ÌâÄ¿ÄÚÈÝ
¾ÝÓйØ×ÊÁϽéÉÜ£º½«±´¿Ç£¨Ö÷Òª³É·Ö̼Ëá¸Æ£©×ÆÈÈÓë²Ýľ»Ò£¨Ö÷Òª³É·Ö̼Ëá¼Ø£©ÔÚË®ÖÐ×÷Ó㬿ÉÒԵõ½ÇâÑõ»¯¼Ø¡£Ä³»¯Ñ§ÐËȤС×飬ΪÁËÖÆÈ¡ÇâÑõ»¯¼Ø£¬½«ÉÙÁ¿Ì¼Ëá¸Æ³ä·Ö×ÆÈȺóµÄ¹ÌÌå·ÅÈëÉÕ±ÖУ¬ÔÙÏòÆäÖмÓÈëÒ»¶¨Á¿µÄ10%µÄ̼Ëá¼ØÈÜÒº£¬³ä·Ö·´Ó¦ÀäÈ´ºó¹ýÂË¡¢Ï´µÓµÃµ½¹ÌÌåAºÍÈÜÒºB¡£Çë»Ø´ðÏÂÁÐÓйØÎÊÌ⣺
£¨1£©È¡ÉÙÁ¿¹ÌÌåAÓÚÊÔ¹ÜÖУ¬¼ÓÈëÏ¡ÑÎËᣬÓÐÆøÅݲúÉú£¬Ö¤Ã÷AÖк¬ÓÐ__________¡£
£¨2£©È¡ÉÙÁ¿¹ÌÌåAÓÚÊÔ¹ÜÖУ¬¼ÓË®³ä·ÖÈܽ⡢¹ýÂË£¬ÏòÂËÒºÖеμÓ______________£¬ÈÜÒº±äºìÉ«£¬Ö¤Ã÷AÖл¹º¬ÓÐÇâÑõ»¯¸Æ¡£
£¨3£©½áºÏ£¨1£©£¨2£©¹ÌÌåA³É·ÖµÄÈ·¶¨£¬¿ÉÒÔÍƶÏÈÜÒºBµÄ³É·ÖÓÐÇâÑõ»¯¼ØºÍÇâÑõ»¯¸Æ¡£ÎªÁ˵õ½´¿¾»µÄÇâÑõ»¯¼Ø£¬ÐèÏòÈÜÒºBÖеμÓÊÊÁ¿µÄ_____________ÈÜÒº£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________________________________________£¬³ä·Ö·´Ó¦ºó¹ýÂË£¬½«ÂËÒºÕô¸É¿ÉÖƵýϴ¿¾»µÄÇâÑõ»¯¼Ø¡£
£¨1£©È¡ÉÙÁ¿¹ÌÌåAÓÚÊÔ¹ÜÖУ¬¼ÓÈëÏ¡ÑÎËᣬÓÐÆøÅݲúÉú£¬Ö¤Ã÷AÖк¬ÓÐ__________¡£
£¨2£©È¡ÉÙÁ¿¹ÌÌåAÓÚÊÔ¹ÜÖУ¬¼ÓË®³ä·ÖÈܽ⡢¹ýÂË£¬ÏòÂËÒºÖеμÓ______________£¬ÈÜÒº±äºìÉ«£¬Ö¤Ã÷AÖл¹º¬ÓÐÇâÑõ»¯¸Æ¡£
£¨3£©½áºÏ£¨1£©£¨2£©¹ÌÌåA³É·ÖµÄÈ·¶¨£¬¿ÉÒÔÍƶÏÈÜÒºBµÄ³É·ÖÓÐÇâÑõ»¯¼ØºÍÇâÑõ»¯¸Æ¡£ÎªÁ˵õ½´¿¾»µÄÇâÑõ»¯¼Ø£¬ÐèÏòÈÜÒºBÖеμÓÊÊÁ¿µÄ_____________ÈÜÒº£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________________________________________£¬³ä·Ö·´Ó¦ºó¹ýÂË£¬½«ÂËÒºÕô¸É¿ÉÖƵýϴ¿¾»µÄÇâÑõ»¯¼Ø¡£
£¨1£©CaCO3 £¨2£©·Ó̪ÊÔÒº
£¨3£©K2 CO3£¨2·Ö£©£» Ca(OH)2 + Na2 CO3 = CaCO3¡ý+ 2 Na OH
