ÌâÄ¿ÄÚÈÝ

ÒÑ֪ǦÐîµç³ØµÄ¹¤×÷Ô­ÀíΪPb+PbO2+2H2SO4 2PbSO4+2H2O£¬ÏÖÓÃÈçͼװÖýøÐеç½â(µç½âÒº×ãÁ¿)£¬²âµÃµ±Ç¦Ðîµç³ØÖÐתÒÆ0.4 mol µç×ÓʱÌúµç¼«µÄÖÊÁ¿¼õÉÙ11.2 g¡£Çë»Ø´ðÏÂÁÐÎÊÌâ¡£

(1)AÊÇǦÐîµç³ØµÄ     ¼«£¬Ç¦Ðîµç³ØÕý¼«·´Ó¦Ê½Îª            £¬·Åµç¹ý³ÌÖеç½âÒºµÄÃܶȠ      (Ìî¡°¼õС¡±¡¢¡°Ôö´ó¡±»ò¡°²»±ä¡±)¡£
(2)Agµç¼«µÄµç¼«·´Ó¦Ê½ÊÇ        £¬¸Ãµç¼«µÄµç¼«²úÎï¹²        g¡£
(3)Cuµç¼«µÄµç¼«·´Ó¦Ê½ÊÇ        £¬CuSO4ÈÜÒºµÄŨ¶È        (Ìî¡°¼õС¡±¡¢¡°Ôö´ó¡±»ò¡°²»±ä¡±)¡£
(4)Èçͼ±íʾµç½â½øÐйý³ÌÖÐij¸öÁ¿(×Ý×ø±êx)Ëæʱ¼äµÄ±ä»¯ÇúÏߣ¬Ôòx±íʾ            ¡£

a.¸÷UÐιÜÖвúÉúµÄÆøÌåµÄÌå»ý
b.¸÷UÐιÜÖÐÑô¼«ÖÊÁ¿µÄ¼õÉÙÁ¿
c.¸÷UÐιÜÖÐÒõ¼«ÖÊÁ¿µÄÔö¼ÓÁ¿

(1)¸º  PbO2+4H++SO42-+2e-=PbSO4+2H2O  ¼õС
(2)2H++2e-=H2¡ü  0.4
(3)Cu-2e-=Cu2+     ²»±ä  (4)b

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÄϺ£Ä³Ð¡µºÉÏ,½â·Å¾üսʿΪÁËÑ°ÕÒºÏÊʵÄÒûÓÃˮԴ,¶ÔµºÉÏɽȪˮ½øÐзÖÎö»¯Ñé,½á¹ûÏÔʾˮµÄÓ²¶ÈΪ28¡ã(ÊôÓÚӲˮ),Ö÷Òªº¬¸ÆÀë×Ó¡¢Ã¾Àë×Ó¡¢ÂÈÀë×ÓºÍÁòËá¸ùÀë×Ó¡£Çë˼¿¼ÏÂÁÐÎÊÌâ:
(1)¸ÃȪˮÊôÓÚ¡¡¡¡¡¡¡¡Ó²Ë®(Ìîд¡°ÔÝʱ¡±»ò¡°ÓÀ¾Ã¡±)¡£ 
(2)ÈôÒª³ýÈ¥Ca2+¡¢Mg2+¿ÉÒÔÍùË®ÖмÓÈëʯ»ÒºÍ´¿¼î,ÊÔ¼ÁÌí¼ÓʱÏȼӡ¡¡¡¡¡¡¡ºó¼Ó¡¡¡¡¡¡¡¡,Ô­ÒòÊÇ                            ¡£
(3)Ä¿Ç°³£ÓÃÑôÀë×Ó½»»»Ê÷Ö¬ÈçNaR¡¢HRÀ´½øÐÐË®µÄÈí»¯,ÈôʹÓÃHR×÷ΪÑôÀë×Ó½»»»Ê÷Ö¬ÔòË®ÖеÄCa2+¡¢Mg2+Óë½»»»Ê÷Ö¬µÄ¡¡¡¡¡¡¡¡ÆðÀë×Ó½»»»×÷Óá£ÈôʹÓÃNaR×÷ΪÑôÀë×Ó½»»»Ê÷֬ʧЧºó¿É·ÅÈë5%-8%¡¡¡¡¡¡¡¡ÈÜÒºÖÐÔÙÉú¡£
(4)µºÉÏ»¹¿ÉÒÔÓú£Ë®µ­»¯À´»ñµÃµ­Ë®¡£ÏÂÃæÊǺ£Ë®ÀûÓõçÉøÎö·¨»ñµÃµ­Ë®µÄÔ­Àíͼ,ÒÑÖªº£Ë®Öк¬Na+¡¢Cl-¡¢Ca2+¡¢Mg2+¡¢SO42¡ªµÈÀë×Ó,µç¼«Îª¶èÐԵ缫¡£Çë·ÖÎöÏÂÁÐÎÊÌâ:

