ÌâÄ¿ÄÚÈÝ

(15·Ö)·Ï¾É¼îÐÔпÃ̸ɵç³ØÄÚ²¿µÄºÚÉ«ÎïÖÊAÖ÷Òªº¬ÓÐMnO2¡¢NH4Cl¡¢ZnCl2£¬»¹ÓÐÉÙÁ¿µÄFeCl2ºÍÌ¿·Û£¬ÓÃAÖƱ¸¸ß´¿MnCO3µÄÁ÷³ÌͼÈçÏ¡£

£¨1£©¼îÐÔпÃ̸ɵç³ØµÄ¸º¼«²ÄÁÏÊÇ        (Ìѧʽ)¡£
£¨2£© µÚI²½²Ù×÷µÃÂËÔüµÄ³É·ÖÊÇ           £»µÚ¢ò²½²Ù×÷µÄÄ¿µÄÊÇ        ¡£
£¨3£©²½Öè¢óÖÐÖƵÃMnSO4£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ            ¡£
£¨4£©µÚ¢ô²½²Ù×÷ÊǶÔÂËÒºa½øÐÐÉî¶È³ýÔÓ£¬³ýÈ¥Zn2+µÄÀë×Ó·½³ÌʽΪ         ¡£
(ÒÑÖª£ºKsp(MnS)=2.510-13,Ksp(ZnS)=1.610-24)
£¨5£©ÒÑÖª£ºMnCO3ÄÑÈÜÓÚË®ºÍÒÒ´¼£¬³±ÊªÊ±Ò×±»¿ÕÆøÑõ»¯£¬l00oCʱ¿ªÊ¼·Ö½â£»Mn(OH)2¿ªÊ¼³ÁµíʱpHΪ7.7¡£
µÚV²½ÏµÁвÙ×÷¿É°´ÒÔϲ½Öè½øÐУº
²Ù×÷l£º¼ÓÈëÊÔ¼ÁX£¬¿ØÖÆpH<7.7£»   ²Ù×÷2£º¹ýÂË£¬ÓÃÉÙÁ¿Ë®Ï´µÓ2~3´Î£»
²Ù×÷3£º¼ì²âÂËÒº£»                 ²Ù×÷4£ºÓÃÉÙÁ¿ÎÞË®ÒÒ´¼Ï´µÓ2~3´Î£»
²Ù×÷5£ºµÍκæ¸É¡£
¢ÙÊÔ¼ÁXÊÇ      £»
¢Ú²Ù×÷3ÖУ¬ËµÃ÷SO42-Òѳý¸É¾»µÄ·½·¨ÊÇ        ¡£
£¨1£©Zn¡££¨2£©MnO2ºÍÌ¿·Û£»³ýȥ̿·Û¡££¨3£©MnO2+H2O2+H2SO4MnSO4+2O2+H2O£»
£¨4£©MnS+Zn2+ZnS+Mn2+¡££¨5£©¢ÙNa2CO3£»¢ÚÈ¡×îºóÒ»´ÎÏ´µÓÒºÉÙÁ¿ÖÃÓÚÊÔ¹ÜÖУ¬¼ÓÈëÏõËá±µ£¬ÈôÎÞ°×É«³ÁµíÉú³É£¬ËµÃ÷SO42-Òѳý¸É¾»¡£

