»¯Ì¼ÓëÑõ»¯Ìú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
 
£®
£¨3£©µÚÈýС×éµÄͬѧ˵£ºÀûÓÃÕâÌ××°Öû¹¿ÉÒԲⶨÑõ»¯ÌúÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£®ËûÃǵIJⶨ·½·¨ÊÇ£º³ÆÈ¡Ñõ»¯ÌúÑùÆ·10.0g£¬ÍêÈ«·´Ó¦²¢ÀäÈ´ºóÔÙ³ÆÁ¿Ê£Óà¹ÌÌåµÄ×ÜÖÊÁ¿Îª7.6g£®¼ÆËãʵÑé²âµÃÑõ»¯ÌúÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊýΪ
 
£®
£¨4£©µÚËÄС×éͬѧ˵£ºÀûÓÃÕâÌ××°Öû¹ÓÐÆäËü²â¶¨ÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊýµÄ·½·¨£ºÏȳÆÁ¿Ñõ»¯ÌúÑùÆ·µÄÖÊÁ¿£¬ÔÙ·Ö±ð³ÆÁ¿E×°ÖÃÔÚ·´Ó¦Ç°ºóµÄ×ÜÖÊÁ¿£¬¼´¿É¼ÆËãÇóµÃÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£®µ«°´´Ë·½·¨Êµ¼Ê²â¶¨½á¹ûÈ´Æ«´ó£¬·ÖÎöÔì³ÉÕâÖÖÎó²îµÄÔ­Òò¿ÉÄÜÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?ËÄ´¨Ä£Ä⣩ijУ»¯Ñ§¿ÎÍâÐËȤС×éµÄͬѧÔÚÑо¿ÐÔѧϰ¿ÎÉÏչʾÁËÒ»Ì×ÈçÏÂͼËùʾµÄʵÑé×°Ö㨼ÙÉèÿ²½»¯Ñ§·´Ó¦¶¼ÍêÈ«£¬Ñõ»¯ÌúÑùÆ·ÖеÄÔÓÖʲ»²Î¼Ó·´Ó¦£©£®
²éÔÄ×ÊÁÏ£º²ÝËáÔÚŨÁòËá´æÔÚʱ¼ÓÈÈ·¢ÉúÈçÏ·´Ó¦£º
H2C2O4 
 Å¨H2SO4 
.
¡÷
CO¡ü+CO2¡ü+H2O

ͨ¹ýÌÖÂÛ£¬Í¬Ñ§ÃǶÔÕâÌ××°ÖÃÓÐÁ˶àÖÖÈÏʶ£®
£¨1£©µÚһС×éͬѧ˵£º´ÓʵÑ鰲ȫºÍʵÑé²Ù×÷³ÌÐò¿´£¬Ê×ÏÈÒª
¼ì²é×°ÖÃÆøÃÜÐÔ
¼ì²é×°ÖÃÆøÃÜÐÔ
£¬ÊµÑ鿪ʼÏȼÓÈÈA´¦£¬ÊµÑé½áÊøʱ£¬Ó¦ÏÈÍ£Ö¹D´¦µÄ¼ÓÈÈ£»´Ó»·¾³±£»¤½Ç¶È·ÖÎö£¬´Ë×°ÖôæÔÚµÄÃ÷ÏÔ²»×ãÊÇ
ûÓÐβÆø´¦Àí£¬ÎÛȾ»·¾³
ûÓÐβÆø´¦Àí£¬ÎÛȾ»·¾³
£»
£¨2£©µÚ¶þС×éµÄͬѧ˵£ºÆäÖÐB×°ÖõÄ×÷ÓóýÈ¥CO2£»×°ÖÃCµÄ×÷ÓÃÊÇ
³ýȥˮÕôÆø
³ýȥˮÕôÆø
£»ÓøÃʵÑé×°ÖÿÉÒÔ¼ìÑéÒ»Ñõ»¯Ì¼ÓëÑõ»¯Ìú·´Ó¦µÄ²úÎE×°ÖõÄ×÷ÓÃ
¼ìÑéCOÓëFe2O3·´Ó¦Éú³ÉµÄ¶þÑõ»¯Ì¼
¼ìÑéCOÓëFe2O3·´Ó¦Éú³ÉµÄ¶þÑõ»¯Ì¼
£»Ò»Ñõ»¯Ì¼ÓëÑõ»¯Ìú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
3CO+Fe2O3
 ¸ßΠ
.
 