¢ÙÑôÀë×Ó½»»»Ä¤ÊÇÖ¸¡¡¡¡¡¡¡¡(ÌîA»òB)¡£
¢Úд³öͨµçºóÑô¼«ÇøµÄµç¼«·´Ó¦Ê½:                       ;Òõ¼«ÇøµÄÏÖÏóÊÇ:¡¡                         ¡£

ijÖÐѧ¿ÎÍâÐËȤС×éÓöèÐԵ缫µç½â±¥ºÍʳÑÎË®£¨º¬ÉÙÁ¿Ca2£« ¡¢Mg2£«£©×÷ϵÁÐ̽¾¿£¬×°ÖÃÈçͼËùʾ£º

£¨1£©µç½âʱ£¬¼×ͬѧ·¢Ïֵ缫a¸½½üÈÜÒº³öÏÖ»ë×Ç£¬ÇëÓÃÀë×Ó·½³Ìʽ±íʾԭÒò________________________________________________________________________¡£
£¨2£©Ò»¶Îʱ¼äºó£¬ÄãÈÏΪCÖÐÈÜÒº¿ÉÄܳöÏÖµÄÏÖÏóÊÇ________________________£¬ÇëÓÃÀë×Ó·½³Ìʽ±íʾԭÒò______________________________________¡£
£¨3£©ÊµÑé½áÊøºó£¬ÒÒͬѧ½«AÖеÄÎïÖÊÀäÈ´ºó¼ÓÈëµ½H2SÈÜÒºÖз¢ÏÖÓÐÆøÅݳöÏÖ£¬µ«¼ÓÈ뵽ϡÑÎËáÖÐȴûÓÐÈκÎÏÖÏó¡£ÇëÓû¯Ñ§·½³ÌʽºÍ¼òÒªµÄÎÄ×Ö½âÊÍÔ­Òò£º________________________________________________________________________________________________________________________________________________¡£
£¨4£©Ëæ×Å·´Ó¦µÄ½øÐУ¬ÐËȤС×éµÄͬѧÃǶ¼Ìرð×¢Òâµ½DÖÐÈÜÒººìÉ«Öð½¥ÍÊÈ¥¡£ËûÃǶÔÈÜÒººìÉ«ÍÊÈ¥µÄÖ÷ÒªÔ­ÒòÌá³öÁËÈçϼÙÉ裬ÇëÄãÍê³É¼ÙÉè¶þ¡£
¼ÙÉèÒ»£ºBÖÐÒݳöµÄÆøÌåÓëË®·´Ó¦Éú³ÉµÄÎïÖÊÓÐÇ¿Ñõ»¯ÐÔ£¬Ê¹ºìÉ«Öð½¥ÍÊÈ¥£»
¼ÙÉè¶þ£º___________________________________________________¡£
£¨5£©ÇëÄãÉè¼ÆʵÑéÑéÖ¤ÉÏÊö¼ÙÉèÒ»£¬Ð´³öʵÑé²½Öè¼°½áÂÛ£º________________________________________________________________________________________________________________________________________________________¡£

µç»¯Ñ§Ô­ÀíÔÚ¹¤ÒµÉú²úÖÐÓÐ×ÅÖØÒªµÄ×÷Óã¬ÇëÀûÓÃËùѧ֪ʶ»Ø´ðÓйØÎÊÌâ¡£
£¨1£©Óõç½âµÄ·½·¨½«Áò»¯ÄÆÈÜÒºÑõ»¯Îª¶àÁò»¯ÎïµÄÑо¿¾ßÓÐÖØÒªµÄʵ¼ÊÒâÒ壬½«Áò»¯Îïת±äΪ¶àÁò»¯ÎïÊǵç½â·¨´¦ÀíÁò»¯Çâ·ÏÆøµÄÒ»¸öÖØÒªÄÚÈÝ¡£Èçͼ£¬Êǵç½â²úÉú¶àÁò»¯ÎïµÄʵÑé×°Öãº

