ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿¾ÛºÏÁòËáÌúÊÇÒ»ÖÖÐÂÐ͸ßЧµÄÎÞ»ú¸ß·Ö×ÓÐõÄý¼Á¡£ÓÃÁòËáÑÇÌú¾§Ìå¼°ÁòËáΪÔÁÏ´ß»¯Ñõ»¯·¨Éú³ÉÁòËáÌú£¬ÔÙË®½â¡¢¾ÛºÏ³É²úÆ·¡£ÊµÑéÊÒÄ£ÄâÉú²ú¹ý³ÌÈçÏ£º
(1)¢ÙÓÃÔÁÏÅäÖÆ2.50 mol/LµÄFeSO4ÈÜҺʱÓõ½µÄ¶¨Á¿ÒÇÆ÷ÓУº____________________
¢Úд³öÑõ»¯¹ý³ÌÖеÄÀë×Ó·½³Ìʽ£º_________________________________________
(2)×ۺϿ¼ÂÇʵ¼ÊͶÁÏÁòËáÑÇÌúÓëÁòËáµÄÎïÖʵÄÁ¿Ö®±ÈΪ×óÓÒ×î¼Ñ¡£ ¼ÓÈëµÄÁòËá±ÈