ËÄÑõ»¯ÈýǦË×Ãû¡°Ç¦µ¤¡±»ò¡°ºìµ¤¡±£¬ÓÉÓÚÓÐÑõ»¯ÐÔ±»´óÁ¿µØÓÃÓÚÓÍÆá´¬²°ºÍÇÅÁº¸Ö¼Ü·ÀÐ⣬Æ仯ѧʽ¿ÉдΪ2PbO¡¤PbO2¡£Óû²â¶¨Ä³ÑùÆ·ÖÐËÄÑõ»¯ÈýǦº¬Á¿£¬½øÐÐÈçϲÙ×÷£º

¢Ù³ÆÈ¡ÑùÆ·0.100 0 g£¬¼ÓËáÈܽ⣬µÃµ½º¬Pb2+µÄÈÜÒº¡£

¢ÚÔÚ¼ÓÈÈÌõ¼þÏÂÓùýÁ¿K2Cr2O7½«Pb2+³ÁµíΪPbCrO4£¬ÀäÈ´ºó¹ýÂËÏ´µÓ³Áµí¡£

¢Û½«PbCrO4³ÁµíÓÃËáÈÜÒºÈܽâ(³ÁµíÈܽâµÄÀë×Ó·½³ÌʽΪ£º2PbCrO4+2H+2Pb2+++H2O)£¬¼ÓÈë¹ýÁ¿KI£¬ÔÙÓÃ0.100 0 mol¡¤L-1 Na2S2O3ÈÜÒºµÎ¶¨£¬µ½µÎ¶¨ÖÕµãʱÓÃÈ¥12.00 mL(µÎ¶¨¹ý³ÌÖÐÀë×Ó·½³ÌʽΪ£ºI2+22I-+)¡£Ôò£º

(1)д³ö²½Öè¢ÛÖмÓÈë¹ýÁ¿KIºóÈÜÒºÖз¢ÉúµÄÀë×Ó·´Ó¦·½³Ìʽ______________________¡£

(2)Óñê×¼ÈÜÒºµÎ¶¨Ê±ËùÓõÄָʾ¼ÁÊÇ___________¡£(дÊÔ¼ÁÃû³Æ)

(3)¼ÆËãÊÔÑùÖÐPb3O4µÄÖÊÁ¿·ÖÊý¡£(PbµÄÏà¶ÔÔ­×ÓÖÊÁ¿Îª207.2)

(1)+6I-+14H+2Cr3++3I2+7H2O

(2)µí·Û

(3)Pb3O4¡ª3Pb2+¡ª3/2Cr2O72-¡ª3¡Á3/2I2¡ª3¡Á3/2¡Á2

     685.6                                                              9

     x g                                             0.100 0 mol¡¤L-1¡Á12.00 mL¡Á10-3 L¡¤mL-1

x=0.091 41 g  Pb3O4%£½91.41%

½âÎö£º±¾ÌâÊÇÀûÓÃÑõ»¯»¹Ô­·´Ó¦Ô­Àí½øÐеĵζ¨²â¶¨£¬2 mol PbCr2O4תΪ1 mol £¬ °ÑI-Ñõ»¯ÎªI2£¬I2°ÑÑõ»¯Îª¡£ÒòΪI2ʹµí·Û±äÀ¶£¬Ö¸Ê¾¼ÁΪµí·ÛÈÜÒº£¬µÎ¶¨ÖÁÀ¶É«ÍÊÈ¥´ïµÎ¶¨Öյ㡣¼ÆËãPb3O4µÄÖÊÁ¿·ÖÊý¿É¸ù¾ÝÏÂÁйØϵʽÍÆËã¡£

