ÌâÄ¿ÄÚÈÝ

£¨8·Ö£©Ê³ÑÎÖк¬ÓÐÒ»¶¨Á¿µÄþ¡¢ÌúµÈÔÓÖÊÔªËØ£¬Ä³Ñ§Ï°Ð¡×é¶Ô¼ÓµâÑνøÐÐÈçÏÂʵÑ飨ÒÑÖª£ºÑõ»¯ÐÔ£ºIO-3>Fe3+>I2£©£»È¡Ò»¶¨Á¿Ä³¼ÓµâÑΣ¨¿ÉÄܺ¬ÓÐKIO3¡¢KI¡¢Mg2+¡¢Fe3+£©£¬ÓÃÊÊÁ¿ÕôÁóË®ÈÜÒº½â£¬²¢¼ÓÏ¡ÑÎËáËữ£¬½«ËùµÃÈÜÒº·ÖΪ3·Ý¡£µÚÒ»·ÝÊÔÒºÖеμÓKSCNÈÜÒººóÏÔºìÉ«£»µÚ¶þ·ÝÊÔÒºÖмÓ×ãÁ¿KI¹ÌÌ壬ÈÜÒºÏÔµ­»ÆÉ«£¬ÓÃCCl4ÝÍÈ¡£¬Ï²ãÈÜÒºÏÔ×ϺìÉ«£»µÚÈý·ÝÊÔÒºÖмÓÈëÊÊÁ¿KIO3¹ÌÌåºó£¬µÎ¼Óµí·ÛÊÔ¼Á£¬ÈÜÒº²»±ä»¯¡£
£¨1£©ÓÉ´Ë¿ÉÖª£¬¸Ã¼ÓµâÑÎÖе嬵âÎïÖÊÊÇ         £¨Ìѧʽ£©£»
£¨2£©µÚÒ»·ÝÊÔÒº¼ÓKSCNÈÜÒºÏÔºìÉ«£¬Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ                £»
£¨3£©µÚ¶þ·ÝÊÔÒº¼ÓCCl4ÝÍÈ¡ºóϲãÈÜÒºÖÐÏÔ×ϺìÉ«µÄÎïÖÊÊÇ      £¨Ìѧʽ£©£º
£¨4£©µÚ¶þ·ÝÊÔÒºÖмÓÈë×ãÁ¿KI¹ÌÌ壬ÏȺó·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ             ¡£

£¨1£© KIO3  £¨2£©  Fe3++3SCN-=Fe(SCN)3    £¨3£©I2
£¨4£©  IO3£­+5I£­+6H+=3I2+3H2O     2Fe3+ 2I£­=" 2" Fe2++ I2  

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2013?À³Îßһģ£©Ê³ÑÎÖк¬ÓÐÒ»¶¨Á¿µÄþ¡¢ÌúµÈÔÓÖÊ£¬¼ÓµâÑÎÖеâµÄËðʧÖ÷ÒªÊÇÓÉÓÚÔÓÖÊ¡¢Ë®·Ö¡¢¿ÕÆøÖеÄÑõÆøÒÔ¼°¹âÕÕ¡¢ÊÜÈȶøÒýÆðµÄ£®
ÒÑÖª£º¢ÙÑõ»¯ÐÔ£º
IO
-
3
£¾Fe3+£¾I2£»»¹Ô­ÐÔ£º
S2O
2-
3
£¾I-£»¢ÚKI+I2?KI3
£¨1£©Ä³Ñ§Ï°Ð¡×é¶Ô¼ÓµâÑνøÐÐÈçÏÂʵÑ飺ȡһ¶¨Á¿Ä³¼ÓµâÑΣ¨¿ÉÄܺ¬ÓÐKIO3¡¢KI¡¢Mg2+¡¢Fe2+¡¢Fe3+£©£¬ÓÃÊÊÁ¿ÕôÁóË®Èܽ⣬²¢¼ÓÏ¡ÑÎËáËữ£¬½«ËùµÃÈÜÒº·ÖΪ4·Ý£®µÚÒ»·ÝÊÔÒºÖеμÓKSCNÈÜÒººóÏÔºìÉ«£»µÚ¶þ·ÝÊÔÒºÖмÓ×ãÁ¿KI¹ÌÌ壬ÈÜÒºÏÔµ­»ÆÉ«£¬ÓÃCCl4ÝÍÈ¡£¬Ï²ãÈÜÒºÏÔ×ϺìÉ«£»µÚÈý·ÝÊÔÒºÖмÓÈëÊÊÁ¿KIO3¹ÌÌåºó£¬µÎ¼Óµí·ÛÊÔ¼Á£¬ÈÜÒº²»±äÉ«£®ÏòµÚ¶þ·ÝÊÔÒºÖмÓÈë×ãÁ¿KI¹ÌÌåºó£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º
IO3-+5I-+6H+=3I2+3H2O
IO3-+5I-+6H+=3I2+3H2O
¡¢
2Fe3++2I-=2Fe2++I2
2Fe3++2I-=2Fe2++I2

