ÌâÄ¿ÄÚÈÝ

̼ËáÄƺÍ̼ËáÇâÄÆÊÇÈÕ³£Éú»îÖг£¼ûµÄÑΣ¬¿ÉÓÃ×÷È¥¹¸ºÍʳÓüijÑо¿ÐÔѧϰС×éµÄͬѧÉè¼ÆÁËÏÂÁÐʵÑéÓÃÀ´Ì½¾¿¡¢±È½Ï̼ËáÄÆ¡¢Ì¼ËáÇâÄƵÄÐÔÖÊ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢Å¸ÃС×éµÄͬѧÉè¼ÆÁËÈçÏÂ×°ÖýøÐÐʵÑ飺

¢Ù¸ÃʵÑéµÄÄ¿µÄÊÇ                                     £»

¢ÚʵÑé¹ý³ÌÖпÉÒÔ¿´µ½ÉÕ±­IIÖеÄʵÑéÏÖÏóÊÇ                             £»

¢Û¼ÓÈȹý³ÌÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£º                                  £»

¢Æ»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙÈôÏòNa2CO3µÄ±¥ºÍÈÜÒºÖÐͨÈë×ãÁ¿µÄCO2ÆøÌ壬³öÏÖ»ì×Ç£¬Ð´³öÕâ¸ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ

                                                £»

¢ÚÔÚŨ¶È¾ùΪ1mol£¯LµÄNa2CO3ºÍNaHCO3µÄÈÜÒºÖзֱðµÎ¼Ó1mol£¯LµÄÑÎËᣬÆð³õ²úÉúÆøÌåËÙÂÊ¿ìµÄÊÇ        £»

¢Ç¸ÃС×éµÄͬѧÔÚ²éÔÄ×ÊÁÏʱ»¹µÃÖª£¬Î¸ËᣨÖ÷ÒªÊÇŨ¶È¼«Ï¡µÄÑÎËᣩ¹ý¶àÄܵ¼ÖÂÈ˲úÉú²»ÊÊ£¬Î¸Ëá¹ý¶àµÄ²¡ÈË¿É·þÓÃÊÊÁ¿µÄСËÕ´ò¼õÇá²»ÊʸС£

¢Ùд³öСËÕ´òºÍÑÎËá·´Ó¦µÄÀë×Ó·½³Ìʽ£º                                         £»

¢ÚAl(OH)3Ò²ÊÇÒ»ÖÖθËáÖкͼÁ£¬Î÷Ò©¡°Î¸Êæƽ¡±µÄÖ÷Òª³É·Ö¾ÍÊÇAl(OH)3¡£Ð´³öAl(OH)3ÖкÍθËáµÄÀë×Ó·½³Ìʽ£º                                                           £»

¢ÛÈç¹ûÄãÊÇÄÚ¿ÆÒ½Éú¸øθËá¹ý¶àµÄθÀ£Ññ²¡ÈË£¨ÆäÖ¢×´Ö®Ò»ÊÇθ±ÚÊÜËðÉ˶ø±ä±¡£©¿ªÒ©·½Ê±£¬Ó¦Ñ¡ÓÃСËÕ´òºÍÇâÑõ»¯ÂÁÖеĠ     £¬ÀíÓÉÊÇ                                     £»

  ±È½Ï̼ËáÄÆ¡¢Ì¼ËáÇâÄƵÄÎȶ¨ÐÔ      

   ÈÜÒº±ä»ì×Ç                                        

¢Û»¯Ñ§·½³ÌʽÊÇ£º 2NaHCO3  Na2CO3 + CO2 ¡ü+  H20                          £»

¢Æ¢Ùд³öÕâ¸ö·´Ó¦µÄ»¯Ñ§·½³ÌʽNa2CO3 + CO2 +  H20  =   2NaHCO3          £»

¢Ú²úÉúÆøÌåËÙÂÊ¿ìµÄÊÇNaHCO3£»

