ÌâÄ¿ÄÚÈÝ

Ëæ×ŲÄÁÏ¿ÆÑ§µÄ·¢Õ¹£¬½ðÊô·°¼°Æä»¯ºÏÎïµÃµ½ÁËÔ½À´Ô½¹ã·ºµÄÓ¦Ó㬲¢±»ÓþΪ¡°ºÏ½ðµÄάÉúËØ¡±£®·°ÔªËع㷺·ÖÉ¢ÓÚ¸÷ÖÖ¿óÎïÖУ¬¼Ø·°ÓË¿óµÄÖ÷Òª³É·Ö¿ÉÓû¯Ñ§Ê½±íʾΪK2H6U2V2O15£¬²â¶¨ÆäÖз°ÔªËغ¬Á¿µÄ·½·¨ÊÇ£ºÏȰѿóʯÖеķ°ÔªËØ×ª»¯ÎªV2O5(·°ÔªËصϝºÏ¼Û²»±ä)£¬V2O5ÔÚËáÐÔÈÜÒºÀïת»¯ÎªVO2+£¬ÔÙÓòÝËáµÈ²â¶¨·°£®×Ü·´Ó¦¿É±íʾΪ£º

¡õVO2+£«¡õH2C2O4£«¡õH+¡õVO2+£«¡õCO2£«¡õH2O

(1)

Ç뽫ÉÏÊö·´Ó¦Å䯽£®

(2)

ÏÖÓмط°ÓË¿óÑùÆ·10.2g£¬ÓÃÉÏÊö·½·¨À´²â¶¨·°µÄº¬Á¿£¬½á¹ûÏûºÄ0.9g²ÝËᣬÄÇô´Ë¼Ø·°ÓË¿óÖз°ÔªËصÄÖÊÁ¿·ÖÊýÊÇ________£®ÈôÓÃÑõ»¯ÎïµÄÐÎʽ±íʾ¼Ø·°ÓË¿óµÄÖ÷Òª³É·Ö£¬Æä»¯Ñ§Ê½Îª________£®

´ð°¸£º
½âÎö£º

(1)

2VO2+£«H2C2O4£«2H+2VO2+£«2CO2£«2H2O

(2)

10%£»K2O¡¤V2O5¡¤U2O6¡¤3H2O»òK2O¡¤V2O5¡¤2UO3¡¤3H2O


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
Ëæ×ŲÄÁÏ¿ÆÑ§µÄ·¢Õ¹£¬½ðÊô·°¼°Æä»¯ºÏÎïµÃµ½ÁËÔ½À´Ô½¹ã·ºµÄÓ¦Ó㬲¢±»ÓþΪ¡°ºÏ½ðµÄάÉúËØ¡±£®Îª»ØÊÕÀûÓú¬·°´ß»¯¼Á£¨º¬ÓÐV2O5¡¢VOSO4¼°²»ÈÜÐÔ²ÐÔü£©£¬¿ÆÑÐÈËÔ±×îÐÂÑÐÖÆÁËÒ»ÖÖÀë×Ó½»»»·¨»ØÊÕ·°µÄй¤ÒÕ£¬»ØÊÕÂÊ´ï91.7%ÒÔÉÏ£®
²¿·Öº¬·°ÎïÖÊÔÚË®ÖеÄÈܽâÐÔÈçϱíËùʾ£º
ÎïÖÊ VOSO4 V2O5 NH4VO3 £¨VO2£©2SO4
ÈܽâÐÔ ¿ÉÈÜ ÄÑÈÜ ÄÑÈÜ Ò×ÈÜ

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Çëд³ö¼ÓÈëNa2SO3ÈÜÒº·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ
V2O5+SO32-+4H+=2VO2++SO42-+2H2O
V2O5+SO32-+4H+=2VO2++SO42-+2H2O
£®
£¨2£©´ß»¯Ñõ»¯ËùʹÓõĴ߻¯¼Á·°´¥Ã½£¨V2O5£©Äܼӿì¶þÑõ»¯ÁòÑõ»¯ËÙÂÊ£¬´Ë¹ý³ÌÖвúÉúÁËÒ»Á¬´®µÄÖмäÌ壨ÈçÏÂ×óͼ£©£®ÆäÖÐa¡¢c¶þ²½µÄ»¯Ñ§·½³Ìʽ¿É±íʾΪ
SO2+V2O5?SO3+V2O4
SO2+V2O5?SO3+V2O4
£¬
4VOSO4+O2?2V2O5+4SO3
4VOSO4+O2?2V2O5+4SO3
£®
£¨3£©¸Ã¹¤ÒÕÖгÁ·¯ÂÊÊÇ»ØÊÕ·°µÄ¹Ø¼üÖ®Ò»£¬³Á·°ÂʵĸߵͳýÊÜÈÜÒºpHÓ°ÏìÍ⣬»¹ÐèÒª¿ØÖÆÂÈ»¯ï§ÏµÊý£¨NH4Cl¼ÓÈëÖÊÁ¿ÓëÁÏÒºÖÐV2O5µÄÖÊÁ¿±È£©ºÍζȣ®
¸ù¾ÝͼÊÔ½¨Òé¿ØÖÆÂÈ»¯ï§ÏµÊýºÍζȣº
4
4
¡¢
80¡æ
80¡æ
£®

