ÌâÄ¿ÄÚÈÝ

1£®Ï±íÊǶÔÇâÑõ»¯ÄÆ¡¢ÇâÑõ»¯¸ÆµÄÏà¹ØÖªÊ¶µÄ¹éÄÉ£º
ÎïÖÊÇâÑõ»¯ÄÆÇâÑõ»¯¸Æ
ÑÕÉ«¡¢×´Ì¬°×É«¹ÌÌå°×É«¹ÌÌå
Ë×Ãû»ð¼î¡¢Éռ¿ÁÐÔ¼îÊìʯ»Ò
ÈܽâÐÔ¼«Ò×ÈÜÓÚˮ΢ÈÜÓÚË®
ÖÆ·¨Ca£¨OH£©2+Na2CO3=CaCO3¡ý+2NaOHCaO+H2O=Ca£¨OH£©2
£¨1£©¹¤ÒµÉÏÓô¿¼îºÍÇâÑõ»¯¸Æ·´Ó¦ÖÆÈ¡Éռд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£ºCa£¨OH£©2+Na2CO3=CaCO3¡ý+2NaOH£®
£¨2£©Ð´³öÇâÑõ»¯¸ÆµÄÒ»¸öË×Ãû£ºÊìʯ»Ò£®
£¨3£©Ð´³öʯ»Òʯת»¯³ÉÉúʯ»ÒµÄ»¯Ñ§·½³Ìʽ£ºCaCO3$\frac{\underline{\;¸ßÎÂ\;}}{\;}$CaO+CO2¡ü£®

·ÖÎö £¨1£©¸ù¾ÝÇâÑõ»¯¸ÆºÍ̼ËáÄÆ·´Ó¦Éú³É̼Ëá¸Æ³ÁµíºÍÇâÑõ»¯ÄƽøÐзÖÎö£»
£¨2£©¸ù¾ÝÇâÑõ»¯¸ÆµÄË׳ÆÊÇÊìʯ»Ò»òÏûʯ»Ò½øÐзÖÎö£»
£¨3£©¸ù¾Ý̼Ëá¸ÆÔÚ¸ßεÄÌõ¼þÏÂÉú³ÉÑõ»¯¸ÆºÍ¶þÑõ»¯Ì¼½øÐзÖÎö£®

½â´ð ½â£º£¨1£©ÇâÑõ»¯¸ÆºÍ̼ËáÄÆ·´Ó¦Éú³É̼Ëá¸Æ³ÁµíºÍÇâÑõ»¯ÄÆ£¬»¯Ñ§·½³ÌʽΪ£ºCa£¨OH£©2+Na2CO3=CaCO3¡ý+2NaOH£»
£¨2£©ÇâÑõ»¯¸ÆµÄË׳ÆÊÇÊìʯ»Ò»òÏûʯ»Ò£»
£¨3£©Ì¼Ëá¸ÆÔÚ¸ßεÄÌõ¼þÏÂÉú³ÉÑõ»¯¸ÆºÍ¶þÑõ»¯Ì¼£¬»¯Ñ§·½³ÌʽΪ£ºCaCO3$\frac{\underline{\;¸ßÎÂ\;}}{\;}$CaO+CO2¡ü£®
¹Ê´ð°¸Îª£º£¨1£©Ca£¨OH£©2+Na2CO3=CaCO3¡ý+2NaOH£»
£¨2£©Êìʯ»Ò£»
£¨3£©CaCO3$\frac{\underline{\;¸ßÎÂ\;}}{\;}$CaO+CO2¡ü£®

µãÆÀ ÔÚ½â´ËÀàÌâʱ£¬Ê×ÏÈ·ÖÎöÓ¦ÓõÄÔ­Àí£¬È»ºóÕÒ³ö·´Ó¦Îï¡¢Éú³ÉÎ×îºó½áºÏ·½³ÌʽµÄÊéд¹æÔòÊéд·½³Ìʽ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø