ÌâÄ¿ÄÚÈÝ

¸õÌú¿óÖ÷Òª³É·ÖΪFeO¡¤Cr2O3£¬»¹º¬ÓÐÔÓÖÊAl2O3¡£Ò»°ã¸õÌú¿óÖÐCr2O3ÖÊÁ¿·ÖÊýԼΪ40£¥¡£ÓɸõÌú¿óÖƱ¸ÖظõËá¼ØµÄ·½·¨ÈçÏ£º

ÒÑÖª£º
¢Ù4FeO¡¤Cr2O3£«8Na2CO3£«7O28Na2CrO4£«2 Fe2O3£«8CO2¡ü£»
¢ÚNa2CO3£«Al2O32NaAlO2£«CO2¡ü£»
¢Û Cr2O72£­£«H2O 2CrO42£­£«2H+
¸ù¾ÝÌâÒâ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²Ù×÷IΪ              £¬¹ÌÌåXÖÐÖ÷Òªº¬ÓР          £¨Ìîд»¯Ñ§Ê½£©£»Òª¼ì²âËữ²Ù×÷ÖÐÈÜÒºµÄpHÊÇ·ñµÈÓÚ4.5£¬Ó¦¸ÃʹÓà             £¨ÌîдÒÇÆ÷»òÊÔ¼ÁÃû³Æ£©¡£
£¨2£©Ëữ²½ÖèÓô×Ëáµ÷½ÚÈÜÒºpH<5£¬ÆäÄ¿µÄÊÇ                              ¡£
£¨3£©²Ù×÷¢óÓжಽ×é³É£¬»ñµÃK2Cr2O7¾§ÌåµÄ²Ù×÷ÒÀ´ÎÊÇ£º¼ÓÈëKCl¹ÌÌå¡¢Õô·¢Å¨Ëõ¡¢        ¡¢¹ýÂË¡¢        ¡¢¸ÉÔï¡£
£¨4£©¹ÌÌåYÖÐÖ÷Òªº¬ÓÐÇâÑõ»¯ÂÁ£¬Çëд³öµ÷½ÚÈÜÒºµÄpH=7~8ʱÉú³ÉÇâÑõ»¯ÂÁµÄÀë×Ó·½³Ìʽ                                    ¡£

£¨1£©¹ýÂË£¨2·Ö£©£»    Fe2O3£¨2·Ö£©£»         pH¼Æ»ò¾«ÃÜpHÊÔÖ½ºÍ²£Á§°ô£¨2·Ö£©£»
£¨2£©Ôö¼ÓH+Ũ¶È¿Éʹ»¯Ñ§Æ½ºâCr2O72£­£«H2O 2CrO42£­£«2H+Ïò×óÒƶ¯£¬Ê¹CrO42£­×ª»¯ÎªCr2O72£­£¨ºÏÀí»Ø´ð¾ù¸ø·Ö£©£¨2·Ö£©£»
£¨3£©ÀäÈ´½á¾§£¨2·Ö£©£»        Ï´µÓ£¨2·Ö£©£»
£¨4£©CH3COOH£«AlO2£­£«H2O = Al(OH)3¡ý£«CH3COO£­ £¨2·Ö£©

ÊÔÌâ·ÖÎö£º¸õÌú¿ó±ºÉÕºóµÄ²úÎïÓÐNa2CrO4¡¢Fe2O3¡¢NaAlO2£¬¼ÓË®½þÈ¡£¬²»ÈÜÓÚË®µÄÓÐFe2O3£¬Ò²¾ÍÊÇX£¬¹ýÂË·ÖÀ룬¼Ó´×Ëáµ÷PHµ½7~8£¬Ê¹AlO2-ת»¯ÎªAl(OH)3³Áµí£¬¹ýÂË·ÖÀ룬Ҳ¾ÍÊÇY£¬¼ÌÐøµ÷PH<5£¬Ê¹CrO42£­×ª»¯ÎªCr2O72£­¼ÓÈëKClÕô·¢½á¾§¼ÈµÃÖظõËá¼Ø
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
·Ï¾ÉÎïµÄ»ØÊÕÀûÓüÈÓÐÀûÓÚ½ÚÔ¼×ÊÔ´£¬ÓÖÓÐÀûÓÚ±£»¤»·¾³¡£Ä³Ñо¿Ð¡×éͬѧÒԷϾÉпÃ̸ɵç³ØΪԭÁÏ£¬½«·Ï¾Éµç³Øº¬Ð¿²¿·Öת»¯³ÉZnSO4¡¤7H2O£¬º¬Ã̲¿·Öת»¯³É´¿¶È½Ï¸ßµÄMnO2£¬½«NH4ClÈÜÒºÓ¦ÓÃÓÚ»¯·ÊÉú²úÖУ¬ÊµÑéÁ÷³ÌÈçÏ£º