Pb3O4¡ª¡ªI2¡ª9

685.6                                       9

   x                               0.100 mol¡¤L-1¡Á0.012 L

x=0.091 41 g

w(Pb3O4)=¡Á100%=91.4%¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ËÄÑõ»¯ÈýǦË×Ãû¡°Ç¦µ¤¡±»ò¡°ºìµ¤¡±£¬ÓÉÓÚÓÐÑõ»¯ÐÔ±»´óÁ¿µØÓÃÓÚÓÍÆá´¬²°ºÍÇÅÁº¸Ö¼Ü·ÀÐ⣬Æ仯ѧʽ¿ÉдΪ2PbO?PbO2£®Óû²â¶¨Ä³ÑùÆ·ÖÐËÄÑõ»¯ÈýǦº¬Á¿£¬½øÐÐÈçϲÙ×÷£º
¢Ù³ÆÈ¡ÑùÆ·0.1000g£¬¼Ó»¹Ô­ÐÔËáÈܽ⣬µÃµ½º¬Pb2+µÄÈÜÒº£®
¢ÚÔÚ¼ÓÈÈÌõ¼þÏÂÓùýÁ¿K2Cr2O7½«Pb2+³ÁµíΪPbCrO4£¬ÀäÈ´ºó¹ýÂËÏ´µÓ³Áµí£®
¢Û½«PbCrO4³ÁµíÓÃËáÈÜÒºÈܽ⣨Àë×Ó·½³ÌʽΪ2PbCrO4+2H+=2Pb2++Cr2O72-+H2O£©£¬¼ÓÈë¹ýÁ¿KIÈÜÒº£¬ÔÙÓÃNa2S2O3±ê×¼ÈÜÒºµÎ¶¨µ½Öյ㣨µÎ¶¨¹ý³ÌÖÐÀë×Ó·½³ÌʽΪ£ºI2+2S2O32-=2I-+S4O62-£©£®
£¨1£©ÒÑÖªPbµÄÔ­×ÓÐòÊýÊÇ82£¬Çëд³öPbλÓÚÖÜÆÚ±íµÄµÚ
Áù
Áù
ÖÜÆÚ
¢ôA
¢ôA
×壮
£¨2£©ÔÚPbCrO4×ÇÒºÖмÓÈëÉÙÁ¿ËᣬÔòKsp£¨PbCrO4£©
²»±ä
²»±ä
£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±¡¢¡°²»±ä¡±£©
£¨3£©Ð´³ö²½Öè¢ÛÖмÓÈë¹ýÁ¿KIºóÈÜÒºÖз¢Éú·´Ó¦µÄÀë×Ó·´Ó¦·½³Ìʽ
Cr2O72-+6I-+14H+=2Cr3++3I2+7H2O
Cr2O72-+6I-+14H+=2Cr3++3I2+7H2O
£®
£¨4£©ÓûÇóÊÔÑùÖÐPb3O4µÄÖÊÁ¿·ÖÊý£¬»¹ÐèÒªµÄÊý¾ÝÓÐ
Na2S2O3±ê×¼ÈÜҺŨ¶ÈºÍµÎ¶¨ÈÜÒºÌå»ý
Na2S2O3±ê×¼ÈÜҺŨ¶ÈºÍµÎ¶¨ÈÜÒºÌå»ý
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ËÄÑõ»¯ÈýǦË×Ãû¡°Ç¦µ¤¡±»ò¡°ºìµ¤¡±£¬ÓÉÓÚÓÐÑõ»¯ÐÔ±»´óÁ¿µØÓÃÓÚÓÍÆá´¬²°ºÍÇÅÁº¸Ö¼Ü·ÀÐ⣬Æ仯ѧʽ¿ÉдΪ2PbO?PbO2£®Óû²â¶¨Ä³ÑùÆ·ÖÐËÄÑõ»¯ÈýǦº¬Á¿£¬½øÐÐÈçϲÙ×÷£º
¢Ù³ÆÈ¡ÑùÆ·0.1000g£¬¼Ó»¹Ô­ÐÔËáÈܽ⣬µÃµ½º¬Pb2+µÄÈÜÒº£®
¢ÚÔÚ¼ÓÈÈÌõ¼þÏÂÓùýÁ¿K2Cr2O7½«Pb2+³ÁµíΪPbCrO4£¬ÀäÈ´ºó¹ýÂËÏ´µÓ³Áµí£®
¢Û½«PbCrO4³ÁµíÓÃËáÈÜÒºÈܽ⣨Àë×Ó·½³ÌʽΪ2PbCrO4+2H+=2Pb2++Cr2O72-+H2O£©£¬
¼ÓÈë¹ýÁ¿KIÈÜÒº£¬ÔÙÓÃ0.1000mol?L-1 Na2S2O3ÈÜÒºµÎ¶¨£¬µ½µÎ¶¨ÖÕµãʱÓÃÈ¥12.00mL£¨µÎ¶¨¹ý³ÌÖÐÀë×Ó·½³ÌʽΪ£ºI2+2S2O32-=2I-+S4O62-£©£®
Çë»Ø´ð£º
£¨1£©ÒÑÖªPbµÄÔ­×ÓÐòÊýÊÇ82£¬Çëд³öPbλÓÚÖÜÆÚ±íµÄµÚ
 
