ÌâÄ¿ÄÚÈÝ
½ÚÈÕÆÚ¼äÒòȼ·Å±ÞÅÚ»áÒýÆð¿ÕÆøÖÐSO2º¬Á¿Ôö¸ß£¬Ôì³É´óÆøÎÛȾ¡£Ä³ÊµÑéС×éͬѧÓû̽¾¿SO2µÄÐÔÖÊ£¬²¢²â¶¨¿ÕÆøÖÐSO2µÄº¬Á¿¡£
£¨1£©ËûÃÇÉè¼ÆÈçÏÂʵÑé×°Öã¬ÇëÄã²ÎÓë̽¾¿£¬²¢»Ø´ðÎÊÌ⣺

¢Ù×°ÖÃA1Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ £»
¢Ú×°ÖÃBÓÃÓÚ¼ìÑéSO2µÄƯ°×ÐÔ£¬ÆäÖÐËùÊ¢ÊÔ¼ÁΪ £¬×°ÖÃDÓÃÓÚ¼ìÑéSO2µÄ ÐÔÖÊ£»
¢Û×°ÖÃCÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ £»
¢ÜΪÁËʵÏÖÂÌÉ«»·±£µÄÄ¿±ê£¬¼×ͬѧÓûÓÃ×°ÖÃA2´úÌæ×°ÖÃA1£¬ÄãÈÏΪװÖÃA2µÄÓŵãÊÇ£¨Ð´¶þµã£© ¡¢ £»
£¨2£©ËûÃÇÄâÓÃÒÔÏ·½·¨²â¶¨¿ÕÆøÖÐSO2º¬Á¿£¨¼ÙÉè¿ÕÆøÖÐÎÞÆäËû»¹ÔÐÔÆøÌ壩¡£

¢ÙÄãÈÏΪÄĸö×°ÖÿÉÐУ¨ÌîÐòºÅ£© £¬Ê¹ÓÃÄãËùÑ¡ÓõÄ×°ÖòⶨSO2º¬Á¿Ê±£¬»¹ÐèÒª²â¶¨µÄÎïÀíÁ¿ÊÇ £»
¢ÚÄãÈÏΪÄĸö×°Öò»¿ÉÐУ¨ÌîÐòºÅ£© £¬ËµÃ÷ÀíÓÉ ¡£
£¨1£©ËûÃÇÉè¼ÆÈçÏÂʵÑé×°Öã¬ÇëÄã²ÎÓë̽¾¿£¬²¢»Ø´ðÎÊÌ⣺
¢Ù×°ÖÃA1Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ £»
¢Ú×°ÖÃBÓÃÓÚ¼ìÑéSO2µÄƯ°×ÐÔ£¬ÆäÖÐËùÊ¢ÊÔ¼ÁΪ £¬×°ÖÃDÓÃÓÚ¼ìÑéSO2µÄ ÐÔÖÊ£»
¢Û×°ÖÃCÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ £»
¢ÜΪÁËʵÏÖÂÌÉ«»·±£µÄÄ¿±ê£¬¼×ͬѧÓûÓÃ×°ÖÃA2´úÌæ×°ÖÃA1£¬ÄãÈÏΪװÖÃA2µÄÓŵãÊÇ£¨Ð´¶þµã£© ¡¢ £»
£¨2£©ËûÃÇÄâÓÃÒÔÏ·½·¨²â¶¨¿ÕÆøÖÐSO2º¬Á¿£¨¼ÙÉè¿ÕÆøÖÐÎÞÆäËû»¹ÔÐÔÆøÌ壩¡£
¢ÙÄãÈÏΪÄĸö×°ÖÿÉÐУ¨ÌîÐòºÅ£© £¬Ê¹ÓÃÄãËùÑ¡ÓõÄ×°ÖòⶨSO2º¬Á¿Ê±£¬»¹ÐèÒª²â¶¨µÄÎïÀíÁ¿ÊÇ £»
¢ÚÄãÈÏΪÄĸö×°Öò»¿ÉÐУ¨ÌîÐòºÅ£© £¬ËµÃ÷ÀíÓÉ ¡£
£¨1£©¢Ù Cu + 2H2SO4(Ũ)
CuSO4 + 2H2O + SO2¡ü£»¢ÚÆ·ºìÈÜÒº£»Ñõ»¯£»¢ÛSO2 + I2 + 2H2O £½SO42- + 2I£+ 4H+ £» ¢Ü²»ÓüÓÈÈ£¨»ò£º½ÚÔ¼ÄÜÔ´£¬½ÚÔ¼Ò©Æ·£©£»Ïà¶Ô°²È«£»Ò×ÓÚ¿ØÖÆ·´Ó¦½øÐУ»·´Ó¦¸ü³ä·Ö£» £¨2£©¢Ù a £»µ±KMnO4ÈÜÒº¸ÕÍÊɫʱ£¬²â¶¨Í¨Èë¿ÕÆøµÄÌå»ýV£» ¢Ú b £»¿ÕÆøÖк¬ÓеÄCO2Ò²ÄÜÓë¼îʯ»Ò·´Ó¦£¬Ôì³É²âÁ¿²»×¼È·¡£
ÊÔÌâ·ÖÎö£º£¨1£©¢ÙÔÚ×°ÖÃA1ÖÐCuÓëŨÁòËá¹²ÈÈ·¢Éú·´Ó¦£ºCu + 2H2SO4(Ũ)
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