£¨4£©¾­¹ýÈÈÖØ·ÖÎö²âµÃ£ºNH4VO3ÔÚ±ºÉÕ¹ý³ÌÖУ¬¹ÌÌåÖÊÁ¿µÄ¼õÉÙÖµ£¨×Ý×ø±ê£©ËæÎ¶ȱ仯µÄÇúÏßÈçÓÒͼËùʾ£®ÔòNH4VO3ÔÚ·Ö½â¹ý³ÌÖÐ
B
B
£®
A£®ÏÈ·Ö½âʧȥH2O£¬ÔÙ·Ö½âʧȥNH3
B£®ÏÈ·Ö½âʧȥNH3£¬ÔÙ·Ö½âʧȥH2O
C£®Í¬Ê±·Ö½âʧȥH2OºÍNH3
D£®Í¬Ê±·Ö½âʧȥH2¡¢N2ºÍH2O£®
£¨2012?½¹×÷һģ£©Ëæ×ŲÄÁÏ¿ÆÑ§µÄ·¢Õ¹£¬½ðÊô·°¼°Æä»¯ºÏÎïµÃµ½ÁËÔ½À´Ô½¹ã·ºµÄÓ¦Ó㬲¢±»ÓþΪ¡°ºÏ½ðµÄάÉúËØ¡±£®Îª»ØÊÕÀûÓú¬·°´ß»¯¼Á£¨º¬ÓÐV2O5¡¢VOSO4¼°²»ÈÜÐÔ²ÐÔü£©£¬¿ÆÑÐÈËÔ±×îÐÂÑÐÖÆÁËÒ»ÖÖÀë×Ó½»»»·¨»ØÊÕ·°µÄй¤ÒÕ£¬»ØÊÕÂÊ´ï91.7%ÒÔÉÏ£®²¿·Öº¬·°ÎïÖÊÔÚË®ÖеÄÈܽâÐÔÈçϱíËùʾ£º
ÎïÖÊ VOSO4 V2O5 NH4VO3 £¨VO2£©2SO4
ÈܽâÐÔ ¿ÉÈÜ ÄÑÈÜ ÄÑÈÜ Ò×ÈÜ

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©23VÔÚÔªËØÖÜÆÚ±íλÓÚµÚ
ËÄ
ËÄ
ÖÜÆÚ
VB
VB
×壮¹¤ÒµÉÏÓÉV2O5Ò±Á¶½ðÊô·°³£ÓÃÂÁÈȼÁ·¨£¬ÆäÓû¯Ñ§·½³Ìʽ±íʾΪ
3V2O5+10Al
 ¸ßΠ
.
 
6V+5Al2O3
3V2O5+10Al
 ¸ßΠ
.
 
6V+5Al2O3
£®
£¨2£©·´Ó¦¢ÙµÄÄ¿µÄÊÇ
½«V2O5ת»¯Îª¿ÉÈÜÐÔµÄVOSO4
½«V2O5ת»¯Îª¿ÉÈÜÐÔµÄVOSO4
£®
£¨3£©¸Ã¹¤ÒÕÖз´Ó¦¢ÛµÄ³ÁµíÂÊ£¨ÓֳƳÁ·¯ÂÊ£©ÊÇ»ØÊÕ·°µÄ¹Ø¼üÖ®Ò»£¬Ð´³ö¸Ã²½·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ
NH4++VO3-=NH4VO3¡ý
NH4++VO3-=NH4VO3¡ý
£®
£¨4£©ÓÃËữµÄH2C2O4ÈÜÒºµÎ¶¨£¨VO2£©2SO4ÈÜÒº£¬ÒԲⶨ·´Ó¦¢ÚºóÈÜÒºÖк¬·°Á¿£¬Ð´³öÅ䯽ºóÍêÕûµÄÀë×Ó·½³Ìʽ£®
2
2
VO2++
1
1
H2C2O4+
2
2
H+¡ú
2
2
VO2++
2
2
 CO2+
2H2O
2H2O
£®
£¨5£©¾­¹ýÈÈÖØ·ÖÎö²âµÃ£ºNH4VO3ÔÚ±ºÉÕ¹ý³ÌÖУ¬¹ÌÌåÖÊÁ¿µÄ¼õÉÙÖµ£¨×Ý×ø±ê£©ËæÎ¶ȱ仯µÄÇúÏßÈçÓÒͼËùʾ£®ÔòNH4VO3ÔÚ·Ö½â¹ý³ÌÖÐ
B
B
£®
A£®ÏÈ·Ö½âʧȥH2O£¬ÔÙ·Ö½âʧȥNH3
B£®ÏÈ·Ö½âʧȥNH3£¬ÔÙ·Ö½âʧȥH2O
C£®Í¬Ê±·Ö½âʧȥH2OºÍNH3
D£®Í¬Ê±·Ö½âʧȥH2¡¢N2ºÍH2O£®
Ëæ×ŲÄÁÏ¿ÆÑ§µÄ·¢Õ¹£¬½ðÊô·°¼°Æä»¯ºÏÎïµÃµ½ÁËÔ½À´Ô½¹ã·ºµÄÓ¦Óã®
£¨1£©·°Ö÷ÒªÓÃÀ´ÖÆÔì·°¸Ö£®·°¸Ö¾ßÓкܸߵÄÄÍÄ¥ËðÐԺͿ¹×²»÷ÐÔ£¬Ô­ÒòÊÇ
 
