Question

18£®ÂÁÊÇÈËÀàÉú»îÖмÌÍ­¡¢ÌúÖ®ºóÓÖÒ»¸öÖØÒªµÄ½ðÊô£®¹¤ÒµÉÏÒ±Á¶½ðÊôÂÁµÄÔ­ÁÏÀ´Ô´ÓÚ×ÔÈ»½çÖÐÖØÒªµÄ¿óÎïÖÊ·°ÍÁ£¨Ö÷Òª³É·Ö£ºAl₂O₃£»»¹ÓÐSiO₂¡¢Fe₂O₃¡¢FeCO₃¡¢MgCO₃µÈÔÓÖÊ£©£®´Ó·°ÍÁÖÐÌáÈ¡µÃµ½Al₂O₃µÄ¹¤ÒÕÁ÷³ÌÈçͼËùʾ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¹ÌÌåAËùº¬ÎïÖʵĻ¯Ñ§Ê½£¨»ò·Ö×Óʽ£©ÊÇFe£¨OH£©₂¡¢Fe£¨OH£©₃¡¢Mg£¨OH£©₂£®
£¨2£©Ð´³öÈÜÒºAÓë×ãÁ¿ÆøÌåB·´Ó¦µÄÀë×Ó·½³ÌʽCO₂+OH^-¨THCO₃^-¡¢AlO₂^-+CO₂+2H₂O=Al£¨OH£©₃¡ý+HCO₃^-£®
£¨3£©¹¤ÒµÒ±Á¶½ðÊôÂÁͨ³£ÓÃʯī̼¿é×÷µç½â²ÛµÄÑô¼«£¬ÇëÄã¸ù¾Ýµç½âÔ­Àí½âÊ͵ç½âÒ±Á¶ÂÁµÄ¹ý³ÌÖУ¬ÐèÒª¶¨ÆÚ²¹³äÑô¼«Ì¼¿éµÄÔ­Òòµç½âÒ±Á¶ÂÁµÄÑô¼«²úÎïΪO₂£¬ÔÚ¸ßÎÂÌõ¼þÏ£¬O₂ÓëÑô¼«Ì¼·¢Éú·´Ó¦ÏûºÄÑô¼«Ì¼¿é£®
£¨4£©Al₂O₃µÄÈÛµãºÜ¸ß£¬Òò¶øÔÚ¹¤ÒµÒ±Á¶Ê±£¬Ð轫Al₂O₃ÈÛÓÚÈÛ»¯µÄ±ù¾§Ê¯£¨Na₃AlF₆£©ÖнøÐеç½â£®Çëд³öµç½â¹ý³ÌÖÐÒõ¼«µÄµç¼«·´Ó¦Ê½Al³⁺+3e^-¨TAl£®¹¤ÒµÉÏͨ³£½«Al£¨OH£©₃ºÍNa₂CO₃һͬÈÜÓÚÇâ·úËáÀ´ÖÆÈ¡±ù¾§Ê¯£¬·´Ó¦Ê±·Å³öCO₂ÆøÌ壬д³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ2Al£¨OH£©₃+3Na₂CO₃+12HF¨T2Na₃AlF₆+3CO₂¡ü+9H₂O£®
£¨5£©ÓÐһλͬѧ²éÔÄ×ÊÁÏ·¢ÏÖ£¬AlCl₃µÄÈÛµãºÜµÍ£®ËûÌá³ö£º¿Éͨ¹ýµç½âÈÛÈÚ״̬µÄAlCl₃ÖÆÈ¡½ðÊôÂÁ£®ÄãÈÏΪËûÌá³öµÄ·½°¸ÊÇ·ñ¿ÉÐУ¿ÎªÊ²Ã´£¿²»¿ÉÐУ¬AlCl₃Êǹ²¼Û»¯ºÏÎÈÛÈÚ״̬µÄAlCl₃²»µ¼µç£®
£¨6£©Ä³ÆóÒµÓÃÉÏÊö¹¤ÒÕÁ÷³Ì½øÐÐÉú²ú£¬Ã¿10.0t·°ÍÁ¿ÉµÃµ½½ðÊôÂÁ1.35t£®Èç¹û²»¿¼ÂÇÉú²ú¹ý³ÌµÄËðºÄ£¬ÇëÄã¼ÆËã·°ÍÁÖÐAl₂O₃µÄÖÊÁ¿·ÖÊý25.5%£®