£¨2£©KI×÷Ϊ¼Óµâ¼ÁµÄʳÑÎÔÚ±£´æ¹ý³ÌÖУ¬ÓÉÓÚ¿ÕÆøÖÐÑõÆøµÄ×÷Óã¬ÈÝÒ×ÒýÆðµâµÄËðʧд³ö³±Êª»·¾³ÏÂKIÓëÑõÆø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
O2+4I-+2H2O=2I2+4KOH
O2+4I-+2H2O=2I2+4KOH
£®
£¨3£©Ä³Í¬Ñ§ÎªÌ½¾¿Î¶Ⱥͷ´Ó¦ÎïŨ¶È¶Ô·´Ó¦2IO3-+5SO32-+2H+¨TI2+5SO42-+H2OµÄËÙÂʵÄÓ°Ï죬Éè¼ÆʵÑéÈçϱíËùʾ£º
0.01mol?L-1KIO3
ËáÐÔÈÜÒº£¨º¬µí·Û£©
µÄÌå»ý/mL
0.01mol?L-Na2SO3
ÈÜÒºµÄÌå»ý/mL
H2OµÄ
Ìå»ý/mL
ʵÑé
ζÈ
/¡æ
ÈÜÒº³öÏÖ
À¶É«Ê±Ëù
Ðèʱ¼ä/s
ʵÑé1 5 V1 35 25 t1
ʵÑé2 5 5 40 25 t2
ʵÑé3 5 5 V2 0 t3
±íÖÐÊý¾Ý£ºt1
£¼
£¼
t2£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£»±íÖÐV2=
40
40
mL£®
£¨2011?Õã½­£©Ê³ÑÎÖк¬ÓÐÒ»¶¨Á¿µÄþ¡¢ÌúµÈÔÓÖÊ£¬¼ÓµâÑÎÖеâµÄËðʧÖ÷ÒªÊÇÓÉÓÚÔÓÖÊ¡¢Ë®·Ö¡¢¿ÕÆøÖеÄÑõÆøÒÔ¼°¹âÕÕ¡¢ÊÜÈȶøÒýÆðµÄ£®ÒÑÖª£ºÑõ»¯ÐÔ£ºIO3-£¾Fe3+£¾I2£»»¹Ô­ÐÔ£ºS2O32-£¾I-3I2+6OH-¨TIO3-+5I-+3H2O£»
KI+I2KI3
£¨1£©Ä³Ñ§Ï°Ð¡×é¶Ô¼ÓµâÑνøÐÐÈçÏÂʵÑ飺ȡһ¶¨Á¿Ä³¼ÓµâÑΣ¨¿ÉÄܺ¬ÓÐKIO3¡¢KI¡¢Mg2+¡¢Fe3+£©£¬ÓÃÊÊÁ¿ÕôÁóË®Èܽ⣬²¢¼ÓÏ¡ÑÎËáËữ£¬½«ËùµÃÈÜÒº·ÖΪ3·Ý£®µÚÒ»·ÝÊÔÒºÖеμÓKSCNÈÜÒººóÏÔºìÉ«£»µÚ¶þ·ÝÊÔÒºÖмÓ×ãÁ¿KI¹ÌÌ壬ÈÜÒºÏÔµ­»ÆÉ«£¬ÓÃCCl4ÝÍÈ¡£¬Ï²ãÈÜÒºÏÔ×ϺìÉ«£»µÚÈý·ÝÊÔÒºÖмÓÈëÊÊÁ¿KIO3¹ÌÌåºó£¬µÎ¼Óµí·ÛÊÔ¼Á£¬ÈÜÒº²»±äÉ«£®
¢Ù¼ÓKSCNÈÜÒºÏÔºìÉ«£¬¸ÃºìÉ«ÎïÖÊÊÇ
Fe£¨SCN£©3
Fe£¨SCN£©3
£¨Óû¯Ñ§Ê½±íʾ£©£»CCl4ÖÐÏÔ×ϺìÉ«µÄÎïÖÊÊÇ
