ÌâÄ¿ÄÚÈÝ


ʵÑéÊÒ´Óº¬µâ·ÏÒº£¨³ýH2OÍ⣬º¬ÓÐCCl4¡¢I2¡¢I£­µÈ£©ÖлØÊյ⣬ÆäʵÑé¹ý³ÌÈçÏ£º

£¨1£©Ïòº¬µâ·ÏÒºÖмÓÈëÉÔ¹ýÁ¿µÄNa2SO3ÈÜÒº£¬½«·ÏÒºÖеÄI2»¹Ô­ÎªI£­£¬ÆäÀë×Ó·½³ÌʽΪ                £»¸Ã²Ù×÷½«I2»¹Ô­ÎªI£­µÄÄ¿µÄÊÇ                ¡£

£¨2£©²Ù×÷XµÄÃû³ÆÎª            ¡£

£¨3£©Ñõ»¯Ê±£¬ÔÚÈý¾±ÉÕÆ¿Öн«º¬I£­µÄË®ÈÜÒºÓÃÑÎËáµ÷ÖÁpHԼΪ2£¬»ºÂýͨÈëCl2£¬ÔÚ400C×óÓÒ·´Ó¦£¨ÊµÑé×°ÖÃÈçͼËùʾ£©¡£ÊµÑé¿ØÖÆÔڽϵÍζÈϽøÐеÄÔ­ÒòÊÇ                £»×¶ÐÎÆ¿ÀïÊ¢·ÅµÄÈÜҺΪ               ¡£

£¨4£©ÒÑÖª£º5SO32—+2IO3+2H£«I2+5SO42—+H2O

ijº¬µâ·ÏË®£¨pHԼΪ8£©ÖÐÒ»¶¨´æÔÚI2£¬¿ÉÄÜ´æÔÚI£­¡¢IO3ÖеÄÒ»ÖÖ»òÁ½ÖÖ¡£Çë²¹³äÍêÕû¼ìÑ麬µâ·ÏË®ÖÐÊÇ·ñº¬ÓÐI£­¡¢IO3µÄʵÑé·½°¸£ºÈ¡ÊÊÁ¿º¬µâ·ÏË®ÓÃCCl4¶à´ÎÝÍÈ¡¡¢·ÖÒº£¬Ö±µ½Ë®²ãÓõí·ÛÈÜÒº¼ìÑé²»³öµâµ¥ÖÊ´æÔÚ£»                                     ¡£

ʵÑéÖпɹ©Ñ¡ÔñµÄÊÔ¼Á£ºÏ¡ÑÎËá¡¢µí·ÛÈÜÒº¡¢FeCl3ÈÜÒº¡¢Na2SO3ÈÜÒº


¡¾¿¼µã¶¨Î»¡¿±¾Ì⿼²é¹¤ÒµÁ÷³ÌÖÐʵÑé×ÛºÏÎÊÌâ¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Na2S2O3ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬Ò×ÈÜÓÚË®¡£ÔÚÖÐÐÔ»ò¼îÐÔ»·¾³ÖÐÎȶ¨¡£

I£®ÖƱ¸Na2S2O3•5H2O

·´Ó¦Ô­Àí£ºNa2SO3£¨aq£©+S£¨s£©Na2S2O3(aq)

ʵÑé²½Ö裺

 ¢Ù³ÆÈ¡15g Na2SO3¼ÓÈëÔ²µ×ÉÕÆ¿ÖУ¬ÔÙ¼ÓÈë80mlÕôÁóË®¡£ÁíÈ¡5gÑÐϸµÄÁò·Û£¬ÓÃ3ml     ÒÒ´¼Èóʪ£¬¼ÓÈëÉÏÊöÈÜÒºÖС£

¢Ú°²×°ÊµÑé×°Öã¨ÈçͼËùʾ£¬²¿·Ö¼Ó³Ö×°ÖÃÂÔÈ¥£©£¬Ë®Ô¡¼ÓÈÈ£¬Î¢·Ð60·ÖÖÓ¡£

¢Û³ÃÈȹýÂË£¬½«ÂËҺˮԡ¼ÓÈÈŨËõ£¬ÀäÈ´Îö³öNa2S2O3•5H2O£¬¾­¹ýÂË£¬Ï´µÓ£¬¸ÉÔµÃµ½²úÆ·¡£

»Ø´ðÎÊÌ⣺

£¨1£©Áò·ÛÔÚ·´Ó¦Ç°ÓÃÒÒ´¼ÈóʪµÄÄ¿µÄÊÇ                                                ¡£

£¨2£©ÒÇÆ÷aµÄÃû³ÆÊÇ            £¬Æä×÷ÓÃÊÇ                                          ¡£

£¨3£©²úÆ·ÖгýÁËÓÐδ·´Ó¦µÄNa2SO3Í⣬×î¿ÉÄÜ´æÔÚµÄÎÞ»úÔÓÖÊÊÇ                               £¬¼ìÑéÊÇ·ñ´æÔÚ¸ÃÔÓÖʵķ½·¨ÊÇ                                                        ¡£

£¨4£©¸ÃʵÑéÒ»°ã¿ØÖÆÔÚ¼îÐÔ»·¾³Ï½øÐУ¬·ñÔò²úÆ··¢»Æ£¬ÓÃÀë×Ó·½³Ìʽ±íʾÆäÔ­Òò£º                

                                                                                      ¡£

II.²â¶¨²úÆ·´¿¶È

   ׼ȷ³ÆÈ¡Wg²úÆ·£¬ÓÃÊÊÁ¿ÕôÁóË®Èܽ⣬ÒÔµí·Û×÷ָʾ¼Á£¬ÓÃ0.1000 mol•L‾1µâµÄ±ê×¼ÈÜÒºµÎ¶¨¡£

   ·´Ó¦Ô­ÀíΪ£º2S2O32‾+I2=S4O62-+2I‾

£¨5£©µÎ¶¨ÖÁÖÕµãʱ£¬ÈÜÒºÑÕÉ«µÄ±ä»¯£º                                    ¡£

£¨6£©µÎ¶¨ÆðʼºÍÖÕµãµÄÒºÃæÎ»ÖÃÈçͼ£¬ÔòÏûºÄµâµÄ±ê×¼ÈÜÒºÌå»ýΪ         mL¡£²úÆ·µÄ´¿¶ÈΪ£¨ÉèNa2S2O3•5H2OÏà¶Ô·Ö×ÓÖÊÁ¿ÎªM£©            ¡£

III.Na2S2O3µÄÓ¦ÓÃ

£¨7£©Na2S2O3»¹Ô­ÐÔ½ÏÇ¿£¬ÔÚÈÜÒºÖÐÒ×±»Cl2Ñõ»¯³ÉSO42‾£¬³£ÓÃ×÷ÍÑÑõ¼Á£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                                              ¡£

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

¾«Ó¢¼Ò½ÌÍø