ÌâÄ¿ÄÚÈÝ

µªÑõ»¯ÎïÊÇ´óÆøÎÛȾÎïÖ®Ò»£¬Ïû³ýµªÑõ»¯ÎïµÄ·½·¨ÓжàÖÖ¡£

£¨1£©ÀûÓü×Íé´ß»¯»¹Ô­µªÑõ»¯Îï¡£ÒÑÖª£º
CH4 (g)£«4NO2(g)£½4NO(g)£«CO2(g)£«2H2O(g)   ¡÷H £½£­574 kJ/mol
CH4(g)£«4NO(g) £½ 2N2(g)£«CO2(g)£«2H2O(g)  ¡÷H £½£­1160 kJ/mol
ÔòCH4 ½«NO2 »¹Ô­ÎªN2 µÄÈÈ»¯Ñ§·½³ÌʽΪ                               ¡£       
£¨2£©ÀûÓÃNH3´ß»¯»¹Ô­µªÑõ»¯ÎSCR¼¼Êõ)¡£¸Ã¼¼ÊõÊÇÄ¿Ç°Ó¦ÓÃ×î¹ã·ºµÄÑÌÆøµªÑõ»¯ÎïÍѳý¼¼Êõ¡£ ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2NH3(g)£«NO(g)£«NO2(g)   2N2(g)£«3H2O(g)¡¡¦¤H < 0
ΪÌá¸ßµªÑõ»¯ÎïµÄת»¯ÂʿɲÉÈ¡µÄ´ëÊ©ÊÇ                 £¨Ð´³ö1Ìõ¼´¿É£©¡£
£¨3£©ÀûÓÃClO2Ñõ»¯µªÑõ»¯Îï¡£Æäת»¯Á÷³ÌÈçÏ£º
NONO2N2
ÒÑÖª·´Ó¦¢ñµÄ»¯Ñ§·½³ÌʽΪ2NO+ ClO2 + H2O £½ NO2 + HNO3 + HCl£¬Ôò·´Ó¦¢òµÄ»¯Ñ§·½³ÌʽÊÇ                £»ÈôÉú³É11.2 L N2£¨±ê×¼×´¿ö£©£¬ÔòÏûºÄClO2          g ¡£
£¨4£©ÀûÓÃCO´ß»¯»¹Ô­µªÑõ»¯ÎïÒ²¿ÉÒÔ´ïµ½Ïû³ýÎÛȾµÄÄ¿µÄ¡£
ÒÑÖªÖÊÁ¿Ò»¶¨Ê±£¬Ôö´ó¹ÌÌå´ß»¯¼ÁµÄ±íÃæ»ý¿ÉÌá¸ß»¯Ñ§·´Ó¦ËÙÂÊ¡£ÈçͼÊÇ·´Ó¦2NO(g) + 2CO(g)2CO2(g)+ N2(g) ÖÐNOµÄŨ¶ÈËæζÈ(T)¡¢µÈÖÊÁ¿´ß»¯¼Á±íÃæ»ý(S)ºÍʱ¼ä(t)µÄ±ä»¯ÇúÏß¡£¾Ý´ËÅжϸ÷´Ó¦µÄ¡÷H     0 (Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°ÎÞ·¨È·¶¨¡±)£»´ß»¯¼Á±íÃæ»ýS1     S2 (Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°ÎÞ·¨È·¶¨¡±)¡£

£¨1£©CH4 (g)£«2NO2(g)£½N2(g)£«CO2(g)£«2H2O(g)   ¡÷H £½£­867 kJ/mol    £¨3·Ö£©
£¨2£©Ôö´óNH3µÄŨ¶È»ò¼õС·´Ó¦ÌåϵµÄѹǿ»ò½µµÍ·´Ó¦ÌåϵµÄζȵȣ¨ºÏÀí´ð°¸Ò²¼Æ·Ö£©
£¨3£©2NO2 + 4 Na2SO3 = N2 + 4 Na2SO4     67.5
£¨4£©£¼      £¾