2Fe+3CO2
3CO+Fe2O3
 ¸ßΠ
.
 
2Fe+3CO2
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2005?Õò½­£©Ä³Ð£»¯Ñ§¿ÎÍâÐËȤС×éµÄͬѧÔÚÑо¿ÐÔѧϰ¿ÎÉÏչʾÁËÒ»Ì×ÈçͼËùʾµÄʵÑé×°Ö㨼ÙÉèÿ²½»¯Ñ§·´Ó¦¶¼ÍêÈ«£¬Ñõ»¯ÌúÑùÆ·ÖеÄÔÓÖʲ»²Î¼Ó·´Ó¦£©£®

²éÔÄ×ÊÁÏ£º²ÝËáÔÚŨÁòËá´æÔÚʱ¼ÓÈÈ·¢ÉúÈçÏ·´Ó¦£º
H2C2O4
 Å¨ÁòËá 
.
 
CO¡ü+CO2¡ü+H2O
ͨ¹ýÌÖÂÛ£¬Í¬Ñ§ÃǶÔÕâÌ××°ÖÃÓÐÁ˶àÖÖÈÏʶ£®
£¨1£©µÚһС×éͬѧ˵£º´ÓʵÑ鰲ȫºÍʵÑé²Ù×÷³ÌÐò¿´£¬Ê×ÏÈÒª¼ì²é×°ÖõÄÆøÃÜÐÔ£¬ÊµÑ鿪ʼÏȼÓÈÈ
A
A
£¨ÌîA»òD£©´¦£¬ÊµÑé½áÊøʱ£¬Ó¦
ÏÈ
ÏÈ
£¨ÌîÏÈ»òºó£©Í£Ö¹D´¦µÄ¼ÓÈÈ£»´Ó»·±£µÄ½Ç¶È½²£¬ÔÚE×°Öúó»¹Ó¦¶ÔβÆø½øÐд¦Àí£¬Æä·½·¨ÊÇ£º
µãȼ»òÊÕ¼¯ÆðÀ´
µãȼ»òÊÕ¼¯ÆðÀ´
£®
£¨2£©µÚ¶þС×éµÄͬѧ˵£ºÓøÃʵÑé×°ÖÿÉÒÔ¼ìÑéÒ»Ñõ»¯Ì¼ÓëÑõ»¯Ìú·´Ó¦µÄ²úÎÆäÖÐB×°ÖõÄ×÷ÓÃ
³ýÈ¥AÖÐÉú³ÉµÄ¶þÑõ»¯Ì¼
³ýÈ¥AÖÐÉú³ÉµÄ¶þÑõ»¯Ì¼
£»E×°ÖõÄ×÷ÓÃ
¼ìÑé²¢ÎüÊÕD×°ÖÃÉú³ÉµÄ¶þÑõ»¯Ì¼
¼ìÑé²¢ÎüÊÕD×°ÖÃÉú³ÉµÄ¶þÑõ»¯Ì¼
£»Ò»Ñõ»¯Ì¼ÓëÑõ»¯Ìú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
3CO+Fe2O3
 ¸ßΠ
.
 
3CO2+2Fe
3CO+Fe2O3
 ¸ßΠ
.
 