¢Ç¢ÙСËÕ´òºÍÑÎËá·´Ó¦µÄÀë×Ó·½³Ìʽ£ºHCO3- + H+H20 + CO2¡ü £»

¢ÚAl(OH)3ÖкÍθËáµÄÀë×Ó·½³Ìʽ£ºAl(OH)3+ 3H+H20 + Al3+ £»

¢ÛӦѡÓà Al(OH)3 £¬ÀíÓÉÊÇ Ð¡ËÕ´òÓëθËá·´Ó¦²úÉúCO2ÆøÌåÒ×ʹθ±Ú´©¿×     £»

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
̼ËáÄƺÍ̼ËáÇâÄÆÊdz£¼ûµÄº¬Ì¼»¯ºÏÎÔÚÉú²ú¡¢Éú»îºÍ»¯Ñ§ÊµÑéÖÐÓÐ×ÅÖØÒªµÄÓÃ;£®
£¨1£©Í¨³£Óñ¥ºÍNaHCO3ÈÜÒº³ýÈ¥»ìÔÚCO2ÖеÄÉÙÁ¿HClÆøÌ壬Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
NaHCO3+HCl¨TNaCl+CO2¡ü+H2O
NaHCO3+HCl¨TNaCl+CO2¡ü+H2O
£¬²»ÓÃNaOHÈÜÒºµÄÔ­ÒòÊÇ£¨Óû¯Ñ§·½³Ìʽ±íʾ£¬ÏÂͬ£©
CO2+2NaOH¨TNa2CO3+H2O
CO2+2NaOH¨TNa2CO3+H2O
£¬²»Óñ¥ºÍNa2CO3ÈÜÒºµÄÔ­ÒòÊÇ
Na2CO3+CO2+H2O¨T2NaHCO3
Na2CO3+CO2+H2O¨T2NaHCO3
£®Çë´ÓÒÔÉÏÈý¸ö»¯Ñ§·´Ó¦·½³ÌʽÖÐÈÎÑ¡Ò»¸ö£¬Ð´³öÆä¶ÔÓ¦µÄÀë×Ó·´Ó¦·½³Ìʽ£º
HCO3-+H+¨TCO2¡ü+H2O
HCO3-+H+¨TCO2¡ü+H2O
£®
£¨2£©Ð¡»ªÍ¬Ñ§ÏòNaOHÈÜÒºÖÐÖ±½ÓͨÈëCO2ÆøÌåµÄ·½·¨ÖƱ¸Na2CO3ÈÜÒº£¬ÓÉÓÚͨÈëCO2µÄÁ¿ÄÑÒÔ¿ØÖÆ£¬³£»ìÓÐNaOH»òNaHCO3£®ÇëÄã¸ù¾Ý·´Ó¦Ô­Àí£ºNaHCO3+NaOH¨TNa2CO3+H2O£¬ÔÚ²»¸Ä±ä·´Ó¦ÎïµÄÇ°ÌáÏÂÉè¼ÆÒ»¸öÖƱ¸·½°¸£º
½«NaOHÈÜÒº·Ö³ÉÁ½µÈ·Ý£¬ÏòÆäÖÐÒ»·ÝÖÐͨÈë¹ýÁ¿µÄCO2£¬È»ºó΢ÈȺóÔÙ½«ÁíÒ»·ÝNaOHÈÜÒº¼ÓÈëµ½µÚÒ»·ÝÖУ¬¼´µÃµ½ËùÐèµÄNa2CO3ÈÜÒº
½«NaOHÈÜÒº·Ö³ÉÁ½µÈ·Ý£¬ÏòÆäÖÐÒ»·ÝÖÐͨÈë¹ýÁ¿µÄCO2£¬È»ºó΢ÈȺóÔÙ½«ÁíÒ»·ÝNaOHÈÜÒº¼ÓÈëµ½µÚÒ»·ÝÖУ¬¼´µÃµ½ËùÐèµÄNa2CO3ÈÜÒº
£®

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

¾«Ó¢¼Ò½ÌÍø