£¨ÌîÐòºÅ£©
A£®·°ÔÚ³£ÎÂÏ£¬»¯Ñ§»îÆÃÐÔ½ÏÈõ
B£®·°¸Ö±íÃæÐγÉÁËÖÂÃÜÇÒ¼á¹ÌµÄÑõ»¯Ä¤
C£®·°¸Ö½á¹¹½ôÃÜ£¬¾ßÓнϸߵÄÈÍÐÔ¡¢µ¯ÐÔºÍÇ¿¶È
£¨2£©ÔÚζȽϵÍʱ£¬ÈÜÒºÖз°ËáÑλáת»¯Îª½¹·°ËáÑΣº
2VO43-+H2O?V2O74-+2OH-¢Ù
ÔÚζȽϸßʱ£¬½¹·°ËáÑÎÓÖת»¯ÎªÆ«·°ËáÑΣº
V2O74-+H2O?2VO3-+2OH-¢Ú
·´Ó¦¢Úƽºâ³£ÊýµÄ±í´ïʽK=
 
£»
£¨3£©ÒÑ֪ijζÈÏ£º4V£¨s£©+5O2£¨g£©=2V2O5£¨s£©¡÷H=-1551kJ?mol-1
4V£¨s£©+3O2£¨g£©=2V2O3£¨s£©¡÷H=-1219kJ?mol-1
2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¡÷H=-571.6kJ?mol-1
д³öH2»¹Ô­V2O5µÃµ½V2O3µÄÈÈ»¯Ñ§·½³Ìʽ£º
 
£»
£¨4£©Ó÷°ËáÄÆ£¨Na3VO4£©ÈÜÒº£¨º¬PO43-¡¢SiO32-µÈÔÓÖÊÀë×Ó£©ÖƱ¸¸ß´¿V2O5Á÷³ÌÈçÏ£º¾«Ó¢¼Ò½ÌÍø
¢Ù¼ÓÈëþÑΣ¬¼ÓÈȽÁ°è£¬ÆäÖмÓÈȵÄ×÷ÓÃÊÇ
 
£»
¢ÚÔÚ25¡æÊ±£¬Ksp[Mg3£¨PO4£©2]=1.0¡Á10-26£¬Èô¾»»¯ºóÈÜÒºÖеÄMg2+µÄƽºâŨ¶ÈΪ 1.0¡Á10-4 mol?L-1£¬ÔòÈÜÒºÖÐc£¨P043-£©=
 
£»
¢Û¼ÓÈëNH4Cl£¬¼ÓÈȽÁ°è£¬¸Ã²½Öè·´Ó¦µÄÀë×Ó·½³ÌʽΪ
 
£»
¢ÜΪ²â¶¨²úÆ·´¿¶È£¬³ÆÈ¡²úÆ·mg£¬Èܽâºó¶¨ÈÝÔÚ100mLÈÝÁ¿Æ¿ÖУ¬Ã¿´ÎÈ¡5.00mLÈÜÒºÓÚ×¶ÐÎÆ¿ÖУ¬¼ÓÈëÒ»¶¨Á¿µÄÏ¡ÑÎËáºÍKIÈÜÒº£¬ÓÃa mol?L-1Na2S2O3±ê×¼ÈÜ ÒºµÎ¶¨£¬·¢ÉúµÄ·´Ó¦Îª£º
V2O5+6HCl+2KI=2VOCl2+2KCl+I2+3H2O
I2+2Na2S2O3=Na2S4O6+2NaI
ÈôÈý´ÎµÎ¶¨ÏûºÄ±ê×¼ÈÜÒºµÄƽ¾ùÌå»ýΪbmL£¬Ôò¸Ã²úÆ·µÄ´¿¶ÈΪ
 
£¨Óú¬m¡¢a¡¢bµÄ´úÊýʽ±íʾ£©£®

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

¾«Ó¢¼Ò½ÌÍø