ÖÜÆÚ
 
×壮
£¨2£©ÔÚPbCrO4×ÇÒºÖзֱð¼ÓÈëÏÂÁÐÊÔ¼Á£¬Ôò·¢ÉúµÄ±ä»¯ÊÇ£º
¢Ù¼ÓÈëÉÙÁ¿K2CrO4¹ÌÌ壬Ôòc£¨CrO42-£©
 
£¬c£¨Pb2+£©
 
£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±£©£®
¢Ú¼ÓÈëÉÙÁ¿ËᣬÔòKsp£¨PbCrO4£©
 
£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©
£¨3£©Ð´³ö²½Öè¢ÛÖмÓÈë¹ýÁ¿KIºóÈÜÒºÖз¢Éú·´Ó¦µÄÀë×Ó·´Ó¦·½³Ìʽ
 
£®
£¨4£©Óñê×¼ÈÜÒºµÎ¶¨Ê±ËùÓõÄָʾ¼ÁÊÇ
 
£¬µÎ¶¨ÖÕµãʱµÄÏÖÏóÊÇ
 
£®
£¨5£©ÊÔÑùÖÐPb3O4µÄÖÊÁ¿·ÖÊý=
 
£®£¨Pb3O4µÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª685£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ÔÚ21¡æºÍ³ä·Ö½Á°èÏ£¬½«²»Í¬Ìå»ý1.0 mol¡¤L¡ª1 HClÈÜÒººÍδ֪Ũ¶ÈµÄNaOHÈÜÒº»ìºÏ¾ùÔȺó²âÁ¿²¢¼Ç¼ÈÜҺζȣ¬ÊµÑé½á¹ûÈçÏ£º

ÑÎËáµÄÌå»ýV£¨mL£©

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

NaOHµÄÌå»ý£¨mL)

45.0

40.0

35.0

30.0

25.0

20.0

15.0

10.0

5.0

ÈÜҺζÈt£¨¡æ£©

22.2

23.3

24.6

25.8

27.0

27.8

26.1

24.4

22.8

£¨1£©ÔÚ¸ø¶¨µÄ×ø±êͼÉÏ»æ³öÈÜҺζÈÓëÑÎËáÌå»ýµÄ¹Øϵͼ¡£

£¨2£©¼Ù¶¨Ëá¼îÇ¡ºÃÍêÈ«·´Ó¦Ç°ºó£¬ÈÜҺζÈÓëÑÎËáÌå»ý¿ÉÒÔ½üËÆ   

µØÈÏΪ³ÊÏßÐÔ¹Øϵ¡£Çëд³öÈÜҺζÈtÓëÑÎËáÌå»ýVµÄÏßÐÔ 

¹Øϵʽ£¨ÇëÓú¬ÓÐtºÍVµÄʽ×Ó±íʾ£©      ¡¢        ¡£

£¨3£©ËùÓÃNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È=                 ¡£

±¸Óà . ËÄÑõ»¯ÈýǦË×Ãû¡°Ç¦µ¤¡±»ò¡°ºìµ¤¡±£¬ÓÉÓÚÓÐÑõ»¯ÐÔ±»´óÁ¿µØÓÃÓÚÓÍÆá´¬²°ºÍÇÅÁº¸Ö¼Ü·ÀÐ⣬Æ仯ѧʽ¿ÉдΪ2PbO¡¤PbO2¡£Óû²â¶¨Ä³ÑùÆ·ÖÐËÄÑõ»¯ÈýǦº¬Á¿£¬½øÐÐÈçϲÙ×÷£º