Answer 1

·ÖÎö ·°ÍÁÖ÷Òª³É·ÖÊÇAl₂O₃£¬»¹ÓÐSiO₂¡¢Fe₂O₃¡¢FeCO₃¡¢MgCO₃µÈÔÓÖÊ£¬³ýSiO₂²»·´Ó¦Í⣬ÓëÑÎËá·´Ó¦Éú³ÉµÄ½ðÊôÑôÀë×ÓÓÐAl³⁺¡¢Fe³⁺¡¢Fe²⁺¡¢Mg²⁺µÈ£¬ÕâЩ½ðÊôÑôÀë×ÓÔÙÓë×ãÁ¿ÇâÑõ»¯ÄÆ·´Ó¦ºóÉú³ÉµÄ¿ÉÈÜÎïΪNaAlO₂£¬²»ÈÜÎï¹ÌÌåAΪFe£¨OH£©₂¡¢Fe£¨OH£©₃¡¢Mg£¨OH£©₂£¬ÈÜÒºAΪNaAlO₂ÈÜÒººÍÇâÑõ»¯ÄÆÈÜÒºµÄ»ìºÏÎÈÜÒºÖÐͨÈë×ãÁ¿BΪ¶þÑõ»¯Ì¼ÆøÌ壬¶þÑõ»¯Ì¼ºÍÇâÑõ»¯ÄÆ·´Ó¦Éú³É̼ËáÇâÄÆ£¬¶þÑõ»¯Ì¼ºÍÆ«ÂÈËáÄÆÈÜÒº·´Ó¦Éú³ÉÇâÑõ»¯ÂÁºÍ̼ËáÇâÄÆ£¬¹ýÂ˵õ½¹ÌÌåBΪÇâÑõ»¯ÂÁ£¬ÈÜÒºBΪ̼ËáÇâÄÆÈÜÒº£¬¹ÌÌåBÊÜÈÈ·Ö½âÉú³ÉÑõ»¯ÂÁ£¬
£¨1£©·¯ÍÁÓëÏ¡ÑÎËá·´Ó¦Éú³É½ðÊôÑôÀë×ÓºóÔÙºÍÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉµÄ³Áµí¾ÍÊǹÌÌåA£»
£¨2£©ÈÜÒºAµÄÈÜÖʾÍÊÇ·¯ÍÁÓëÏ¡ÑÎËá·´Ó¦Éú³É½ðÊôÑôÀë×ÓºóÔÙºÍÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉµÄ¿ÉÈÜÐÔÎïÖʺ¬ÓÐAlO₂^-£»
£¨3£©µç½âʱÑô¼«Ê§µç×ÓÉú³ÉÑõÆø£¬ÑõÆøÔÚ¸ßÎÂʱ»áÑõ»¯Ì¼¿é£¬Ê¹Ì¼¿éÏûºÄ£»
£¨4£©µç½âÑõ»¯ÂÁ£¬Òõ¼«ÉÏÊÇÂÁÀë×ӵõç×Ó·¢Éú»¹Ô­·´Ó¦Éú³ÉÂÁ£¬½«Al£¨OH£©₃ºÍNa₂CO₃һͬÈÜÓÚÇâ·úËáÀ´ÖÆÈ¡±ù¾§Ê¯£¬·´Ó¦Ê±·Å³öCO₂ÆøÌ壬¸ù¾ÝÔªËØÊغã¿Éд³ö»¯Ñ§·½³Ìʽ£»
£¨5£©AlCl₃Êǹ²¼Û»¯ºÏÎÔÚÈÛÈÚ״̬ϲ»µ¼µç£¬²»ÄÜÓÃÀ´µç½â£»
£¨6£©¸ù¾ÝAlÔªËØÖÊÁ¿Êغ㣬1molAlÉú³É0.5molAl₂O₃¼ÆË㣮

