Question

14£®ÂÁÊÇÖØÒªµÄ½ðÊô£¬ÔÚ¹¤ÒµÉú²úÖÐÓÃ;¹ã·º£®
£¨1£©ÂÁÓëÑõ»¯Ìú·¢ÉúÂÁÈÈ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Al+Fe₂O₃$\frac{\underline{\;¸ßÎÂ\;}}{\;}$Al₂O₃+2Fe£»
£¨2£©¹¤ÒµÉÏÓÃÂÁÍÁ¿ó£¨º¬Ñõ»¯ÂÁ¡¢Ñõ»¯Ìú£©ÖÆÈ¡ÂÁµÄ¹ý³ÌÈçÏ£º
¢Ù²Ù×÷¢ñ¡¢²Ù×÷¢òºÍ III²Ù×÷¶¼ÊǹýÂË£¨Ìî²Ù×÷Ãû³Æ£©£¬ÊµÑéÊÒ½øÐиòÙ×÷ËùÐèµÄ²£Á§ÒÇÆ÷ÊÇÉÕ±­¡¢Â©¶·¡¢²£Á§°ô£®
¢Úд³öNa₂CO₃ÈÜÒºÓëCaO·´Ó¦µÄÀë×Ó·½³Ìʽ£ºCO₃^2-+CaO+H₂O¨T2OH^-+CaCO₃¡ý£®
¢Ûд³öÈÜÒºBͨCO₂µÃ³ÁµíDµÄÀë×Ó·½³Ìʽ2AlO₂^-+CO₂+3H₂O=2Al£¨OH£©₃¡ý+CO₃^2-£®
¢Üµç½âÈÛÈÚµÄÑõ»¯ÂÁʱ£¬ÈôµÃµ½±ê×¼×´¿öÏÂ11.2L O₂£¬ÔòͬʱÉú³É18gÂÁ£®
£¨3£©ÎÒ¹úÊ×´´µÄº£Ñóµç³ØÒÔAlµÄµ¥ÖÊΪ¸º¼«£¬²¬ÍøΪÕý¼«£¬¿ÕÆø×öÑõ»¯¼Á£¬º£Ë®×öµç½âÖÊÈÜÒº£®µç³ØµÄÕý¼«·´Ó¦Ê½Îª2H₂O+O₂+4e^-=4OH^-£®

Answer 1

·ÖÎö £¨1£©·´Ó¦Éú³ÉÑõ»¯ÂÁºÍFe£»
£¨2£©ÂÁÍÁ¿óÖÐÑõ»¯ÂÁºÍÑõ»¯Ìú£¬ÏòÂÁÍÁ¿óÖмÓÈëNaOHÈÜÒº£¬·¢Éú·´Ó¦Al₂O₃+2OH^-¨T2AlO₂^-+H₂O£¬Ñõ»¯Ìú²»ÈÜÓÚNaOHÈÜÒº£¬È»ºó¹ýÂË£¬µÃµ½ÈÜÒºBΪNaOH¡¢NaAlO₂»ìºÏÈÜÒº£¬³ÁµíCΪFe₂O₃£¬ÏòÈÜÒºBÖÐͨÈëCO₂£¬·¢Éú·´Ó¦2AlO₂^-+CO₂+3H₂O=2Al£¨OH£©₃¡ý+CO₃^2-¡¢2OH^-+CO₂=CO₃^2-+H₂O£¬È»ºó¹ýÂ˵õ½³ÁµíAl£¨OH£©₃ºÍNa₂CO₃ÈÜÒº£¬ÏòÈÜÒºÖмÓÈëCaO£¬·¢Éú·´Ó¦Na₂CO₃+CaO+H₂O¨T2NaOH+CaCO₃¡ý£¬È»ºó¹ýÂË£¬½«NaOHÑ­»·ÀûÓ㻽«Al£¨OH£©₃¼ÓÈȵõ½Al₂O₃£¬µç½âÈÛÈÚAl₂O₃µÃµ½Al£»
£¨3£©º£Ë®×öµç½âÖÊÈÜÒº£¬ÔòAlΪ¸º¼«Ê§È¥µç×Ó£¬Õý¼«ÉÏÑõÆøµÃµ½µç×Ó£®