½âÎöÊÔÌâ·ÖÎö£º£¨1£©½«ÒÑÖªµÄÁ½¸öÈÈ»¯Ñ§·½³ÌʽÏà¼Ó³ýÒÔ2¼´µÃ£¬´ð°¸ÎªCH4 (g)£«2NO2(g)£½N2(g)£«CO2(g)£«2H2O(g)   ¡÷H £½£­867 kJ/mol 
£¨2£©Ìá¸ßµªÑõ»¯ÎïµÄת»¯Âʼ´Ê¹·´Ó¦ÕýÏò½øÐУ¬¸ù¾ÝÀÕÏÄÌØÁÐÔ­Àí£¬¿É²ÉÈ¡Ôö´óNH3µÄŨ¶È»ò¼õС·´Ó¦ÌåϵµÄѹǿ»ò½µµÍ·´Ó¦ÌåϵµÄζȵȴëÊ©£»
£¨3£©·´Ó¦¢òÖз´Ó¦ÎïΪNO2¡¢Na2SO3£¬²úÎïÖ®Ò»ÓÐN2£¬¸ù¾ÝÑõ»¯»¹Ô­·´Ó¦ÀíÂÛ£¬¿ÉÍƶÏÁíÒ»²úÎïΪNa2SO4»¯Ñ§·½³ÌʽÊÇ2NO2 + 4 Na2SO3 = N2 + 4 Na2SO4£»ÈôÉú³É11.2 L N2£¨±ê×¼×´¿ö£©£¬ÆäÎïÖʵÄÁ¿Îª0.5mol£¬ÏûºÄNO2µÄÎïÖʵÄÁ¿Îª1mol,´Ó¶ø¿É¼ÆËã³öÏûºÄClO2µÄÖÊÁ¿Îª67.5g;
£¨4£©¸ù¾Ý»¯Ñ§Æ½ºâÖеġ°ÏȹÕÏÈƽ¡±¹æÂÉ£¬Î¶ȸߵÄÏÈ´ïƽºâ£¬±íÃæ»ý´óµÄËÙÂʿ죬ÏÈ´ïƽºâ£¬ËùÒÔT2>T1,S1>S2,ËæζÈÉý¸ß£¬COµÄŨ¶ÈÔö´ó£¬ËµÃ÷ÉýÎÂƽºâÄæÏòÒƶ¯£¬ÕýÏòÊÇ·ÅÈÈ·´Ó¦¡÷H<0¡£
¿¼µã£º¿¼²é¸Ç˹¶¨ÂɵÄÓ¦Óá¢Æ½ºâÒƶ¯ÓëÍâ½çÌõ¼þµÄ¹Øϵ¡¢»¯Ñ§·½³ÌʽµÄÍƶϼ°¼ÆËã¡¢»¯Ñ§Æ½ºâͼÏñµÄ·ÖÎöÄÜÁ¦

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ʵÑéÊÒ²ÉÓÃMgCl2¡¢AlCl3µÄ»ìºÏÈÜÒºÓë¹ýÁ¿°±Ë®·´Ó¦ÖƱ¸MgAl2O4¶þÖ÷ÒªÁ÷³ÌÈçÏÂ:

¢ÅÖƱ¸MgAl2O4¹ý³ÌÖУ¬¸ßαºÉÕʱ·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ                        ¡£
¢ÆÈçͼËùʾ£¬¹ýÂ˲Ù×÷ÖеÄÒ»´¦´íÎóÊÇ                          ¡£ÅжÏÁ÷³ÌÖгÁµíÊÇ·ñÏ´¾»ËùÓõÄÊÔ¼ÁÊÇ           ¡£¸ßαºÉÕʱ£¬ÓÃÓÚÊ¢·Å¹ÌÌåµÄÒÇÆ÷Ãû³ÆÊÇ                      ¡£

¢ÇÔÚ25¡æÏ£¬ÏòŨ¶È¾ùΪ0£®01 mol?L-1µÄMgCl2ºÍAlCl3»ìºÏÈÜÒºÖÐÖðµÎ¼ÓÈ백ˮ£¬ÏÈÉú³É_______³Áµí(Ìѧʽ)£¬Éú³É¸Ã³ÁµíµÄÀë×Ó·½³Ìʽ_______  ______________
£¨ÒÑÖª25¡æʱKsp[Mg(OH)2]=1.8¡Á10-11£¬Ksp[Al(OH)3]=3¡Á10 -34¡££©
¢ÈÎÞË®AlCl3(183¡æÉý»ª)Óö³±Êª¿ÕÆø¼´²úÉú´óÁ¿°×Îí£¬ÊµÑéÊÒ¿ÉÓÃÏÂÁÐ×°ÖÃÖƱ¸¡£