£¨3£©K2 CO3£¨2·Ö£©£» Ca(OH)2 + Na2 CO3 = CaCO3¡ý+ 2 Na OH
¿¼µã£º
·ÖÎö£º¸ù¾ÝÒÑÓеÄ֪ʶ½øÐзÖÎö£¬Ì¼ËáÑÎÄÜÓëËá·´Ó¦Éú³ÉÆøÌ壬·Ó̪ÊÔÒºÔÚ¼îÐÔÈÜÒºÖбäºì£¬ÇâÑõ»¯¸ÆÄÜÓë̼Ëá¼Ø·´Ó¦£¬·´Ó¦½øÐеij̶ÈÓëÎïÖʵÄÁ¿Óйأ¬¿ÉÄÜ´æÔÚÎïÖʹýÁ¿µÄÏÖÏó£®
½â´ð£º½â£º£¨1£©È¡ÉÙÁ¿¹ÌÌåAÓÚÊÔ¹ÜÖУ¬¼ÓÈëÏ¡ÑÎËᣬÓÐÆøÅݲúÉú£¬Ö¤Ã÷¹ÌÌåAÖк¬ÓÐ̼Ëá¸Æ£¬¹ÊÌCaCO3£»
£¨2£©È¡ÉÙÁ¿¹ÌÌåAÓÚÊÔ¹ÜÖУ¬¼ÓË®³ä·ÖÈܽ⡢¹ýÂË£¬AÖл¹º¬ÓÐÇâÑõ»¯¸Æ£¬ËµÃ÷ÂËÒº³Ê¼îÐÔ£¬ÄÜʹ·Ó̪ÊÔÒº±äºì£¬¹ÊµÎ¼ÓµÄÊÔ¼ÁÊÇ·Ó̪ÊÔÒº£¬¹ÊÌ·Ó̪ÊÔÒº£»
£¨3£©ÒªµÃµ½´¿¾»µÄÇâÑõ»¯¼Ø£¬ÐèÒª¼ÓÈë̼Ëá¼ØÈÜÒºÓëÇâÑõ»¯¸Æ·´Ó¦£¬Ì¼Ëá¼ØÄÜÓëÇâÑõ»¯¸Æ·´Ó¦Éú³É̼Ëá¸Æ³ÁµíºÍÇâÑõ»¯¼Ø£¬¹ÊÌK2CO3£¬Ca£¨OH£©2+K2CO3=CaCO3¡ý+2KOH£®
µãÆÀ£º±¾Ì⿼²éÁË»ìºÏÎï³É·ÖµÄÍƶϣ¬Íê³É´ËÌ⣬¿ÉÒÔÒÀ¾ÝÌâ¸ÉÌṩµÄÐÅÏ¢½áºÏÎïÖʵÄÐÔÖʽøÐУ®
·ÖÎö£º¸ù¾ÝÒÑÓеÄ֪ʶ½øÐзÖÎö£¬Ì¼ËáÑÎÄÜÓëËá·´Ó¦Éú³ÉÆøÌ壬·Ó̪ÊÔÒºÔÚ¼îÐÔÈÜÒºÖбäºì£¬ÇâÑõ»¯¸ÆÄÜÓë̼Ëá¼Ø·´Ó¦£¬·´Ó¦½øÐеij̶ÈÓëÎïÖʵÄÁ¿Óйأ¬¿ÉÄÜ´æÔÚÎïÖʹýÁ¿µÄÏÖÏó£®
½â´ð£º½â£º£¨1£©È¡ÉÙÁ¿¹ÌÌåAÓÚÊÔ¹ÜÖУ¬¼ÓÈëÏ¡ÑÎËᣬÓÐÆøÅݲúÉú£¬Ö¤Ã÷¹ÌÌåAÖк¬ÓÐ̼Ëá¸Æ£¬¹ÊÌCaCO3£»
£¨2£©È¡ÉÙÁ¿¹ÌÌåAÓÚÊÔ¹ÜÖУ¬¼ÓË®³ä·ÖÈܽ⡢¹ýÂË£¬AÖл¹º¬ÓÐÇâÑõ»¯¸Æ£¬ËµÃ÷ÂËÒº³Ê¼îÐÔ£¬ÄÜʹ·Ó̪ÊÔÒº±äºì£¬¹ÊµÎ¼ÓµÄÊÔ¼ÁÊÇ·Ó̪ÊÔÒº£¬¹ÊÌ·Ó̪ÊÔÒº£»
£¨3£©ÒªµÃµ½´¿¾»µÄÇâÑõ»¯¼Ø£¬ÐèÒª¼ÓÈë̼Ëá¼ØÈÜÒºÓëÇâÑõ»¯¸Æ·´Ó¦£¬Ì¼Ëá¼ØÄÜÓëÇâÑõ»¯¸Æ·´Ó¦Éú³É̼Ëá¸Æ³ÁµíºÍÇâÑõ»¯¼Ø£¬¹ÊÌK2CO3£¬Ca£¨OH£©2+K2CO3=CaCO3¡ý+2KOH£®
µãÆÀ£º±¾Ì⿼²éÁË»ìºÏÎï³É·ÖµÄÍƶϣ¬Íê³É´ËÌ⣬¿ÉÒÔÒÀ¾ÝÌâ¸ÉÌṩµÄÐÅÏ¢½áºÏÎïÖʵÄÐÔÖʽøÐУ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