¢ÙÒÑÖªÑô¼«µÄ·´Ó¦Îª£¨x£«1£©S2£­=Sx£«S2£­£«2xe£­£¬ÔòÒõ¼«µÄµç¼«·´Ó¦Ê½ÊÇ________________________________________________________________________¡£
µ±·´Ó¦×ªÒÆx molµç×Óʱ£¬²úÉúµÄÆøÌåÌå»ýΪ____________£¨±ê×¼×´¿öÏ£©¡£
¢Ú½«Na2S¡¤9H2OÈÜÓÚË®ÖÐÅäÖÆÁò»¯ÎïÈÜҺʱ£¬Í¨³£ÊÇÔÚµªÆøÆø·ÕÏÂÈܽ⡣ÆäÔ­ÒòÊÇ£¨ÓÃÀë×Ó·´Ó¦·½³Ìʽ±íʾ£©£º___________________________________________________¡£
£¨2£©MnO2ÊÇÒ»ÖÖÖØÒªµÄÎÞ»ú¹¦ÄܲÄÁÏ£¬ÖƱ¸MnO2µÄ·½·¨Ö®Ò»ÊÇÒÔʯīΪµç¼«£¬µç½âËữµÄMnSO4ÈÜÒº£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª______________________¡£ÏÖÒÔǦÐîµç³ØΪµçÔ´µç½âËữµÄMnSO4ÈÜÒº£¬ÈçͼËùʾ£¬Ç¦Ðîµç³ØµÄ×Ü·´Ó¦·½³ÌʽΪ_______________________________________________£¬µ±Ðîµç³ØÖÐÓÐ4 mol H£«±»ÏûºÄʱ£¬Ôòµç·ÖÐͨ¹ýµÄµç×ÓµÄÎïÖʵÄÁ¿Îª________£¬MnO2µÄÀíÂÛ²úÁ¿Îª________g¡£

£¨3£©ÓÃͼµç½â×°ÖÿÉÖƵþßÓо»Ë®×÷ÓõÄFeO42-¡£ÊµÑé¹ý³ÌÖУ¬Á½¼«¾ùÓÐÆøÌå²úÉú£¬Y¼«ÇøÈÜÒºÖð½¥Éú³ÉFeO42-¡£

¢Ùµç½â¹ý³ÌÖУ¬X¼«ÇøÈÜÒºµÄpH________£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£©¡£
¢Úµç½â¹ý³ÌÖУ¬Y¼«·¢ÉúµÄµç¼«·´Ó¦ÎªFe£­6e£­£«8OH£­=FeO42-£«4H2OºÍ________________________________________________________________________£¬
ÈôÔÚX¼«ÊÕ¼¯µ½672 mLÆøÌ壬ÔÚY¼«ÊÕ¼¯µ½168 mLÆøÌ壨¾ùÒÑÕÛËãΪ±ê×¼×´¿öʱÆøÌåÌå»ý£©£¬ÔòYµç¼«£¨Ìúµç¼«£©ÖÊÁ¿¼õÉÙ________g¡£

ijÖÖ̼ËáÃÌ¿óµÄÖ÷Òª³É·ÖÓÐMnCO3¡¢MnO2¡¢FeCO3¡¢MgO¡¢SiO2¡¢Al2O3µÈ¡£ÒÑ֪̼ËáÃÌÄÑÈÜÓÚË®¡£Ò»ÖÖÔËÓÃÒõÀë×ÓĤµç½â·¨µÄм¼Êõ¿ÉÓÃÓÚ´Ó̼ËáÃÌ¿óÖÐÌáÈ¡½ðÊôÃÌ£¬Á÷³ÌÈçÏ£º

ÒõÀë×ÓĤ·¨µç½â×°ÖÃÈçͼËùʾ£º

£¨1£©Ð´³öÓÃÏ¡ÁòËáÈܽâ̼ËáÃÌ·´Ó¦µÄÀë×Ó·½³Ìʽ                               ¡£
£¨2£©ÔÚ½þ³öÒºÀïÃÌÔªËØÖ»ÒÔMn2+µÄÐÎʽ´æÔÚ£¬ÇÒÂËÔüÖÐÒ²ÎÞMnO2£¬Çë½âÊÍÔ­Òò                                         .
£¨3£©£¨5·Ö£©ÒÑÖª²»Í¬½ðÊôÀë×ÓÉú³ÉÇâÑõ»¯Îï³ÁµíËùÐèµÄpHÈçÏÂ±í£º

¼Ó°±Ë®µ÷½ÚÈÜÒºµÄpHµÈÓÚ6£¬ÔòÂËÔüµÄ³É·ÖÊÇ                             £¬ÂËÒºÖк¬ÓеÄÑôÀë×ÓÓÐH+ºÍ                       ¡£
£¨4£©µç½â×°ÖÃÖмýÍ·±íʾÈÜÒºÖÐÒõÀë×ÓÒƶ¯µÄ·½Ïò£¬ÔòAµç¼«ÊÇ    ¼«¡£Êµ¼ÊÉú²úÖУ¬Ñô¼«ÒÔÏ¡ÁòËáΪµç½âÒº£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª                     ¡£
£¨5£©¸Ã¹¤ÒÕÖ®ËùÒÔ²ÉÓÃÒõÀë×Ó½»»»Ä¤£¬ÊÇΪÁË·ÀÖ¹Mn2+½øÈëÑô¼«Çø·¢Éú¸±·´Ó¦Éú³ÉMnO2Ôì³É×ÊÀË·Ñ£¬Ð´³ö¸Ã¸±·´Ó¦µÄµç¼«·´Ó¦Ê½                                  ¡£

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

¾«Ó¢¼Ò½ÌÍø