×°ÖÃBÖÐÊ¢·Å±¥ºÍNaClÈÜÒº,¸Ã×°ÖõÄÖ÷Òª×÷ÓÃÊÇ                 £» FÖÐÊÔ¼ÁµÄ×÷ÓÃÊÇ           £»ÓÃÒ»¼þÒÇÆ÷×°ÌîÊʵ±ÊÔ¼ÁºóÒ²¿ÉÆðµ½FºÍGµÄ×÷Óã¬Ëù×°ÌîµÄÊÔ¼ÁΪ                   ¡£
¢É½«Mg¡¢Cu×é³ÉµÄ3.92g»ìºÏÎïͶÈë¹ýÁ¿Ï¡ÏõËáÖУ¬³ä·Ö·´Ó¦ºó£¬¹ÌÌåÍêÈ«ÈܽâʱÊÕ¼¯µ½»¹Ô­²úÎïNOÆøÌå1.792L£¨±ê×¼×´¿ö£©£¬Ïò·´Ó¦ºóµÄÈÜÒºÖмÓÈë4mol/LµÄNaOHÈÜÒº80mLʱ½ðÊôÀë×ÓÇ¡ºÃÍêÈ«³Áµí¡£ÔòÐγɳÁµíµÄÖÊÁ¿Îª               g¡£

ÖظõËá¼ØÊǹ¤ÒµÉú²úºÍʵÑéÊÒµÄÖØÒªÑõ»¯¼Á£¬¹¤ÒµÉϳ£ÓøõÌú¿ó£¨Ö÷Òª³É·ÝΪFeO¡¤Cr2O3£¬ÒÔ¼°SiO2¡¢Al2O3µÈÔÓÖÊ£©ÎªÔ­ÁÏÉú²ú£¬ÊµÑéÊÒÄ£Ä⹤ҵ·¨ÓøõÌú¿óÖÆK2Cr2O7µÄÖ÷Òª¹¤ÒÕÈçÏ£º

·´Ó¦Æ÷ÖÐÖ÷Òª·¢ÉúµÄ·´Ó¦Îª£º
¢ñ.FeO¡¤Cr2O3£«NaOH£«KClO3¡úNa2CrO4£«Fe2O3£«H2O+KCl£¨Î´Åäƽ£©
¢ò.Na2CO3+SiO2¦¤ Na2SiO3+CO2¡ü
¢ó.Al2O3+2NaOH¦¤ 2NaAlO2+H2O
ÔÚ²½Öè¢ÛÖн«ÈÜÒºpHµ÷½Úµ½7~8¿ÉÒÔ½«SiO32-ºÍAlO2-ת»¯ÎªÏàÓ¦µÄ³Áµí³ýÈ¥¡£
£¨1£©ÔÚ·´Ó¦¢ñÖÐÑõ»¯¼ÁÊÇ________£¬ÈôÓÐ245g KClO3²Î¼Ó·´Ó¦£¬ÔòתÒƵĵç×ÓÊýΪ_____________¡£
£¨2£©·´Ó¦Æ÷ÖÐÉú³ÉµÄFe2O3ÓֿɺÍNa2CO3·´Ó¦µÃµ½Ò»ÖÖĦ¶ûÖÊÁ¿Îª111g/molµÄ»¯ºÏÎÄÜÇ¿ÁÒË®½â£¬ÔÚ²Ù×÷¢ÚÉú³É³Áµí¶ø³ýÈ¥£¬Ð´³öÉú³É¸Ã»¯ºÏÎïµÄ»¯Ñ§·´Ó¦·½³Ìʽ_____________________________
___________________________¡£
£¨3£©²Ù×÷¢ÜÄ¿µÄÊǽ«CrO42-ת»¯ÎªCr2O72-£¬ÆäÏÖÏóΪ__________________________£¬Àë×Ó·½³ÌʽΪ_______________________________________¡£
£¨4£©ÇëÑ¡ÓúÏÊʵķ½·¨½øÒ»²½Ìá´¿´Ö²úÆ·ÖظõËá¼Ø__________£¨Ìî×Öĸ£©
A£®Öؽᾧ        B£®ÝÍÈ¡·ÖÒº        C£®ÕôÁó
£¨5£©·ÖÎö²úÆ·ÖÐK2Cr2O7µÄ´¿¶ÈÊÇÀûÓÃÁòËáËữµÄK2Cr2O7½«KIÑõ»¯³ÉI2,È»ºóÀûÓÃÏà¹ØÎïÖʲâ³öI2µÄÁ¿´Ó¶ø»ñµÃK2Cr2O7µÄÁ¿£¬Ð´³öËữµÄK2Cr2O7ÓëKI·´Ó¦µÄ»¯Ñ§·½³Ìʽ________________________¡£

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

¾«Ó¢¼Ò½ÌÍø