ÓÐ45.45 g KClO3ºÍMnO2µÄ»ìºÏÎ¼ÓÈÈÒ»¶Îʱ¼äºó£¬ÖÊÁ¿±äΪ40.65 g£®½«´ËÊ£Óà¹ÌÌåƽ¾ù·Ö³ÉÁ½·Ý¡£?

(1)Ò»·Ý¼Ó×ãÁ¿µÄË®Èܽâºó£¬¹ýÂË£¬ÔÚÂËÒºÖмÓ×ãÁ¿µÄÏõËáËữÁ˵ÄAgNO3ÈÜÒº£¬¿ÉµÃ³Áµí¶àÉÙ¿Ë£¿?

(2)ÁíÒ»·Ý¼ÓÈë×ãÁ¿µÄŨÑÎËᣬ¼ÓÈÈʹ֮·´Ó¦£®½«·Å³öµÄÆøÌåͨÈ뺬KI¡¢KBrµÄ1 L»ìºÏÈÜÒº£¬Ç¡ºÃʹÈÜÒºÖÐBr£­¡¢I£­ÍêÈ«·´Ó¦£®ÈôKIµÄÎïÖʵÄÁ¿Å¨¶ÈÓëKBrµÄÎïÖʵÄÁ¿Å¨¶ÈÏàµÈ£¬¶¼Îª0.35 mol¡¤L£­1£¬ÊÔÇóKClO3µÄ·Ö½â°Ù·ÖÂÊ¡£

 (1)7.175 g  (2)33%?


½âÎö:

²îÁ¿·¨ÊÇ»¯Ñ§¼ÆËãÖеij£Ó÷½·¨Ö®Ò»¡£¼õÇáµÄÖÊÁ¿¼´ÎªÉú³ÉµÄÑõÆøµÄÖÊÁ¿¡£¾Ý´Ë¿É¼ÆËã³öÉú³ÉµÄKClÖÊÁ¿£¬¼´Çó³öAgCl³ÁµíµÄÁ¿¡£?

(1)»ìºÏÎï¼ÓÈȺóÖÊÁ¿¼õÇá45.45£­40.65£½4.8 g£¬¼´Éú³ÉO2µÄÖÊÁ¿Îª4.8 g£¬»ìºÏÎïµÄÒ»°ë¿É²úÉú2.4 g O2¡£?

Éè·Ö½âÁËKClO3 x mol£¬Éú³ÉKCly mol£¬Éú³É³Áµíz g¡£

2KClO32KCl  £«3O2¡ü

2 mol       2 mol     96 g?

x mol     ?y mol    2.4 g

x£½y£½0£®05?

KCl£«AgNO3 = AgCl¡ý£«KNO3?

1 mol      143.5 g?

0.05 mol   z mol

z£½7£®175 g?

(2)ÓÉÓÚKBr¡¢KIµÄÎïÖʵÄÁ¿¾ùΪ£º0.35 mol¡¤L£­1¡Á1 L£½0.35 mol

Cl2  £«  2KBr = 2KCl£«Br2?

1        2?

n(Cl2)  0.35

n(Cl2)£½0.175 mol?

¼´ÁíÒ»°ë¹ÌÌå»ìºÏÎïÓëŨÑÎËá·´Ó¦£¬¹²·Å³ö0.35 mol Cl2?

6HCl£«KClO3KCl£«3Cl2¡ü£«3H2O?

122.5 g            3 mol?

m g                 mol

4HCl£«MnO2MnCl2£«Cl2¡ü£«2H2O?

87 g               1 mol?

w g                 mol

½âµÃ  ÔòKClO3µÄ·Ö½âÂÊ£½¡Á100%£½33%£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