¢Ù³ÆÈ¡ÑùÆ·0.1000g£¬¼ÓËáÈܽ⣬µÃµ½º¬Pb2£«µÄÈÜÒº¡£

¢ÚÔÚ¼ÓÈÈÌõ¼þÏÂÓùýÁ¿K2Cr2O7½«Pb2£«³ÁµíΪPbCrO4£¬ÀäÈ´ºó¹ýÂËÏ´µÓ³Áµí¡£

¢Û½«PbCrO4³ÁµíÓÃËáÈÜÒºÈܽ⣨³ÁµíÈܽâµÄÀë×Ó·½³ÌʽΪ£º2PbCrO4+2H£«£½

2Pb2+Cr2O72£­+H2O£©£¬¼ÓÈë¹ýÁ¿KI£¬ÔÙÓÃ0.1000mol¡¤L¨D1 Na2S2O3ÈÜÒºµÎ¶¨£¬µ½µÎ¶¨ÖÕµãʱÓÃ

È¥12.00mL£¨µÎ¶¨¹ý³ÌÖÐÀë×Ó·½³ÌʽΪ£ºI2£«2S2O32£­£½2I£­£«S4O62£­£©¡£

Ôò£º(1)д³ö²½Öè¢ÛÖмÓÈë¹ýÁ¿KIºóÈÜÒºÖз¢ÉúµÄÀë×Ó·´Ó¦·½³Ìʽ                 ¡£

£¨2£©Óñê×¼ÈÜÒºµÎ¶¨Ê±ËùÓõÄָʾ¼ÁÊÇ                   ¡££¨Ð´ÊÔ¼ÁÃû³Æ£©

£¨3£©¼ÆËãÊÔÑùÖÐPb3O4µÄÖÊÁ¿·ÖÊý¡£(PbµÄÏà¶ÔÔ­×ÓÖÊÁ¿Îª207.2)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010¡ª2011ѧÄêºÚÁú½­Ê¡´óÇìʵÑéÖÐѧ¸ß¶þÏÂѧÆÚ¿ªÑ§²âÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

£¨9·Ö£©ËÄÑõ»¯ÈýǦË×Ãû¡°Ç¦µ¤¡±»ò¡°ºìµ¤¡±£¬ÓÉÓÚÓÐÑõ»¯ÐÔ±»´óÁ¿µØÓÃÓÚÓÍÆá´¬²°ºÍÇÅÁº¸Ö¼Ü·ÀÐ⣬Æ仯ѧʽ¿ÉдΪÑõ»¯Îï2PbO¡¤PbO2»òÑÎPb2PbO4¡£Óû²â¶¨Ä³ÑùÆ·ÖÐËÄÑõ»¯ÈýǦº¬Á¿£¬½øÐÐÈçϲÙ×÷£º
¢Ù³ÆÈ¡ÑùÆ·0.1000 g£¬¼Ó»¹Ô­ÐÔËáÈܽ⣬µÃµ½º¬Pb2£«µÄÈÜÒº¡£
¢ÚÔÚ¼ÓÈÈÌõ¼þÏÂÓùýÁ¿K2Cr2O7½«Pb2£«³ÁµíΪPbCrO4£¬ÀäÈ´ºó¹ýÂËÏ´µÓ³Áµí¡£
¢Û½«PbCrO4³ÁµíÓÃËáÈÜÒºÈܽ⣨Àë×Ó·½³ÌʽΪ2PbCrO4£«2H£«£½2Pb2£«£«Cr2O72£­£«H2O£©£¬¼ÓÈë¹ýÁ¿KI
ÈÜÒº£¬ÔÙÓÃNa2S2O3±ê×¼ÈÜÒºµÎ¶¨µ½Öյ㣨µÎ¶¨¹ý³ÌÖÐÀë×Ó·½³ÌʽΪ£ºI2£«2S2O32£­£½2I£­£«S4O62£­£©¡£
(1)ÒÑÖªPbµÄÔ­×ÓÐòÊýÊÇ82£¬Çëд³öPbλÓÚÖÜÆÚ±íµÄµÚ_____ÖÜÆÚ______×å¡£
£¨2£©ÔÚPbCrO4×ÇÒºÖмÓÈëÉÙÁ¿ËᣬÔòKsp(PbCrO4)        £¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±¡¢¡°²»±ä¡±£©
£¨3£©Ð´³ö²½Öè¢ÛÖмÓÈë¹ýÁ¿KIºóÈÜÒºÖз¢Éú·´Ó¦µÄÀë×Ó·´Ó¦·½³Ìʽ               ¡£
£¨4£©ÓûÇóÊÔÑùÖÐPb3O4µÄÖÊÁ¿·ÖÊý£¬»¹ÐèÒªµÄÊý¾ÝÓР                                        ¡£
£¨5£©Fe3O4ÓëPb3O4ÏàËÆ£¬ÆäÑõ»¯ÎïºÍÑεÄÐÎʽ·Ö±ðÊÇ                  ¡¢                ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìºÚÁú½­Ê¡¸ß¶þÏÂѧÆÚ¿ªÑ§²âÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