ÊÔÌâ·ÖÎö£º¸ù¾ÝÌâ¸øÐÅÏ¢ºÍ¹¤ÒÕÁ÷³ÌÖª£¬ºÚÉ«ÎïÖÊAÖ÷Òªº¬ÓÐMnO2¡¢NH4Cl¡¢ZnCl2£¬»¹ÓÐÉÙÁ¿µÄFeCl2ºÍÌ¿·Û£¬MnO2ºÍÌ¿·Û²»ÈÜÓÚË®£¬¾­Ë®½þ¡¢¹ýÂË¡¢Ï´µÓµÃÂËÔüµÄ³É·ÖÊÇMnO2ºÍÌ¿·Û£»MnO2ºÍÌ¿·Û¼ÓÈÈÖÁºãÖØ£¬Ì¼·ÛÓë¿ÕÆøÖеÄÑõÆø·´Ó¦Éú³É̼µÄÑõ»¯Îï¶ø³ýÈ¥£¬ºÚÉ«ÎïÖÊMµÄÖ÷Òª³É·ÖΪMnO2£¬MnO2ÓëÁòËᡢ˫ÑõË®·´Ó¦Éú³ÉÁòËáÃÌ¡¢ÑõÆøºÍË®£¬¾­¹ýϵÁвÙ×÷µÄ¸ß´¿MnCO3¡££¨1£©¸ù¾Ý½Ì²Ä֪ʶ֪£¬¼îÐÔпÃ̸ɵç³ØµÄ¸º¼«²ÄÁÏÊÇZn¡££¨2£© ÓÉÉÏÊö·ÖÎöÖª£¬µÚI²½²Ù×÷µÃÂËÔüµÄ³É·ÖÊÇMnO2ºÍÌ¿·Û£»µÚ¢ò²½²Ù×÷µÄÄ¿µÄÊdzýȥ̿·Û¡££¨3£©MnO2ÓëÁòËᡢ˫ÑõË®·´Ó¦Éú³ÉÁòËáÃÌ¡¢ÑõÆøºÍË®£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪMnO2+H2O2+H2SO4MnSO4+2O2+H2O¡££¨4£©ÓÉKsp(MnS)=2.510-13,Ksp(ZnS)=1.610-24Öª£¬ZnS±ÈMnS¸üÄÑÈÜ£¬ÀûÓóÁµíµÄת»¯³ýÈ¥Zn2+µÄÀë×Ó·½³ÌʽΪMnS+Zn2+ZnS+Mn2+¡£ £¨5£©µÚV²½ÏµÁвÙ×÷ÓÉÁòËáÃÌÖƵøߴ¿MnCO3¡£¢ÙÊÔ¼ÁXÊÇNa2CO3£»¢Ú²Ù×÷3Ϊ¼ìÑé³ÁµíÊÇ·ñÏ´µÓ¸É¾»£¬ËµÃ÷SO42-Òѳý¸É¾»µÄ·½·¨ÊÇÈ¡×îºóÒ»´ÎÏ´µÓÒºÉÙÁ¿ÖÃÓÚÊÔ¹ÜÖУ¬¼ÓÈëÏõËá±µ£¬ÈôÎÞ°×É«³ÁµíÉú³É£¬ËµÃ÷SO42-Òѳý¸É¾»¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨19·Ö£©Ì¼¼°Æ仯ºÏÎïÓй㷺µÄÓÃ;¡£
£¨1£©½«Ë®ÕôÆøͨ¹ýºìÈȵÄ̼¼´¿É²úÉúˮúÆø¡£·´Ó¦Îª£º
C(s)£« H2O(g)  CO(g) £«H2(g) ¦¤H=" +131.3" kJ?mol-1£¬ÒÔÉÏ·´Ó¦´ïµ½Æ½ºâºó£¬ÔÚÌå»ý²»±äµÄÌõ¼þÏ£¬ÒÔÏ´ëÊ©¼Ó¿ì·´Ó¦ËÙÂÊÇÒÓÐÀûÓÚÌá¸ßH2OµÄƽºâת»¯ÂʵÄÊÇ         ¡£(ÌîÐòºÅ)
A£®Éý¸ßζȡ¡B£®Ôö¼Ó̼µÄÓÃÁ¿C£®¼ÓÈë´ß»¯¼ÁD£®ÓÃCOÎüÊÕ¼Á³ýÈ¥CO E.Ôö´óѹǿ
£¨2£©ÓÖÖª£¬C£¨s£©+ CO2£¨g£© 2CO£¨g£© ¡÷H=+172.5kJ?mol-1
д³öC£¨s£©ÓëH2O£¨g£©·´Ó¦Éú³ÉCO2£¨g£©ºÍH2£¨g£©µÄÈÈ»¯Ñ§·½³Ìʽ                            ¡£  