£¨Óõç×Óʽ±íʾ£©£®
¢ÚµÚ¶þ·ÝÊÔÒºÖмÓÈë×ãÁ¿KI¹ÌÌåºó£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ
IO3-+5I-+6H+¨T3I2+3H2O
IO3-+5I-+6H+¨T3I2+3H2O
¡¢
2Fe3++2I-¨T2Fe2++I2
2Fe3++2I-¨T2Fe2++I2
£®
£¨2£©KI×÷Ϊ¼Óµâ¼ÁµÄʳÑÎÔÚ±£´æ¹ý³ÌÖУ¬ÓÉÓÚ¿ÕÆøÖÐÑõÆøµÄ×÷Óã¬ÈÝÒ×ÒýÆðµâµÄËðʧ£®Ð´³ö³±Êª»·¾³ÏÂKIÓëÑõÆø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
4KI+O2+2H2O¨T2I2+4KOH
4KI+O2+2H2O¨T2I2+4KOH
£®½«I2ÈÜÓÚKIÈÜÒº£¬ÔÚµÍÎÂÌõ¼þÏ£¬¿ÉÖƵÃKI3?H2O£®¸ÃÎïÖÊ×÷ΪʳÑμӵâ¼ÁÊÇ·ñºÏÊÊ£¿
·ñ
·ñ
£¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©£¬²¢ËµÃ÷ÀíÓÉ
KI3ÊÜÈÈ£¨»ò³±Êª£©Ìõ¼þϲúÉúKIºÍI2£¬KI±»ÑõÆøÑõ»¯£¬I2Ò×Éý»ª
KI3ÊÜÈÈ£¨»ò³±Êª£©Ìõ¼þϲúÉúKIºÍI2£¬KI±»ÑõÆøÑõ»¯£¬I2Ò×Éý»ª
£®
£¨3£©ÎªÁËÌá¸ß¼ÓµâÑΣ¨Ìí¼ÓKI£©µÄÎȶ¨ÐÔ£¬¿É¼ÓÎȶ¨¼Á¼õÉÙµâµÄËðʧ£®ÏÂÁÐÎïÖÊÖÐÓпÉÄÜ×÷ΪÎȶ¨¼ÁµÄÊÇ
AC
AC
£®
A£®Na2S2O3 B£®AlCl3 C£®Na2CO3 D£®NaNO2
£¨4£©¶Ôº¬Fe2+½Ï¶àµÄʳÑΣ¨¼ÙÉè²»º¬Fe3+£©£¬¿ÉÑ¡ÓÃKI×÷Ϊ¼Óµâ¼Á£®ÇëÉè¼ÆʵÑé·½°¸£¬¼ìÑé¸Ã¼ÓµâÑÎÖеÄFe2+£®
È¡×ãÁ¿¸Ã¼ÓµâÑÎÈÜÓÚÕôÁóË®ÖУ¬ÓÃÑÎËáËữ£¬µÎ¼ÓÊÊÁ¿Ñõ»¯¼Á£¨È磺ÂÈË®¡¢¹ýÑõ»¯ÇâµÈ£©£¬ÔٵμÓKSCNÈÜÒº£¬ÈôÏÔѪºìÉ«£¬Ôò¸Ã¼ÓµâÑÎÖдæÔÚFe2+
È¡×ãÁ¿¸Ã¼ÓµâÑÎÈÜÓÚÕôÁóË®ÖУ¬ÓÃÑÎËáËữ£¬µÎ¼ÓÊÊÁ¿Ñõ»¯¼Á£¨È磺ÂÈË®¡¢¹ýÑõ»¯ÇâµÈ£©£¬ÔٵμÓKSCNÈÜÒº£¬ÈôÏÔѪºìÉ«£¬Ôò¸Ã¼ÓµâÑÎÖдæÔÚFe2+
£®

