½â£ºÓÉKÓëJ¾ùÓÉ2ÖÖÏàͬµÄÔªËØ×é³É£¬KΪµ»ÆÉ«¹ÌÌ壬ÔòKΪNa
2O
2£¬JΪNa
2O£¬C¡¢E¡¢L¡¢IΪ³£¼û¹ÌÌåµ¥ÖÊ£¬C¡¢E¡¢L¶¼ÄÜÓëË®ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉÆøÌåµ¥ÖÊR£¬D¡¢FÖÐËùº¬µÄ·Ç½ðÊôÔªËØËùÐγɵĵ¥Öʳ£ÎÂÏÂΪ»ÆÂÌÉ«ÆøÌ壬ÔòLΪNa£¬RΪH
2£¬EΪFe£¬D£¨F£©ÎªFeCl
2£¬F£¨D£©ÎªFeCl
3£¬CΪC£¬AΪCO£¬BΪCO
2£¬ÓÖC¡¢L¡¢IÈýÔªËØÔ×ÓµÄÔ×ÓÐòÊýÖ®ºÍΪ30£¬IµÄÔ×ÓÐòÊýΪ30-11-6=13£¬ÔòIΪAl£¬GΪAl£¨OH£©
3£¬HΪNaAlO
2£¬
£¨1£©ÓÉÉÏÊö·ÖÎö¿ÉÖª£¬RΪÇâÆø£¬KΪNa
2O
2£¬Æäµç×ÓʽΪ
£¬¹Ê´ð°¸Îª£ºÇâÆø£»
£»
£¨2£©ÓÉBÉú³ÉCµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ2Mg+CO
22MgO+C£¬¹Ê´ð°¸Îª£º2Mg+CO
22MgO+C£»
£¨3£©IºÍÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ2Al+2OH
-+2H
2O=2AlO
2-+3H
2¡ü£¬HË®ÈÜҺˮ½â³Ê¼îÐÔ£¬ÔòÀë×Ó·´Ó¦ÎªAlO
2-+2H
2O?Al£¨OH£©
3+OH
-£¬
¹Ê´ð°¸Îª£º2Al+2OH
-+2H
2O=2AlO
2-+3H
2¡ü£»AlO
2-+2H
2O?Al£¨OH£©
3+OH
-£»
£¨4£©¶èÐԵ缫µç½âNaClÈÜÒº£¬ÓÉ2NaCl¡«2NaOH¡«2e
-£¬Ôòͨ¹ýµç×ÓΪ0.2mol£¬n£¨NaOH£©=0.2mol£¬c£¨NaOH£©=
=0.1mol/L£¬ËùÒÔpH=13£¬¹Ê´ð°¸Îª£º13£»
£¨5£©ÈôDÈÜÒºÖк¬ÓÐÔÓÖÊF£¬³ýÔÓ·½·¨ÎªÈôDΪFeCl
2£¬ÔòͨÈë×ãÁ¿µÄCl
2¼´¿É£®ÈôDΪFeCl
3£¬Ôò¼ÓÈë×ãÁ¿µÄFe£¬ÔÙ¹ýÂË£¬
¹Ê´ð°¸Îª£ºÈôDΪFeCl
2£¬ÔòͨÈë×ãÁ¿µÄCl
2¼´¿É£®ÈôDΪFeCl
3£¬Ôò¼ÓÈë×ãÁ¿µÄFe£¬ÔÙ¹ýÂË£®
·ÖÎö£ºÓÉKÓëJ¾ùÓÉ2ÖÖÏàͬµÄÔªËØ×é³É£¬KΪµ»ÆÉ«¹ÌÌ壬ÔòKΪNa
2O
2£¬JΪNa
2O£¬C¡¢E¡¢L¡¢IΪ³£¼û¹ÌÌåµ¥ÖÊ£¬C¡¢E¡¢L¶¼ÄÜÓëË®ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉÆøÌåµ¥ÖÊR£¬D¡¢FÖÐËùº¬µÄ·Ç½ðÊôÔªËØËùÐγɵĵ¥Öʳ£ÎÂÏÂΪ»ÆÂÌÉ«ÆøÌ壬ÔòLΪNa£¬RΪH
2£¬EΪFe£¬D£¨F£©ÎªFeCl
2£¬F£¨D£©ÎªFeCl
3£¬CΪC£¬AΪCO£¬BΪCO
2£¬ÓÖC¡¢L¡¢IÈýÔªËØÔ×ÓµÄÔ×ÓÐòÊýÖ®ºÍΪ30£¬IµÄÔ×ÓÐòÊýΪ30-11-6=13£¬ÔòIΪAl£¬GΪAl£¨OH£©
3£¬HΪNaAlO
2£¬È»ºó½áºÏµ¥Öʼ°»¯ºÏÎïµÄÐÔÖÊÀ´½â´ð£®
µãÆÀ£º±¾Ì⿼²éÎÞ»úÎïµÄÍƶϣ¬K¼°C¡¢E¡¢L¶¼ÄÜÓëË®ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉÆøÌåµ¥ÖÊRΪ½â´ð±¾ÌâµÄÍ»ÆÆ¿Ú£¬×¢ÒâÀûÓÃת»¯¹ØϵÍƶϳö¸÷ÎïÖÊÊǽâ´ðµÄ¹Ø¼ü£¬ÊìϤ³£¼ûµÄµ¥Öʼ°»¯ºÏÎïµÄÐÔÖʼ´¿É½â´ð£¬×¢Òâ»ù´¡ÖªÊ¶µÄ»ýÀۺ͹éÄÉ£¬ÌâÄ¿ÄѶÈÖеȣ®