3CO2+2Fe
£®
£¨3£©µÚÈýС×éµÄͬѧ˵£ºÀûÓÃÕâÌ××°Öû¹¿ÉÒԲⶨÑõ»¯ÌúÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£®ËûÃǵIJⶨ·½·¨ÊÇ£º³ÆÁ¿Ñõ»¯ÌúÑùÆ·µÄÖÊÁ¿10.0g£¬ÑùÆ·Óë²£Á§¹ÜµÄ×ÜÖÊÁ¿Îª60.0g£¬ÍêÈ«·´Ó¦²¢ÀäÈ´ºóÔÙ³ÆÁ¿²£Á§¹ÜÓëÊ£Óà¹ÌÌåµÄ×ÜÖÊÁ¿Îª57.6g£®¼ÆËãʵÑé²âµÃÑõ»¯ÌúÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊýΪ
80%
80%
£®
£¨4£©µÚËÄС×éͬѧ˵£ºÀûÓÃÕâÌ××°Öû¹ÓÐÁíÍâ²â¶¨ÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊýµÄ·½·¨£ºÏȳÆÁ¿Ñõ»¯ÌúÑùÆ·µÄÖÊÁ¿£¬ÔÙ·Ö±ð³ÆÁ¿E×°ÖÃÔÚ·´Ó¦Ç°ºóµÄ×ÜÖÊÁ¿£¬¼´¿É¼ÆËãÇóµÃÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£®µ«°´´Ë·½·¨Êµ¼ÊʵÑé²â¶¨½á¹ûÈ´Æ«´ó£¬·ÖÎöÔì³ÉÆ«´óµÄÔ­Òò¿ÉÄÜÊÇ
B×°ÖÃÎüÊÕCO2²»³ä·Ö£»EÎüÊÕÁË×°ÖÃÀï¿ÕÆøÖеÄCO2»òÕßÍâ½çCO2½øÈëµÄ¸ÉÈÅ
B×°ÖÃÎüÊÕCO2²»³ä·Ö£»EÎüÊÕÁË×°ÖÃÀï¿ÕÆøÖеÄCO2»òÕßÍâ½çCO2½øÈëµÄ¸ÉÈÅ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ijУ»¯Ñ§¿ÎÍâÐËȤС×éµÄͬѧÔÚÑо¿ÐÔѧϰ¿ÎչʾÁËÒ»Ì×ÈçͼËùʾµÄʵÑé×°Ö㨼ÙÉèÿ²½»¯Ñ§·´Ó¦¶¼ÍêÈ«£¬Ñõ»¯ÌúÑùÆ·ÖеÄÔÓÖʲ»·´Ó¦£©£®²éÔÄ×ÊÁÏ£º²ÝËáÔÚŨÁòËá´æÔÚʱ¼ÓÈÈ·¢ÉúÈçÏ·´Ó¦£º
H2C2O4
ŨÁòËá
.
.
¼ÓÈÈ
 CO¡ü+CO2¡ü+H2O£¬Í¨¹ýÌÖÂÛ£¬Í¬Ñ§ÃǶÔÕâÌ××°ÖÃÓÐÁ˶àÖÖÈÏʶ£®
£¨1£©µÚһС×éͬѧ˵£º´ÓʵÑ鰲ȫºÍʵÑé²Ù×÷³ÌÐò¿´£¬Ê×ÏÈÒª¼ì²é×°ÖõÄÆøÃÜÐÔ£¬ÊµÑ鿪ʼÏȼÓÈÈ
A
A
£¨ÌîA»òD£©´¦£»ÊµÑé½áÊøʱ£¬Ó¦¸Ã
ÏÈ
ÏÈ
£¨ÌîÏÈ»òºó£©Í£Ö¹D´¦µÄ¼ÓÈÈ£»´Ó»·±£µÄ½Ç¶È½²£¬´Ë×°Öû¹´æÔÚµÄÎÊÌâÊÇ£º
ûÓжԷ´Ó¦ÖÐÊ£ÓàµÄÒ»Ñõ»¯Ì¼½øÐд¦Àí
ûÓжԷ´Ó¦ÖÐÊ£ÓàµÄÒ»Ñõ»¯Ì¼½øÐд¦Àí
£®
£¨2£©µÚ¶þС×éµÄͬѧ˵£ºÓøÃʵÑé×°ÖÿÉÒÔ¼ìÑéÒ»Ñõ»¯Ì¼ÓëÑõ»¯Ìú·´Ó¦µÄ²úÎÆäÖÐB×°ÖõÄ×÷ÓÃÊÇ
ÎüÊÕAÖз´Ó¦Éú³ÉµÄ¶þÑõ»¯Ì¼
ÎüÊÕAÖз´Ó¦Éú³ÉµÄ¶þÑõ»¯Ì¼
£»E×°ÖõÄ×÷ÓÃÊÇ
ÎüÊÕDÖз´Ó¦Éú³ÉµÄ¶þÑõ»¯Ì¼
ÎüÊÕDÖз´Ó¦Éú³ÉµÄ¶þÑõ»¯Ì¼
£»Ò»Ñõ»¯Ì¼ÓëÑõ»¯Ìú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
3CO+Fe2O3
 ¸ßΠ
.
 
3CO2+2Fe
3CO+Fe2O3
 ¸ßΠ
.
 