¹¤ÒµºÏ³É°±ÓëÖƱ¸ÏõËáÒ»°ã¿ÉÁ¬ÐøÉú²ú£¬Á÷³ÌÈçÏÂ

£¨1£©¢Ù¹¤ÒµÉú²úʱ£¬ÖÆÈ¡ÇâÆøµÄÒ»¸ö·´Ó¦Îª£ºCO+H2O£¨g£©?CO2+H2£¬850¡æʱ£¬Íù1LÃܱÕÈÝÆ÷ÖгäÈë0.3mol COºÍ0.2molH2O£¨g£©£®·´Ó¦4minºó½¨Á¢Æ½ºâ£¬ÌåϵÖÐc£¨H2£©=0.12mol?L-1£®COµÄƽºâŨ¶ÈΪ
0.18mol/L
0.18mol/L
 ×ª»¯ÂÊΪ
40%
40%
 ¸ÃζÈÏ´˷´Ó¦µÄƽºâ³£ÊýK=
1
1
 £¨Ìî¼ÆËã½á¹û£©£®
¢ÚÔÚ850¡æʱ£¬ÒÔ±íÖеÄÎïÖʵÄÁ¿Í¶ÈëºãÈÝ·´Ó¦Æ÷ÖУ¬ÆäÖÐÏòÄæ·´Ó¦·½Ïò½øÐеÄÓÐ
A
A
 £¨Ñ¡ÌîA¡¢B¡¢C¡¢D¡¢E£©
A B C D E
n£¨CO2£© 3 l 0 1 l
n£¨H2£© 2 l 0 1 2
n£¨CO£© 1 2 3 0.5 3
n£¨H2O£© 5 2 3 2 l
£¨2£©ºÏ³ÉËþÖз¢Éú·´Ó¦N2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H£¼0£®Ï±íΪ²»Í¬Î¶Èϸ÷´Ó¦µÄƽºâ³£Êý£®ÓÉ´Ë¿ÉÍÆÖª£¬±íÖÐT1
£¼
£¼
300¡æ£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®
 T/¡ãC  T1  300  T2
 K  1.00¡Á107  2.45¡Á105  1.88¡Á103
£¨3£©N2ºÍH2ÔÚÌú×÷´ß»¯¼Á×÷ÓÃÏ´Ó145¡æ¾Í¿ªÊ¼·´Ó¦£¬²»Í¬Î¶ÈÏÂNH3²úÂÊͼ1Ëùʾ£®Î¶ȸßÓÚ900¡æʱ£¬NH3²úÂÊϽµ£¬Ô­ÒòÊÇ
900¡æʱ·´Ó¦´ïµ½Æ½ºâ״̬£¬ÔÙÉý¸ßζÈƽºâÏò×óÒƶ¯
900¡æʱ·´Ó¦´ïµ½Æ½ºâ״̬£¬ÔÙÉý¸ßζÈƽºâÏò×óÒƶ¯
£®

£¨4£©ÔÚ»¯Ñ§·´Ó¦ÖÐÖ»Óм«ÉÙÊýÄÜÁ¿±Èƽ¾ùÄÜÁ¿¸ßµÃ¶àµÄ·´Ó¦Îï·Ö×Ó·¢ÉúÅöײʱ²Å¿ÉÄÜ·¢Éú»¯Ñ§·´Ó¦£¬ÕâЩ·Ö×Ó±»³ÆΪ»î»¯·Ö×Ó£®Ê¹ÆÕͨ·Ö×Ó±ä³É»î»¯·Ö×ÓËùÐèÌṩµÄ×îµÍÏ޶ȵÄÄÜÁ¿½Ð»î»¯ÄÜ£¬Æ䵥λͨ³£ÓÃkJ?mol-1±íʾ£®ÇëÈÏÕæ¹Û²ìͼ2£¬»Ø´ðÎÊÌ⣮
ͼÖÐËùʾ·´Ó¦ÊÇ
·ÅÈÈ
·ÅÈÈ
£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©·´Ó¦£¬¸Ã·´Ó¦µÄ¡÷H=
-£¨E1-E2£©kJ/mol
-£¨E1-E2£©kJ/mol
 £¨Óú¬E1¡¢E2EµÄ´úÊýʽ±íʾ£©£®ÒÑÖªÈÈ»¯Ñ§·½³Ìʽ£ºH2£¨g£©+
1
2
O2£¨g£©=H2O£¨g£©¡÷H=-241.8kJ?mol-1£¬¸Ã·´Ó¦µÄ»î»¯ÄÜΪ167.2kJ?mol-1£¬ÔòÆäÄæ·´Ó¦µÄ»î»¯ÄÜΪ
409kJ/mol
409kJ/mol
£®
£¨5£©ÏõË᳧µÄβÆøÖ±½ÓÅŷŽ«ÎÛȾ¿ÕÆø£®Ä¿Ç°¿Æѧ¼Ò̽Ë÷ÀûÓÃȼÁÏÆøÌåÖеļ×ÍéµÈ½«µªµÄÑõ»¯ÎﻹԭΪµªÆøºÍË®£¬·´Ó¦»úÀíΪ£º
CH4£¨g£©+4NO2£¨g£©=4NO£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-574kJ?mol-1
CH4£¨g£©+4NO£¨g£©=2N2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-1160kJ?mol-1
Ôò¼×ÍéÖ±½Ó½«NO2»¹Ô­ÎªN2µÄÈÈ»¯Ñ§·½³ÌʽΪ£º
CH4£¨g£©+2NO2£¨g£©=CO2£¨g£©+2H2O£¨g£©+N2£¨g£©¡÷H=-867kJ?mol-1
CH4£¨g£©+2NO2£¨g£©=CO2£¨g£©+2H2O£¨g£©+N2£¨g£©¡÷H=-867kJ?mol-1
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºËÄ´¨Ê¡ÅÊÖ¦»¨ÊÐ2012£­2013ѧÄê¸ßÒ»ÏÂѧÆÚÆÚÄ©µ÷Ñмà²â»¯Ñ§ÊÔÌâ ÌâÐÍ£º022

