ÌâÄ¿ÄÚÈÝ

ÒÑÖªX¡¢Y¡¢Z¶¼ÊÇÖÐѧ»¯Ñ§Öеij£¼ûÈýÖÖÆøÌ壬XΪ»ÆÂÌÉ«£¬ÏòXµÄË®ÈÜÒºÖеÎÈëÆ·ºìÊÔÒº£¬Æ·ºìÍÊÉ«£¬½«ÈÜÒº¼ÓÈȺó£¬ÈÜÒºÑÕÉ«ÎÞÃ÷ÏԱ仯£»ÏòYµÄË®ÈÜÒºÖеÎÈëÆ·ºìÊÔÒº£¬Æ·ºìÍÊÉ«£¬½«ÈÜÒº¼ÓÈȺó£¬ÈÜÒºÓֳʺìÉ«£»ÏòZµÄË®ÈÜÒºÖеÎÈëºìɫʯÈïÊÔÒº£¬ÈÜÒº³ÊÀ¶É«¡£¾Ý´Ë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÊÔд³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£ºY______________¡¢Z________________£»
£¨2£©½«XͨÈëYµÄË®ÈÜÒºÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                        £»
£¨3£©ÊµÑéÊÒÖÆÈ¡YµÄ»¯Ñ§·½³ÌʽÊÇ                                         ¡£

¹²10·Ö
(1)   SO2   NH3   (¸÷2·Ö¹²4·Ö)
(2) C12 + SO2 2H2O===4H£«£«2Cl- + SO42¡ª         £¨3·Ö£©
(3)2 NH4Cl + Ca(OH)2CaC12 +2NH3 ¡ü+2H2O  £¨3·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©XΪ»ÆÂÌÉ«£¬XΪC12£¬ÏòYµÄË®ÈÜÒºÖеÎÈëÆ·ºìÊÔÒº£¬Æ·ºìÍÊÉ«£¬½«ÈÜÒº¼ÓÈȺó£¬ÈÜÒºÓֳʺìÉ«£¬YΪSO2£¬ÏòZµÄË®ÈÜÒºÖеÎÈëºìɫʯÈïÊÔÒº£¬ÈÜÒº³ÊÀ¶É«¡£¾Ý´Ë»Ø´ðÏÂÁÐÎÊÌ⣬ZΪNH3¡£
£¨2£©C12°ÑÑõ»¯SO2£¬Àë×Ó·½³ÌʽΪ£ºC12 + SO2 2H2O===4H£«£«2Cl- + SO42¡ª¡£
£¨3£©ÖÆÈ¡NH3µÄ·¢Éú£¬ÒÀ¾Ý½Ì²Ä·½³ÌʽΪÊÇ£º2 NH4Cl + Ca(OH)2CaC12 +2NH3 ¡ü+2H2O¡£
¿¼µã£ºÎÞ»úÎïµÄÍÆ¶Ï
µãÆÀ£º±¾ÌâÊÇÒ»µÀ¿¼²éÎÞ»úÎïµÄÐÔÖʼ°Íƶϣ¬¶¼ÊÇ»ù´¡ÖªÊ¶£¬¿¼²éѧÉú·ÖÎöºÍ½â¾öÎÊÌâµÄÄÜÁ¦£¬ÄѶȲ»´ó¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø