ÌâÄ¿ÄÚÈÝ

¾«Ó¢¼Ò½ÌÍøÒ»Ì죬ʵÑéÖúÊÖС¾ê×ß½øʵÑéÊÒ£¬ºÍÀÏʦһÆð¼ì²éÿ¸öʵÑé×ÀÉϵÄÒ©Æ·¡¢ÒÇÆ÷ÊÇ·ñÆ뱸£¬×ßµ½Ä³×éµÄʱºò£¬¿´µ½ÁËÒ»¸ö²»ºÍгµÄ¡°Òô·û¡±£¬Èçͼ£º
£¨1£©´ËÇé´Ë¾°ÄãÊ×ÏÈÏëµ½µÄÊÇËü¿ÉÄܱäÖÊÁË£¬¸Ã±äÖÊ·´Ó¦µÄ»¯Ñ§·½³Ìʽ
 
£®
£¨2£©Î§ÈÆ´ËÆ¿NaOHÈÜÒºÊÇ·ñ±äÖʵÄÎÊÌ⣬С¾êÀûÓÃʵÑéÊÒµÄÈýÖÖÊÔ¼Á£¨ÂÈ»¯¸ÆÈÜÒº¡¢Ï¡ÑÎËá¡¢·Ó̪ÊÔÒº£©Õ¹¿ªÁË̽¾¿»î¶¯£®
¡¾²éÔÄ×ÊÁÏ¡¿£ºa£®ÂÈ»¯¸ÆÈÜÒº³ÊÖÐÐÔ£¬Ì¼ËáÄÆÈÜÒº³Ê¼îÐÔ
b£®ÂÈ»¯¸ÆÈÜÒºÓë̼ËáÄÆÈÜÒº·´Ó¦£ºCaCl2+Na2CO3
      
.
.
 CaCO3
¡ý+2NaCl
¢ÙÑéÖ¤¸ÃÈÜÒºÒѾ­±äÖÊ£®ÄãÈÏΪС¾êËù¼ÓµÄÊÔ¼ÁÊÇ
 
£®
¢ÚÓûÖ¤Ã÷±äÖʵÄÈÜÒºÖÐÉдæNaOH£¬ÇëÄã°ïÖúС¾êÍê³ÉÒÔÏÂ̽¾¿·½°¸£º
̽¾¿Ä¿µÄ ̽¾¿²½Öè Ô¤¼ÆÏÖÏó
³ý¾¡ÈÜÒºÖеÄCO32- aÈ¡ÉÙÁ¿ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó×ãÁ¿µÄ
 