(1)ÔÚ200¡æ¡¢101 kPaʱ£¬1 g¡¡H2ÓëµâÕôÆøÍêÈ«·´Ó¦·Å³ö7.45 kJµÄÈÈÁ¿£¬Çëд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ________£®

(2)ÔÚ1¡Á105 PaºÍ298 Kʱ£¬½«²ð¿ª1 mol¹²¼Û¼üËùÐèÒªµÄÄÜÁ¿³ÆΪ¼üÄÜ(kJ/mol)£®ÏÂÃæÊÇһЩ¹²¼Û¼üµÄ¼üÄÜ£º

ÔÚ298 Kʱ£¬ÔÚ´ß»¯¼Á´æÔÚÏ£¬H2(g)£«Cl2(g)£½2HCl(g)¡¡¦¤H£½________kJ/mol

(3)ÒÑÖª£ºCO(g)£«2H2(g)CH3OH(g)¡¡¦¤H1£½£­116 kJ¡¤mol£­1

CO(g)£«O2(g)£½CO2(g)¡¡¦¤H2£½£­283 kJ¡¤mol£­1

H2(g)£«O2(g)£½H2O(g)¡¡¦¤H3£½£­242 kJ¡¤mol£­1

Ôò±íʾ1 molÆø̬¼×´¼ÍêȫȼÉÕÉú³ÉCO2ºÍË®ÕôÆøʱµÄÈÈ»¯Ñ§·½³ÌʽΪ£º

________£®

(4)N2(g)£«3H2(g)2NH3(g)·´Ó¦¹ý³ÌµÄÄÜÁ¿±ä»¯ÈçͼËùʾ£®ÒÑÖª2 mol¡¡N2(g)·´Ó¦Éú³É4 mol¡¡NH3(g)·Å³ö£­188.4 kJ/mol£®Ôò£º

ͼÖÐEµÄ´óС¶Ô¸Ã·´Ó¦µÄ·´Ó¦ÈÈÓÐÎÞÓ°Ï죿________£®(Ìî¡°ÓÐÓ°Ï족»ò¡°ÎÞÓ°Ï족)£®¸Ã·´Ó¦Í¨³£ÓÃÌú´¥Ã½×÷´ß»¯¼Á£¬¼ÓÌú´¥Ã½»áʹͼÖÐBµãÉý¸ß»¹ÊǽµµÍ£¿________(Ìî¡°Éý¸ß¡±»ò¡°½µµÍ¡±)£®Í¼ÖЦ¤H£½________kJ/mol£®¶ÔÓÚ´ïƽºâºóµÄ¸Ã·´Ó¦£¬ÈôÖ»¸Ä±äÏÂÁÐÌõ¼þÖ®Ò»£¬ÄÜʹµ¥Î»Ìå»ýÄڻ·Ö×Ó°Ù·ÖÊýÔö¼ÓµÄÊÇ________£®

A£®ÉýÎÂ

B£®ÔÙ³äÈëN2ºÍH2£¬²¢Ê¹ËüÃǵÄŨ¶È¶¼Ôö¼ÓÒ»±¶

C£®¼õѹ

D£®¼Ó´ß»¯¼Á

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄê±±¾©ÊÐͨÖÝÇø¸ßÈýÄ꼶ÉÏѧÆÚÆÚÄ©Ãþµ×¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A£®2 mol/L KClÈÜÒºÓë1 mol/L K2SO4ÈÜÒº»ìºÏºó£¬c(K£«)Ϊ2 mol/L

B£®120 g NaClÈÜÒºÖÐÈÜÓÐ20 g NaCl£¬¸ÃζÈÏÂNaClµÄÈܽâ¶ÈΪ20g

C£®22.4 L HClÆøÌåÈÜÓÚË®ÖƳÉ1 LÈÜÒº£¬¸ÃÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ1 mo1/L

D£®°Ñ5 gµ¨·¯(CuSO4∙5H2O)ÈÜÓÚ45 gË®ÖУ¬ËùµÃÈÜÒºÈÜÖʵÄÖÊÁ¿·ÖÊýΪ10%

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄê¸ßÈýÉÏѧÆÚµ¥Ôª²âÊÔ£¨6£©»¯Ñ§ÊÔ¾í£¨ÐÂÈ˽̰棩 ÌâÐÍ£ºÊµÑéÌâ

£¨10·Ö£©ÎÒ¹ú»¯¹¤×¨¼ÒºîµÂ°ñµÄ¡°ºîÊÏÖƼ¡±ÔøΪÊÀ½çÖƼҵ×ö³öÁËÍ»³ö¹±Ïס£ËûÀûÓÃNaHCO3¡¢NaCl¡¢NH4C1µÈÎïÖÊÈܽâ¶ÈµÄ²îÒ죬ÒÔʳÑΡ¢°±Æø¡¢¶þÑõ»¯Ì¼µÈΪԭÁÏÖƵÃNaHCO3£¬½ø¶øÉú²ú³ö´¿¼î¡£ÒÔÏÂA¡¢B¡¢C¡¢DËĸö×°ÖÿÉ×é×°³ÉʵÑéÊÒÄ£Äâ ¡°ºîÊÏÖƼ¡±ÖÆÈ¡NaHCO3µÄʵÑé×°Öá£×°ÖÃÖзֱðÊ¢ÓÐÒÔÏÂÊÔ¼Á£ºB£ºÏ¡ÁòË᣻C£ºÑÎËᡢ̼Ëá¸Æ£»D£ºº¬°±µÄ±¥ºÍʳÑÎË®¡¢Ë®

       ËÄÖÖÑÎÔÚ²»Í¬Î¶ÈϵÄÈܽâ¶È£¨g£¯100gË®£©±í

 

0¡æ

10¡æ

20¡æ

30¡æ

40¡æ

50¡æ

60¡æ

100¡æ

NaCl

35£®7

35£®8

36£®0

36_3

36£®6

37£®0

37£®3

39£®8

NH4HCO3

11£®9

15£®8

21£®0

27£®0

¡ª¢Ù

¡ª

¡ª

¡ª

NaHCO3

6£®9

8£®1

9£®6

11£®1

12£®7

14£®5

16£®4

 

NH4Cl

29£®4

33£®3

37£®2[À´Ô´:Z|xx|k.Com]

41£®4

45£®8[À´Ô´:Z*xx*k.Com]

50£®4

55£®3

77£®3

£¨ËµÃ÷£º¢Ù>35¡æNH4HCO3»áÓзֽ⣩

       Çë»Ø´ðÒÔÏÂÎÊÌ⣺

£¨1£©×°ÖõÄÁ¬½Ó˳ÐòÓ¦ÊÇ          £¨Ìî×Öĸ£©¡£

£¨2£©A×°ÖÃÖÐÊ¢·ÅµÄÊÔ¼ÁÊÇ          £¬Æä×÷ÓÃÊÇ               ¡£

£¨3£©ÔÚʵÑé¹ý³ÌÖУ¬ÐèÒª¿ØÖÆDζÈÔÚ30¡æ¡«35¡æ£¬Ô­ÒòÊÇ                   ¡£

£¨4£©·´Ó¦½áÊøºó£¬½«×¶ÐÎÆ¿½þÔÚÀäË®ÖУ¬Îö³öNaHCO3¾§Ìå¡£ÓÃÕôÁóˮϴµÓNaHCO3¾§ÌåµÄÄ¿µÄÊdzýÈ¥             ÔÓÖÊ£¨ÒÔ»¯Ñ§Ê½±íʾ£©

£¨5£©½«×¶ÐÎÆ¿ÖеIJúÎï¹ýÂ˺ó£¬ËùµÃµÄĸҺÖк¬ÓР        £¨ÒÔ»¯Ñ§Ê½±íʾ£©£¬¼ÓÈëÂÈ»¯Ç⣬²¢½øÐР           ²Ù×÷£¬Ê¹NaClÈÜҺѭ»·Ê¹Óã¬Í¬Ê±¿É»ØÊÕNH4C1¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ


  1. A.
    2 mol/L KClÈÜÒºÓë1 mol/L K2SO4ÈÜÒº»ìºÏºó£¬c(K+)Ϊ2 mol/L
  2. B.
    120 g NaClÈÜÒºÖÐÈÜÓÐ20 g NaCl£¬¸ÃζÈÏÂNaClµÄÈܽâ¶ÈΪ20g
  3. C.
    22.4 L HClÆøÌåÈÜÓÚË®ÖƳÉ1 LÈÜÒº£¬¸ÃÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ1 mo1/L
  4. D.
    °Ñ5 gµ¨·¯(CuSO4?5H2O)ÈÜÓÚ45 gË®ÖУ¬ËùµÃÈÜÒºÈÜÖʵÄÖÊÁ¿·ÖÊýΪ10%

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