½â´ð ½â£º£¨1£©·°ÍÁÖ÷Òª³É·ÖÊÇAl₂O₃£¬»¹ÓÐSiO₂¡¢Fe₂O₃¡¢FeCO₃¡¢MgCO₃µÈÔÓÖÊ£¬³ýSiO₂²»·´Ó¦Í⣬ÓëÑÎËá·´Ó¦Éú³ÉµÄ½ðÊôÑôÀë×ÓÓÐAl³⁺¡¢Fe³⁺¡¢Fe²⁺¡¢Mg²⁺µÈ£¬ÕâЩ½ðÊôÑôÀë×ÓÔÙÓëÇâÑõ»¯ÄÆ·´Ó¦ºóÉú³ÉµÄ¿ÉÈÜÎïΪNaAlO₂£¬²»ÈÜÎïΪFe£¨OH£©₂¡¢Fe£¨OH£©₃¡¢Mg£¨OH£©₂£»
¹Ê´ð°¸Îª£ºFe£¨OH£©₂¡¢Fe£¨OH£©₃¡¢Mg£¨OH£©₂£»
£¨2£©ÓÉ£¨1£©ÖªÈÜÒºAÖ÷ҪΪNaAlO₂ÈÜÒººÍÊ£ÓàµÄÇâÑõ»¯ÄÆÈÜÒº£¬Í¨Èë¶þÑõ»¯Ì¼ÆøÌåÊ×ÏȺÍÇâÑõ»¯ÄÆ·´Ó¦CO₂+OH^-¨THCO₃^-£¬»¹ÓкÍNaAlO₂µÄ·´Ó¦AlO₂^-+CO₂+2H₂O=Al£¨OH£©₃¡ý+HCO₃^-£»
¹Ê´ð°¸Îª£ºCO₂+OH^-¨THCO₃^-¡¢AlO₂^-+CO₂+2H₂O=Al£¨OH£©₃¡ý+HCO₃^-£»
£¨3£©µç½âʱÑô¼«Ê§µç×ÓÉú³ÉÑõÆø£¬ÑõÆøÔÚ¸ßÎÂÌõ¼þÏ»áÑõ»¯Ì¼¿é£¬Ê¹Ì¼¿éÏûºÄ£¬ËùÒÔÐèÒª¶¨ÆÚ²¹³äÑô¼«Ì¼¿é£»
¹Ê´ð°¸Îª£ºµç½âÒ±Á¶ÂÁµÄÑô¼«²úÎïΪO₂£¬ÔÚ¸ßÎÂÌõ¼þÏ£¬O₂ÓëÑô¼«Ì¼·¢Éú·´Ó¦ÏûºÄÑô¼«Ì¼¿é£»
£¨4£©µç½âÑõ»¯ÂÁ£¬Òõ¼«ÉÏÊÇÂÁÀë×ӵõç×Ó·¢Éú»¹Ô­·´Ó¦Éú³ÉÂÁ£¬µç¼«·´Ó¦Ê½ÎªAl³⁺+3e^-¨TAl£¬½«Al£¨OH£©₃ºÍNa₂CO₃һͬÈÜÓÚÇâ·úËáÀ´ÖÆÈ¡±ù¾§Ê¯£¬·´Ó¦Ê±·Å³öCO₂ÆøÌ壬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Al£¨OH£©₃+3Na₂CO₃+12HF¨T2Na₃AlF₆+3CO₂¡ü+9H₂£¬
¹Ê´ð°¸Îª£ºAl³⁺+3e^-¨TAl£»2Al£¨OH£©₃+3Na₂CO₃+12HF¨T2Na₃AlF₆+3CO₂¡ü+9H₂O£»
£¨5£©AlCl₃Êǹ²¼Û»¯ºÏÎÔÚÈÛÈÚ״̬ϲ»µ¼µç£¬²»ÄÜÓÃÀ´µç½â£¬¶øÑõ»¯ÂÁΪÀë×Ó»¯ºÏÎÔÚÈÛÈÚ״̬Ï¿ɵ¼µç£¬¿ÉÒÔÓÃÀ´µç½â£»
¹Ê´ð°¸Îª£º²»¿ÉÐУ¬AlCl₃Êǹ²¼Û»¯ºÏÎÈÛÈÚ״̬µÄAlCl₃²»µ¼µç£»
£¨6£©¸ù¾ÝÔªËØÊغ㣬·¯ÍÁÖÐÿ1molAl»áÉú³É0.5molAl₂O₃£¬ÔòÒÔAl₂O₃±íʾµÄÖÊÁ¿·ÖÊý=$\frac{\frac{1.35t¡Á10{\;}^{6}g/t}{27g/mol}¡Á102g/mol}{10t¡Á10{\;}^{6}g/t}$=25.5%£¬¹Ê´ð°¸Îª£º25.5%£®

µãÆÀ ±¾ÌâΪÉú²ú¹¤ÒÕÁ÷³ÌÌ⣬Éæ¼°µç½â¡¢¼ÆËãµÈÎÊÌ⣬ÌâÄ¿½ÏΪ×ۺϣ¬×öÌâʱעÒâ×ÐϸÉóÌ⣬´ÓÌâÄ¿ÖлñÈ¡¹Ø¼üÐÅÏ¢£¬±¾ÌâÖеȣ®