ʳÑÎÖк¬ÓÐÒ»¶¨Á¿µÄþ¡¢ÌúµÈÎïÖÊ£¬¼ÓµâÑÎÖеâµÄËðʧÖ÷ÒªÓÉÓÚÔÓÖÊ¡¢Ë®·Ö¡¢¿ÕÆøÖеÄÑõÆøÒÔ¼°¹âÕÕ¡¢ÊÜÈȶøÒýÆðµÄ£¬ÒÑÖª£º

Ñõ»¯ÐÔ£ºIO3-£¾Fe3£«£¾I2£»»¹Ô­ÐÔ£ºS2O32-£¾I£­

3I2£«6OH£­=5I£­£«IO3-£«3H2O¡¡KI£«IKI3

£¨1£©Ä³Ñ§Ï°Ð¡×é¶Ô¼ÓµâÑνøÐÐÁËÈçÏÂʵÑ飺ȡһ¶¨Á¿¼ÓµâÑÎ(¿ÉÄܺ¬ÓÐKIO3¡¢KI¡¢Mg2£«¡¢Fe3£«)£¬ÓÃÊÊÁ¿ÕôÁóË®Èܽ⣬²¢¼ÓÏ¡ÑÎËáËữ£¬½«ËùµÃÈÜÒº·Ö3·Ý¡£µÚÒ»·ÝÊÔÒºÖеμÓKSCNÈÜÒººóÏÔºìÉ«£»µÚ¶þ·ÝÊÔÒºÖмÓ×ãÁ¿KI¹ÌÌ壬ÈÜÒºÏÔʾµ­»ÆÉ«£¬ÓÃCCl4ÝÍÈ¡£¬Ï²ãÈÜÒºÏÔ×ϺìÉ«£»µÚÈý·ÝÊÔÒºÖмÓÈëÊÊÁ¿KIO3¹ÌÌåºó£¬µÎ¼Óµí·ÛÊÔ¼Á£¬ÈÜÒº²»±äÉ«¡£¢Ù  ¼ÓKSCNÈÜÒºÏÔºìÉ«£¬¸ÃÊÔÒºÖк¬ÓеÄÎïÖÊÊÇ________(ÓÃÀë×Ó·ûºÅ±íʾ)£»CCl4ÖÐÏÔʾ×ϺìÉ«µÄÎïÖÊÊÇ________(Óõç×Óʽ±íʾ)¡£

¢Ú µÚ¶þ·ÝÊÔÒºÖмÓÈë×ãÁ¿KI¹ÌÌåºó£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______________________¡£

£¨2£©½«I2ÈÜÓÚKIÈÜÒº£¬ÔÚµÍÎÂÌõ¼þÏ£¬¿ÉÖƵÃKI3¡¤H2O¡£¸ÃÎïÖÊ×÷ΪʳÑμӵâ¼ÁÊÇ·ñºÏÊÊ£¿(Ìî¡°ÊÇ¡±»ò¡°·ñ¡±)£¬²¢ËµÃ÷ÀíÓÉ£º________________________________¡£

£¨3£©ÎªÁËÌá¸ß¼ÓµâÑÎ(Ìí¼ÓKI)µÄÎȶ¨ÐÔ£¬¿É¼ÓÎȶ¨¼Á¼õÉÙµâµÄËðʧ¡£ÏÂÁÐÎïÖÊÖÐÓпÉÄÜ×÷ΪÎȶ¨¼ÁµÄÊÇ________¡£