½â´ð ½â£º£¨1£©ÂÁÓëÑõ»¯Ìú·¢ÉúÂÁÈÈ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Al+Fe₂O₃$\frac{\underline{\;¸ßÎÂ\;}}{\;}$Al₂O₃+2Fe£¬¹Ê´ð°¸Îª£º2Al+Fe₂O₃$\frac{\underline{\;¸ßÎÂ\;}}{\;}$Al₂O₃+2Fe£»
£¨2£©ÂÁÍÁ¿óÖÐÑõ»¯ÂÁºÍÑõ»¯Ìú£¬ÏòÂÁÍÁ¿óÖмÓÈëNaOHÈÜÒº£¬·¢Éú·´Ó¦Al₂O₃+2OH^-¨T2AlO₂^-+H₂O£¬Ñõ»¯Ìú²»ÈÜÓÚNaOHÈÜÒº£¬È»ºó¹ýÂË£¬µÃµ½ÈÜÒºBΪNaOH¡¢NaAlO₂»ìºÏÈÜÒº£¬³ÁµíCΪFe₂O₃£¬ÏòÈÜÒºBÖÐͨÈëCO₂£¬·¢Éú·´Ó¦2AlO₂^-+CO₂+3H₂O=2Al£¨OH£©₃¡ý+CO₃^2-¡¢2OH^-+CO₂=CO₃^2-+H₂O£¬È»ºó¹ýÂ˵õ½³ÁµíAl£¨OH£©₃ºÍNa₂CO₃ÈÜÒº£¬ÏòÈÜÒºÖмÓÈëCaO£¬·¢Éú·´Ó¦Na₂CO₃+CaO+H₂O¨T2NaOH+CaCO₃¡ý£¬È»ºó¹ýÂË£¬½«NaOHÑ­»·ÀûÓ㻽«Al£¨OH£©₃¼ÓÈȵõ½Al₂O₃£¬µç½âÈÛÈÚAl₂O₃µÃµ½Al£»
¢Ù²Ù×÷¢ñ¡¢²Ù×÷¢òºÍ III²Ù×÷¶¼ÊǹýÂË£¬ÊµÑéÊÒ½øÐиòÙ×÷ËùÐèµÄ²£Á§ÒÇÆ÷ÊÇÉÕ±­¡¢Â©¶·¡¢²£Á§°ô£¬¹Ê´ð°¸Îª£º¹ýÂË£»ÉÕ±­¡¢Â©¶·¡¢²£Á§°ô£»
¢ÚNa₂CO₃ÈÜÒºÓëCaO·´Ó¦µÄÀë×Ó·½³ÌʽΪCO₃^2-+CaO+H₂O¨T2OH^-+CaCO₃¡ý£¬¹Ê´ð°¸Îª£ºCO₃^2-+CaO+H₂O¨T2OH^-+CaCO₃¡ý£»
¢ÛÈÜÒºBͨCO₂µÃ³ÁµíDµÄÀë×Ó·½³ÌʽΪ2AlO₂^-+CO₂+3H₂O=2Al£¨OH£©₃¡ý+CO₃^2-£¬¹Ê´ð°¸Îª£º2AlO₂^-+CO₂+3H₂O=2Al£¨OH£©₃¡ý+CO₃^2-£»
¢Üµç½âÈÛÈÚµÄÑõ»¯ÂÁʱ£¬ÈôµÃµ½±ê×¼×´¿öÏÂ11.2L O₂£¬Óɵç×ÓÊغã¿ÉÖªÉú³ÉAlΪ$\frac{\frac{11.2L}{22.4L/mol}¡Á2¡Á£¨2-0£©}{£¨3-0£©}$¡Á27g/mol=18g£¬¹Ê´ð°¸Îª£º18£»
£¨3£©º£Ë®×öµç½âÖÊÈÜÒº£¬ÔòAlΪ¸º¼«Ê§È¥µç×Ó£¬Õý¼«ÉÏÑõÆøµÃµ½µç×Ó£¬Õý¼«·´Ó¦Îª2H₂O+O₂+4e^-=4OH^-£¬¹Ê´ð°¸Îª£º2H₂O+O₂+4e^-=4OH^-£®

µãÆÀ ±¾Ì⿼²é»ìºÏÎï·ÖÀëÌá´¿£¬Îª¸ßƵ¿¼µã£¬°ÑÎÕÁ÷³ÌÖеķ´Ó¦¡¢»ìºÏÎï·ÖÀë·½·¨¡¢Ô­µç³ØµÈΪ½â´ðµÄ¹Ø¼ü£¬²àÖØ·ÖÎöÓëÓ¦ÓÃÄÜÁ¦µÄ¿¼²é£¬×¢ÒâÔªËØ»¯ºÏÎï֪ʶµÄÓ¦Óã¬ÌâÄ¿ÄѶȲ»´ó£®