3CO2+2Fe
£®
£¨3£©µÚÈýС×éµÄͬѧ˵£ºÀûÓÃÕâÌ××°Öû¹¿ÉÒԲⶨÑõ»¯ÌúÑùÆ·µÄÖÊÁ¿·ÖÊý£®ËûÃǵIJⶨ·½·¨ÊÇ£º³ÆÈ¡Ñõ»¯ÌúµÄÖÊÁ¿·ÖÊýΪ10.0g£¬ÍêÈ«·´²¢ÀäÈ´ºóÔÙ³ÆÁ¿Ê£Óà¹ÌÌåµÄ×ÜÖÊÁ¿Îª7.6g£®¼ÆËãʵÑé²âµÃÑõ»¯ÌúÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊýΪ
80%
80%
£®
£¨4£©µÚËÄС×éͬѧ˵£ºÀûÓÃÕâÌ××°Öû¹ÓÐÆäËü²â¶¨ÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊýµÄ·½·¨£ºÏȳÆÁ¿Ñõ»¯ÌúÑùÆ·µÄÖÊÁ¿£¬ÔÙ·Ö±ð³ÆÁ¿E×°ÖÃÔÚ·´Ó¦Ç°ºóµÄ×ÜÖÊÁ¿£¬¼´¿É¼ÆËã³öÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£®µ«°´´Ë·½·¨²â¶¨µÄ½á¹ûÈ´±Èʵ¼ÊµÄÆ«´ó£¬Ôì³ÉÕâÖÖÎó²îµÄÔ­Òò¿ÉÄÜÊÇ
B×°ÖÃÎüÊÕCO2²»³ä·Ö£»EÎüÊÕÁË×°ÖÃÀï¿ÕÆøÖеÄCO2»òÕßÍâ½çCO2½øÈëµÄ¸ÉÈÅ£¨¸ø³öÒ»ÌõºÏÀí´ð°¸¼´¿É£©
B×°ÖÃÎüÊÕCO2²»³ä·Ö£»EÎüÊÕÁË×°ÖÃÀï¿ÕÆøÖеÄCO2»òÕßÍâ½çCO2½øÈëµÄ¸ÉÈÅ£¨¸ø³öÒ»ÌõºÏÀí´ð°¸¼´¿É£©
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2008Äêºþ±±Ê¡Ð¢¸ÐÊÐÓ¦³ÇÊгõÖÐѧÉú»¯Ñ§ÖªÊ¶¾ºÈü³õÈüÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º½â´ðÌâ

ijУ»¯Ñ§¿ÎÍâÐËȤС×éµÄͬѧÔÚÑо¿ÐÔѧϰ¿ÎÉÏչʾÁËÒ»Ì×ÈçÏÂͼËùʾʵÑé×°Ö㨼ÙÉèÿ²½»¯Ñ§·´Ó¦¶¼ÍêÈ«£¬Ñõ»¯ÌúÑùÆ·ÖеÄÔÓÖʲ»²Î¼Ó·´Ó¦£©£®²éÔÄ×ÊÁÏ£º²ÝËáÔÚŨÁòËá´æÔÚʱ¼ÓÈÈ·¢ÉúÈçÏ·´Ó¦£ºH2C2O4CO¡ü+CO2¡ü+H2O
ͨ¹ýÌÖÂÛ£¬Í¬Ñ§ÃǶÔÕâÌ××°ÖÃÓÐÁ˶àÖÖÈÏʶ£®

£¨1£©µÚһС×éͬѧ˵£º´ÓʵÑ鰲ȫºÍʵÑé²Ù×÷³ÌÐò¿´£¬Ê×ÏÈÒª¼ì²é×°ÖõÄÆøÃÜÐÔ£¬ÊµÑ鿪ʼÏȼÓÈÈ______£¨ÌîA»ò
D£©´¦£¬ÊµÑé½áÊøʱ£¬Ó¦______£¨ÌîÏÈ»òºó£©Í£Ö¹¼ÓÈÈ£»´Ó»·±£µÄ½Ç¶È½²£¬ÔÚE×°Öúó»¹Ó¦¶ÔβÆø½øÐд¦Àí£¬Æä·½·¨ÊÇ£º______£®
£¨2£©µÚ¶þС×éµÄͬѧ˵£ºÓÃʵÑé×°ÖÿÉÒÔ¼ìÑéÒ»Ñõ»¯Ì¼ÓëÑõ»¯Ìú·´Ó¦µÄ²úÉú£¬ÆäÖÐB×°ÖõÄ×÷ÓÃ______£»E×°ÖõÄ×÷ÓÃ______£»Ò»Ñõ»¯Ì¼ÓëÑõ»¯Ìú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ______ 3CO2+2Fe

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