A£®Na2S2O3        B£®AlCl3      C£®Na2CO3        D£®NaNO2

 

ʳÑÎÖк¬ÓÐÒ»¶¨Á¿µÄþ¡¢ÌúµÈÎïÖÊ£¬¼ÓµâÑÎÖеâµÄËðʧÖ÷ÒªÓÉÓÚÔÓÖÊ¡¢Ë®·Ý¡¢¿ÕÆøÖеÄÑõÆøÒÔ¼°¹âÕÕ£¬ÊÜÈȶøÒýÆðµÄ£¬ÒÑÖª£º

 

 

 

£¨1£©Ä³Ñ§Ï°Ð¡×é¶Ô¼ÓµâÑνøÐÐÁËÈçÏÂʵÑ飺ȥһ¶¨Á¿Ä³¼ÓµâÑΣ¨¿ÉÄܺ¬ÓÐKIO2¡¢KI¡¢Mg2+¡¢Fe3+£©¡£ÓÃÊÊÁ¿ÕôÁóË®Èܽ⣬²¢¼ÓÏ¡ÑÎËáËữ£¬½«ËùµÃÈÜÒº·Ö3·Ý£¬µÚÒ»·ÝÊÔÒºÖеμÓKSCNÈÜÒººóÏÔºìÉ«£»µÚ¶þ·ÝÊÔÒºÖмÓ×ãÁ¿KI¹ÌÌ壬ÈÜÒºÏÔʾµ­»ÆÉ«£¬ÓÃCCI4ÝÍÈ¡£¬Ï²ãÈÜÒºÏÔ×ϺìÉ«£»µÚÈý·ÝÊÔÒºÖмÓÈë×ãÁ¿µÄKIO£¬¹ÌÌåºó£¬µÎ¼Óµí·ÛÊÔ¼Á£¬ÈÜÒº²»±äÉ«¡£

¢Ù ¸ÃºìÉ«ÎïÖÊÊÇ           £¨Óû¯Ñ§ÊDZíʾ£©£»CCI4ÖÐÏÔʾ×ϺìÉ«µÄÎïÖÊÊÇ       £¨Óõç×Óʽ±íʾ£©¡£

¢Ú µÚ¶þ·ÝÊÔÒºÖмÓÈë×ãÁ¿KI¹ÌÌåºó£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ                      

(2)KI×÷Ϊ¼Óµâ¼ÁµÄʳÑÎÔÚ±£´æ¹ý³ÌÖУ¬ÓÉÓÚ¿ÕÆøÖÐÑõÆøµÄ×÷Óã¬ÈÝÒ×ÒýÆðµâµÄËðʧ¡£Ð´³ö³±Êª»·¾³ÖÐKIÓëÑõÆø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º             ¡£

  ½«I2ÈÜÓÚKIÈÜÒº£¬ÔÚµÍÎÂÌõ¼þÏ£¬¿ÉÖƵ᣸ÃÎïÖÊ×÷ΪʳÑμӵâ¼ÁÊÇ·ñºÏÊÊ£¿

          £¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©£¬²¢ËµÃ÷ÀíÓÉ£º          ¡£

(3)ΪÁËÌá¸ß¼ÓµâÑΣ¨Ìí¼ÓKI£©µÄÎȶ¨ÐÔ£¬¿É¼ÓÎȶ¨¼Á¼õÉÙµâµÄËðʧ¡£ÏÂÁÐÎïÖÊÖÐÓпÉÄÜ×÷ΪÎȶ¨¼ÁµÄÊÇ           ¡£

A. Na2S2O2          B. AICI3         C. Na2CO3           D.NaNO2

(4)¶Ôº¬Fe2+½Ï¶àµÄʳÑΣ¨¼ÙÉè²»º¬Fe3+£©£¬¿ÉÑ¡ÓÃKI×÷Ϊ¼Óµâ¼Á¡£ÇëÉè¼ÆʵÑé·½°¸£¬¼ìÑé¸Ã¼ÓµâÑÎÖеÄFe2+£º           ¡£

 

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

¾«Ó¢¼Ò½ÌÍø