£¨1£©²Ù×÷¢ÚÖÐËùÓõļÓÈÈÒÇÆ÷Ӧѡ          £¨Ñ¡Ìî¡°Õô·¢Ãó¡±»ò¡°ÛáÛö¡±£©¡£
£¨2£©½«ÈÜÒºA´¦ÀíµÄµÚÒ»²½ÊǼÓÈ백ˮµ÷½ÚpHΪ9£¬Ê¹ÆäÖеÄFe3+ºÍZn2+³Áµí£¬Çëд³ö°±Ë®ºÍFe3+·´Ó¦µÄÀë×Ó·½³Ìʽ                        ¡£
£¨3£©²Ù×÷¢ÝÊÇΪÁ˳ýÈ¥ÈÜÒºÖеÄZn2+¡£ÒÑÖª25¡æʱ£¬
NH3¡¤H2OµÄKb
Zn2+ÍêÈ«³ÁµíµÄpH
Zn(OH)2ÈÜÓÚ¼îµÄpH
1.8¡Á10£­5
8.9
£¾11
 
ÓÉÉϱíÊý¾Ý·ÖÎöÓ¦µ÷½ÚÈÜÒºpH×îºÃΪ            £¨ÌîÐòºÅ£©¡£
a£®9                 b£®10               c£®11
£¨4£© MnO2¾«´¦ÀíµÄÖ÷Òª²½Ö裺
²½Öè1£ºÓÃ3%H2O2ºÍ6.0mol/LµÄH2SO4µÄ»ìºÍÒº½«´ÖMnO2Èܽ⣬¼ÓÈȳýÈ¥¹ýÁ¿H2O2£¬µÃMnSO4ÈÜÒº£¨º¬ÉÙÁ¿Fe3+£©¡£·´Ó¦Éú³ÉMnSO4µÄÀë×Ó·½³ÌʽΪ              £»
²½Öè2£ºÀäÈ´ÖÁÊÒΣ¬µÎ¼Ó10%°±Ë®µ÷½ÚpHΪ6£¬Ê¹Fe3+³ÁµíÍêÈ«£¬ÔÙ¼Ó»îÐÔÌ¿½Á°è£¬³éÂË¡£¼Ó»îÐÔÌ¿µÄ×÷ÓÃÊÇ                         £»
²½Öè3£ºÏòÂËÒºÖеμÓ0.5mol/LµÄNa2CO3ÈÜÒº£¬µ÷½ÚpHÖÁ7£¬Â˳ö³Áµí¡¢Ï´µÓ¡¢¸ÉÔ×ÆÉÕÖÁºÚºÖÉ«£¬Éú³ÉMnO2¡£×ÆÉÕ¹ý³ÌÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ               ¡£
£¨5£© ²éÎÄÏ׿ÉÖª£¬´ÖMnO2µÄÈܽ⻹¿ÉÒÔÓÃÑÎËá»òÕßÏõËá½þÅÝ£¬È»ºóÖÆÈ¡MnCO3¹ÌÌå¡£
¢ÙÔÚÑÎËáºÍÏõËáÈÜÒºµÄŨ¶È¾ùΪ5mol/L¡¢Ìå»ýÏàµÈºÍ×î¼Ñ½þÅÝʱ¼äÏ£¬½þÅÝζȶÔMnCO3²úÂʵÄÓ°ÏìÈçͼ4£¬ÓÉͼ¿´³öÁ½ÖÖËáµÄ×î¼Ñ½þÅÝζȶ¼ÔÚ         ¡æ×óÓÒ£»

¢ÚÔÚ×î¼Ñζȡ¢×î¼Ñ½þÅÝʱ¼äºÍÌå»ýÏàµÈÏ£¬ËáµÄŨ¶È¶ÔMnCO3²úÂʵÄÓ°ÏìÈçͼ5£¬ÓÉͼ¿´³öÏõËáµÄ×î¼ÑŨ¶ÈӦѡÔñ            mol/L×óÓÒ¡£

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

¾«Ó¢¼Ò½ÌÍø