£¨9·Ö£©ËÄÑõ»¯ÈýǦË×Ãû¡°Ç¦µ¤¡±»ò¡°ºìµ¤¡±£¬ÓÉÓÚÓÐÑõ»¯ÐÔ±»´óÁ¿µØÓÃÓÚÓÍÆá´¬²°ºÍÇÅÁº¸Ö¼Ü·ÀÐ⣬Æ仯ѧʽ¿ÉдΪÑõ»¯Îï2PbO¡¤PbO2»òÑÎPb2PbO4¡£Óû²â¶¨Ä³ÑùÆ·ÖÐËÄÑõ»¯ÈýǦº¬Á¿£¬½øÐÐÈçϲÙ×÷£º

¢Ù³ÆÈ¡ÑùÆ·0.1000 g£¬¼Ó»¹Ô­ÐÔËáÈܽ⣬µÃµ½º¬Pb2£«µÄÈÜÒº¡£

¢ÚÔÚ¼ÓÈÈÌõ¼þÏÂÓùýÁ¿K2Cr2O7½«Pb2£«³ÁµíΪPbCrO4£¬ÀäÈ´ºó¹ýÂËÏ´µÓ³Áµí¡£

¢Û½«PbCrO4³ÁµíÓÃËáÈÜÒºÈܽ⣨Àë×Ó·½³ÌʽΪ2PbCrO4£«2H£«£½2Pb2£«£«Cr2O72£­£«H2O£©£¬¼ÓÈë¹ýÁ¿KI

ÈÜÒº£¬ÔÙÓÃNa2S2O3±ê×¼ÈÜÒºµÎ¶¨µ½Öյ㣨µÎ¶¨¹ý³ÌÖÐÀë×Ó·½³ÌʽΪ£ºI2£«2S2O32£­£½2I£­£«S4O62£­£©¡£

 (1)ÒÑÖªPbµÄÔ­×ÓÐòÊýÊÇ82£¬Çëд³öPbλÓÚÖÜÆÚ±íµÄµÚ_____ÖÜÆÚ______×å¡£

£¨2£©ÔÚPbCrO4×ÇÒºÖмÓÈëÉÙÁ¿ËᣬÔòKsp(PbCrO4)         £¨Ìî¡°Ôö´ó¡±¡¢ ¡°¼õС¡±¡¢¡°²»±ä¡±£©

£¨3£©Ð´³ö²½Öè¢ÛÖмÓÈë¹ýÁ¿KIºóÈÜÒºÖз¢Éú·´Ó¦µÄÀë×Ó·´Ó¦·½³Ìʽ                ¡£

£¨4£©ÓûÇóÊÔÑùÖÐPb3O4µÄÖÊÁ¿·ÖÊý£¬»¹ÐèÒªµÄÊý¾ÝÓР                                         ¡£

£¨5£©Fe3O4ÓëPb3O4ÏàËÆ£¬ÆäÑõ»¯ÎïºÍÑεÄÐÎʽ·Ö±ðÊÇ                  ¡¢                ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