ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿£¨1£©.FeCl3ÊÇÖÐѧʵÑéÊÒ³£ÓõÄÊÔ¼Á£¬¿ÉÒÔÓÃÀ´ÖƱ¸ÇâÑõ»¯Ìú½ºÌå¡£
ÏÂÁÐÖƱ¸ÇâÑõ»¯Ìú½ºÌåµÄ²Ù×÷·½·¨ÕýÈ·µÄÊÇ____________£¨Ìî×Öĸ£©£»
A£®Ïò±¥ºÍÂÈ»¯ÌúÈÜÒºÖеμÓÉÙÁ¿µÄÇâÑõ»¯ÄÆÏ¡ÈÜÒº
B£®¼ÓÈÈÖó·ÐÂÈ»¯Ìú±¥ºÍÈÜÒº
C£®ÔÚ°±Ë®ÖеμÓÂÈ»¯ÌúŨÈÜÒº
D£®ÔÚ·ÐË®Öеμӱ¥ºÍÂÈ»¯ÌúÈÜÒº£¬Öó·ÐÖÁÒºÌå³ÊºìºÖÉ«¡£
£¨2£©Ð´³öBa(OH)2ÈÜÒºÓëÉÙÁ¿NaHCO3ÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ:___________________¡£
£¨3£©Ð´³ö´ÎÂÈËáµÄµç×Óʽ£º________________£»Ð´³ö¹ýÑõ»¯ÄÆÖдæÔÚµÄËùÓл¯Ñ§¼üÀàÐÍ£º____________________¡£
£¨4£©Ìú·ÛÖк¬ÓÐÂÁ·Û£¬¿É¼Ó______³ýÈ¥ÔÓÖÊ£¬·¢ÉúµÄ»¯Ñ§·½³ÌʽΪ______________¡£
¡¾´ð°¸¡¿D Ba2++OH-+HCO3-=BaCO3¡ý+H2O Àë×Ó¼üºÍ¹²¼Û¼ü£¨»òÀë×Ó¼üºÍ·Ç¼«ÐÔ¼ü£© ÇâÑõ»¯ÄÆÈÜÒº 2Al+2NaOH+2H2O=2NaAlO2+3H2¡ü
¡¾½âÎö¡¿
(1)ʵÑéÊÒÖƱ¸ÇâÑõ»¯Ìú½ºÌåµÄ·½·¨Îª£ºÏò·ÐË®Öеμӱ¥ºÍÂÈ»¯ÌúÈÜÒº¼ÓÈȵ½ºìºÖÉ«£»½ºÌåÊǾùÒ»Îȶ¨µÄ·Öɢϵ£»
(2)ËáʽÑÎÓë¼î·´Ó¦£¬ÒªÒÔ²»×ãÁ¿µÄÎïÖÊΪ±ê׼дÀë×Ó·½³Ìʽ£»
(3)½áºÏÔ×ӽṹ¼°ÎïÖʽáºÏ˳Ðòдµç×Óʽ¡¢Åжϻ¯ºÏ¼Û£»
(4)ÀûÓöþÕßÐÔÖʵIJ»Í¬µã³ýÔÓ¡£
(1)A£®±¥ºÍÂÈ»¯ÌúÈÜÒºÖеμÓÊÊÁ¿µÄÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉÇâÑõ»¯Ìú³Áµí£¬µÃ²»µ½ÇâÑõ»¯Ìú½ºÌ壬ѡÏîA´íÎó£»
B£®ÂÈ»¯ÌúÔÚÈÜÒºÖÐË®½âÉú³ÉÇâÑõ»¯Ìú£¬¼ÓÈÈ´Ù½øË®½â£¬Öó·Ð»á³öÏÖºìºÖÉ«³Áµí£¬µÃµ½µÄÊÇÇâÑõ»¯Ìú³Áµí£¬Ñ¡ÏîB´íÎó£»
C£®ÔÚ°±Ë®ÖеμÓÂÈ»¯ÌúŨÈÜÒº·´Ó¦Éú³ÉÇâÑõ»¯Ìú³Áµí£¬µÃ²»µ½ÇâÑõ»¯Ìú½ºÌ壬ѡÏîC´íÎó£»
D£®ÊµÑéÊÒÖƱ¸ÇâÑõ»¯Ìú½ºÌåµÄ·½·¨ÊÇÔÚ·ÐË®Öеμӱ¥ºÍÂÈ»¯ÌúÈÜÒº¼ÓÈÈ£¬µÃµ½ºìºÖÉ«ÒºÌåΪÇâÑõ»¯Ìú½ºÌ壬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºFeCl3+3H2O(ÈÈ) Fe(OH)3(½ºÌå)+3HCl£¬Ñ¡ÏîDÕýÈ·£»
(2) Ba(OH)2ÈÜÒºÓëÉÙÁ¿NaHCO3ÈÜÒº·´Ó¦Ê±£¬NaHCO3²»×ãÁ¿£¬ÍêÈ«·´Ó¦£¬Àë×Ó·½³ÌʽÊÇ£ºBa2++OH-+HCO3-=BaCO3¡ý+H2O£»
(3)´ÎÂÈËá·Ö×ÓÖÐOÔ×Ó·Ö±ðÓëH¡¢ClÔ×ÓÐγÉÒ»¶Ô¹²Óõç×Ó¶Ô£¬µç×ÓʽΪ£»¹ýÑõ»¯ÄÆÖÐÁ½¸öÑõÔ×Ó¼äÐγÉÒ»¸ö·Ç¼«ÐÔ¹²¼Û¼ü£¬Ã¿¸öOÔ×ÓÔÙ·Ö±ðÓëÄÆÐγÉÐγÉÀë×Ó¼ü£»
(4)Al¡¢Fe¶¼ÊDZȽϻîÆõĽðÊô£¬¶¼¿ÉÒÔÓëËá·¢Éú·´Ó¦£¬¶øAl»¹¿ÉÒÔÓëÇ¿¼îÈÜÒº·´Ó¦£¬Fe²»ÄÜ·´Ó¦£¬ËùÒÔ³ýÈ¥Ìú·ÛÖк¬ÓÐÂÁ·Û¿ÉÒÔÏò»ìºÏÎïÖмÓÈë×ãÁ¿NaOHÈÜÒº³ýÈ¥£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Al+2NaOH+2H2O=2NaAlO2+3H2¡ü¡£