£¨3£©¼×´¼ÊÇÒ»ÖÖȼÁÏ£¬¿ÉÀûÓü״¼Éè¼ÆÒ»¸öȼÁϵç³Ø£¬ÓÃKOHÈÜÒº×÷µç½âÖÊÈÜÒº£¬¶à¿×ʯī×öµç¼«£¬¸Ãµç³Ø¸º¼«·´Ó¦Ê½Îª£º                         ¡£
ÈôÓøõç³ØÌṩµÄµçÄܵç½â600mLNaClÈÜÒº£¬ÉèÓÐ0.01molCH3OHÍêÈ«·Åµç£¬NaCl×ãÁ¿£¬ÇÒµç½â²úÉúµÄCl2È«²¿Òç³ö£¬µç½âÇ°ºóºöÂÔÈÜÒºÌå»ýµÄ±ä»¯£¬Ôòµç½â½áÊøºóËùµÃÈÜÒºµÄpH=                
£¨4£©½«Ò»¶¨Á¿µÄCO(g)ºÍH2O(g)·Ö±ðͨÈëµ½Ìå»ýΪ2.0LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÒÔÏ·´Ó¦£º
CO(g)£«H2O(g)  CO2(g)£«H2(g)£¬µÃµ½ÈçÏÂÊý¾Ý£º
ζÈ/¡æ
ÆðʼÁ¿/mol
ƽºâÁ¿/mol
´ïµ½Æ½ºâËùxÐèʱ¼ä/min
H2O
CO
H2
CO
900
1.0
2.0
0.4
1.6
3.0
ͨ¹ý¼ÆËãÇó³ö¸Ã·´Ó¦µÄƽºâ³£Êý(½á¹û±£ÁôÁ½Î»ÓÐЧÊý×Ö)¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
¸Ä±ä·´Ó¦µÄijһÌõ¼þ£¬·´Ó¦½øÐе½tminʱ£¬²âµÃ»ìºÏÆøÌåÖÐCO2µÄÎïÖʵÄÁ¿Îª0.6 mol¡£ÈôÓÃ200 mL 4.5 mol/LµÄNaOHÈÜÒº½«ÆäÍêÈ«ÎüÊÕ£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£¨ÓÃÒ»¸öÀë×Ó·½³Ìʽ±íʾ£©¡¡¡¡¡¡¡¡¡¡¡£
£¨5£©¹¤ÒµÉú²úÊÇ°ÑˮúÆøÖеĻìºÏÆøÌå¾­¹ý´¦Àíºó»ñµÃµÄ½Ï´¿H2ÓÃÓںϳɰ±¡£ºÏ³É°±·´Ó¦Ô­ÀíΪ£ºN2(g)+3H2(g) 2NH3(g)  ¦¤H£½-92.4kJ?mol-1¡£ÊµÑéÊÒÄ£Ä⻯¹¤Éú²ú£¬·Ö±ðÔÚ²»Í¬ÊµÑéÌõ¼þÏ·´Ó¦£¬N2Ũ¶ÈËæʱ¼ä±ä»¯ÈçÏÂͼ¡£
²»Í¬ÊµÑéÌõ¼þÏ·´Ó¦£¬N2Ũ¶ÈËæʱ¼ä±ä»¯ÈçÏÂͼ1¡£
    
ͼ1                             Í¼2
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÓëʵÑé¢ñ±È½Ï£¬ÊµÑé¢ò¸Ä±äµÄÌõ¼þΪ                 ¡£
¢ÚʵÑé¢ó±ÈʵÑé¢ñµÄζÈÒª¸ß£¬ÆäËüÌõ¼þÏàͬ£¬ÇëÔÚÉÏͼ2Öл­³öʵÑé¢ñºÍʵÑé¢óÖÐNH3Ũ¶ÈËæʱ¼ä±ä»¯µÄʾÒâͼ¡£

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

¾«Ó¢¼Ò½ÌÍø