Question

ÈçͼÊÇÖÐѧ»¯Ñ§Öг£¼ûÎïÖʵÄת»¯¹Øϵ£¬²¿·ÖÎïÖʺͷ´Ó¦Ìõ¼þÂÔÈ¥£®

£¨1£©µ¥ÖÊBµÄ»¯Ñ§Ê½ÊÇ

 

£¬µ¥ÖÊFµÄ»¯Ñ§Ê½ÊÇ

 

£¬Ð´³öÎïÖÊAµÄÒ»ÖÖÓÃ;

 

£®
£¨2£©Ð´³öµ¥ÖÊBÓëÇ¿¼îÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ

 

£®Ð´³öÓɳÁµíJÉú³É HÈÜÒºµÄÀë×Ó·½³Ìʽ

 

£®
£¨3£©ÈÜÒºEÖмÓÈ백ˮʱ£¬ÏÈÉú³É°×É«³ÁµíL£¬Ð´³öÉú³ÉLµÄÀë×Ó·½³Ìʽ

 

£®°×É«³ÁµíL»áѸËÙ±äΪ

 

É«£¬×îÖÕ±äΪºìºÖÉ«³ÁµíG£¬Ð´³öL±äΪGµÄ»¯Ñ§·´Ó¦·½³Ìʽ

 

£®
£¨4£©ÈÜÒºIÖÐËùº¬½ðÊôÀë×ÓÊÇ

 

£®

Answer 1

¿¼µã£ºÎÞ»úÎïµÄÍƶÏ

רÌ⣺ÍƶÏÌâ

·ÖÎö£º½ðÊôBÓëÇ¿¼î·´Ó¦µÃµ½µ¥ÖÊFÓëH£¬ÔòBΪAl¡¢FΪH₂£¬HΪƫÂÁËáÑΣ¬ºì×ØÉ«ÎïÖÊAÓë½ðÊôAl·¢ÉúÖû»·´Ó¦Éú³Éµ¥ÖÊCÓëÑõ»¯ÎïD£¬ÔòAΪFe₂O₃¡¢CΪFe¡¢DΪAl₂O₃£¬ÓÉת»¯¹Øϵ¿ÉÖª£¬EΪFeCl₂£¬HΪNaAlO₂£¬GΪFe£¨OH£©₃£¬ÈÜÒºIΪNaCl¡¢HCl¡¢AlCl₃»ìºÏÎJΪAl£¨OH£©₃£¬¾Ý´Ë½â´ð£®

½â´ð£º ½â£º½ðÊôBÓëÇ¿¼î·´Ó¦µÃµ½µ¥ÖÊFÓëH£¬ÔòBΪAl¡¢FΪH₂£¬HΪƫÂÁËáÑΣ¬ºì×ØÉ«ÎïÖÊAÓë½ðÊôAl·¢ÉúÖû»·´Ó¦Éú³Éµ¥ÖÊCÓëÑõ»¯ÎïD£¬ÔòAΪFe₂O₃¡¢CΪFe¡¢DΪAl₂O₃£¬ÓÉת»¯¹Øϵ¿ÉÖª£¬EΪFeCl₂£¬HΪNaAlO₂£¬GΪFe£¨OH£©₃£¬ÈÜÒºIΪNaCl¡¢HCl¡¢AlCl₃»ìºÏÎJΪAl£¨OH£©₃£¬
£¨1£©ÓÉÉÏÊö·ÖÎö¿ÉÖª£¬µ¥ÖÊBµÄ»¯Ñ§Ê½ÊÇAl£¬µ¥ÖÊFµÄ»¯Ñ§Ê½ÊÇH₂£¬ÎïÖÊAΪFe₂O₃£¬¿ÉÒÔÓÃ×÷ÓÍÆáÑÕÁÏ£¬Á¶ÌúµÈ£¬
¹Ê´ð°¸Îª£ºAl£»H₂£»ÓÃ×÷ÓÍÆáÑÕÁÏ£¬Á¶ÌúµÈ£»
£¨2£©µ¥ÖÊAlÓëÇ¿¼îÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£º2Al+2OH^-+2H₂O=2AlO₂^-+3H₂¡ü£¬ÓɳÁµíJÉú³É HÈÜÒºµÄÀë×Ó·½³ÌʽΪ£ºAl£¨OH£©₃+OH^-=AlO₂^-+2H₂O£¬
¹Ê´ð°¸Îª£º2Al+2OH^-+2H₂O=2AlO₂^-+3H₂¡ü£»Al£¨OH£©₃+OH^-=AlO₂^-+2H₂O£»
£¨3£©ÈÜÒºFeCl₂ÖмÓÈ백ˮʱ£¬ÏÈÉú³É°×É«³ÁµíLΪFe£¨OH£©₂£¬Àë×Ó·½³Ìʽ£ºFe²⁺+2OH^-=Fe£¨OH£©₂¡ý£¬°×É«³Áµí»áѸËÙ±äΪ»ÒÂÌÉ«£¬×îÖÕ±äΪºìºÖÉ«³Áµí£¬L±äΪGµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ£º4Fe£¨OH£©₂+O₂+2H₂O=4Fe£¨OH£©₃£¬
¹Ê´ð°¸Îª£ºFe²⁺+2OH^-=Fe£¨OH£©₂¡ý£»»ÒÂÌ£»4Fe£¨OH£©₂+O₂+2H₂O=4Fe£¨OH£©₃£»
£¨4£©ÈÜÒºIΪNaCl¡¢HCl¡¢AlCl₃»ìºÏÎËùº¬½ðÊôÀë×ÓÊÇNa⁺¡¢Al³⁺£¬¹Ê´ð°¸Îª£ºNa⁺¡¢Al³⁺£®

µãÆÀ£º±¾Ì⿼²éÎÞ»úÎïÍƶϣ¬ÎïÖʵÄÑÕÉ«¼°×ª»¯¹ØϵÖÐÌØÊâ·´Ó¦ÊÇÍƶÏÍ»ÆÆ¿Ú£¬ÔÙ½áºÏת»¯¹ØϵÍƶϣ¬ÐèҪѧÉúÊìÁ·ÕÆÎÕÔªËØ»¯ºÏÎï֪ʶ£¬ÄѶÈÖеȣ®