ÊÔ¼Á
 
Ö¤Ã÷ÈÜÒºÖÐÉдæNaOH bÏòʵÑéaËùµÃÈÜÒºÖеμӷÓ̪ÊÔÒº
 
·ÖÎö£ºÅªÇåÇâÑõ»¯ÄƵıäÖÊÇé¿ö¼°¸ù¾ÝÇâÑõ»¯ÄƺÍ̼ËáÄƵÄÓйصĻ¯Ñ§ÐÔÖʵIJî±ð´ïµ½¼ìÑéÎïÖÊ´æÔÚµÄÄ¿µÄ£®
½â´ð£º½â£º£¨1£©ÓÉÓÚÇâÑõ»¯ÄÆÄܹ»ºÍ¿ÕÆøÖеĶþÑõ»¯Ì¼·´Ó¦Éú³É̼ËáÄƶø±äÖÊ£¬¹Ê´ð°¸Îª£º2NaOH+CO2¨TNa2CO3+H2O
£¨2£©¢ÙÑéÖ¤ÇâÑõ»¯ÄÆÊÇ·ñ±äÖÊÖ»¿´ÊÇ·ñÓÐ̼ËáÄÆ´æÔÚ¼´¿É£¬ÓÉÓÚ̼ËáÄƺÍÏ¡ÑÎËá·´Ó¦ÓÐÆøÅÝÉú³É£¬¹Ê´ð°¸Îª£ºÏ¡ÑÎËá
¢ÚÓÉÓÚÇâÑõ»¯ÄÆÈÜÒººÍ̼ËáÄÆÈÜÒº¶¼³Ê¼îÐÔ£¬Òò´ËÏëÖ¤Ã÷ÇâÑõ»¯ÄƵĴæÔÚ£¬Ê×ÏȵÃÏÈ°Ñ̼ËáÄƳýÈ¥£¬Òò´ËÏÈÓÃÂÈ»¯¸ÆÈÜÒº³ýȥ̼ËáÄÆ£¬ÔÙÓ÷Ó̪ÊÔÒºÖ¤Ã÷ÇâÑõ»¯ÄƵĴæÔÚ£¬¹Ê´ð°¸Îª£ºÂÈ»¯¸ÆÈÜÒº£¬°×É«³Áµí£¬·Ó̪±äºì£®
µãÆÀ£ºÖ÷Òª¿¼²éÁËÇâÑõ»¯ÄƵıäÖʼ°ÇâÑõ»¯ÄƺÍ̼ËáÄƵÄÓйصĻ¯Ñ§ÐÔÖÊ£¬ÅàÑøѧÉú½â¾öÎÊÌâµÄÄÜÁ¦£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
24¡¢Ò»Ì죬ʵÑéÖúÊÖС¾êºÍСÀö×ß½øʵÑéÊÒ£¬ºÍÀÏʦһÆð¼ì²éÿ¸öʵÑé×ÀÉϵÄÒ©Æ·¡¢ÒÇÆ÷ÊÇ·ñÆ뱸£¬×ßµ½Ä³×éµÄʱºò£¬¿´µ½ÁËÒ»¸ö²»ºÍгµÄ¡°Òô·û¡±£¨Èçͼ£©£®
£¨1£©´ËÇé´Ë¾°ÄãÊ×ÏÈÏëµ½µÄÊÇËü¿ÉÄܱäÖÊÁË£¬¸Ã±äÖÊ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
CO2+2NaOH=Na2CO3+H2O
£»
£¨2£©Î§ÈÆ´ËÆ¿NaOHÈÜÒºÊÇ·ñ±äÖʵÄÎÊÌ⣬ËýÃÇÀûÓÃʵÑéÊÒÏÖÓеļ¸ÖÖÖÖÊÔ¼Á£¨ÂÈ»¯¸ÆÈÜÒº¡¢ÇâÑõ»¯¸ÆÈÜÒº¡¢Ï¡ÑÎËᡢʯÈïÊÔÒº¡¢·Ó̪ÊÔÒºµÈ£©Õ¹¿ªÁË̽¾¿»î¶¯£®
¢ÙС¾êÈ¡ÉÙÁ¿ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼ÓijÖÖÊÔ¼Á£¬ÓÐÆøÅݲúÉú£¬ÓÉ´ËÖ¤Ã÷NaOHÈÜÒºÒѾ­±äÖÊ£®
ÄãÈÏΪС¾êËù¼ÓµÄÊÔ¼ÁÊÇ
Ï¡ÑÎËá
£®Ð¡ÀöÈ¡ÉÙÁ¿ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó¼¸µÎ·Ó̪ÊÔÒº£¬¿´µ½ÈÜÒº³ÊºìÉ«£¬ÓÉ´ËÀ´ËµÃ÷NaOHÈÜÒºÒѱäÖÊ£®
ÄãÈÏΪËýµÄ¼ìÑé·½·¨ÊÇ·ñ³É¹¦£¿
²»Äܳɹ¦
£»
ÀíÓÉÊÇ
Na2CO3ÈÜÒº¾ùÏÔ¼îÐÔ£¬¶¼ÄÜʹ·Ó̪ÊÔÒº³ÊºìÉ«
£®
¢ÚС¾êÓû³ýÈ¥±äÖʵÄÈÜÒºÖеÄÔÓÖÊ£¬ÇëÄã°ïÖúС¾êÍê³ÉÒÔÏÂ̽¾¿·½°¸£ºÈ¡±äÖÊÈÜÒºÓÚÉÕ±­ÖУ¬Öð½¥¼ÓÈë
ÇâÑõ»¯¸ÆÈÜÒº
ÊÔ¼Á£¬²¢²»¶ÏÕñµ´£¬ÖÁ²»ÔÙ²úÉú
³Áµí
Ϊֹ£¬ÔÙ½øÐйýÂ˲Ù×÷£¬¼´¿ÉµÃµ½NaOHÈÜÒº£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
Ca£¨OH£©2+Na2CO3=CaCO3¡ý+2NaOH
£®
13¡¢Ò»Ì죬ʵÑéÖúÊÖС¾êºÍСÀö×ß½øʵÑéÊÒ£¬ºÍÀÏʦһÆð¼ì²éÿ¸öʵÑé×ÀÉϵÄÒ©Æ·¡¢ÒÇÆ÷ÊÇ·ñÆ뱸£¬×ßµ½Ä³×éµÄʱºò£¬¿´µ½ÁËÒ»¸ö²»ºÍгµÄ¡°Òô·û¡±£¨Èçͼ£©£®
£¨1£©´ËÇé´Ë¾°ÄãÊ×ÏÈÏëµ½µÄÊÇËü¿ÉÄܱäÖÊÁË£¬¸Ã±äÖÊ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
CO2+2NaOH=Na2CO3+H2O

£¨2£©Î§ÈÆ´ËÆ¿NaOHÈÜÒºÊÇ·ñ±äÖʵÄÎÊÌ⣬ËýÃÇÀûÓÃʵÑéÊÒÏÖÓеļ¸ÖÖÖÖÊÔ¼Á£¨ÂÈ»¯¸ÆÈÜÒº¡¢ÇâÑõ»¯¸ÆÈÜÒº¡¢Ï¡ÑÎËᡢʯÈïÊÔÒº¡¢·Ó̪ÊÔÒºµÈ£©Õ¹¿ªÁË̽¾¿»î¶¯£®
¢ÙС¾êÈ¡ÉÙÁ¿ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼ÓijÖÖÊÔ¼Á£¬ÓÐÆøÅݲúÉú£¬ÓÉ´ËÖ¤Ã÷NaOHÈÜÒºÒѾ­±äÖÊ£®
ÄãÈÏΪС¾êËù¼ÓµÄÊÔ¼ÁÊÇ
Ï¡ÑÎËá
£®Ð¡ÀöÈ¡ÉÙÁ¿ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó¼¸µÎ·Ó̪ÊÔÒº£¬¿´µ½ÈÜÒº³ÊºìÉ«£¬ÓÉ´ËÀ´ËµÃ÷NaOHÈÜÒºÒѱäÖÊ£®ÄãÈÏΪËýµÄ¼ìÑé·½·¨ÊÇ·ñ³É¹¦£¿ÀíÓÉÊÇ
NaOHÈÜÒº¡¢Na2CO3ÈÜÒº¾ùÏÔ¼îÐÔ£¬¶¼ÄÜʹ·Ó̪ÊÔÒº³ÊºìÉ«
£®
¢ÚС¾êÓû³ýÈ¥±äÖʵÄÈÜÒºÖеÄÔÓÖÊ£¬ÇëÄã°ïÖúС¾êÍê³ÉÒÔÏÂ̽¾¿·½°¸£ºÈ¡±äÖÊÈÜÒºÓÚÉÕ±­ÖУ¬Öð½¥¼ÓÈë
ÇâÑõ»¯¸ÆÈÜÒº
ÊÔ¼Á£¬²¢²»¶ÏÕñµ´£¬ÖÁ²»ÔÙ²úÉú
³Áµí
Ϊֹ£¬ÔÙ½øÐйýÂ˲Ù×÷£¬¼´¿ÉµÃµ½NaOHÈÜÒº£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
Ca£¨OH£©2+Na2CO3=CaCO3¡ý+2NaOH
£®

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

¾«Ó¢¼Ò½ÌÍø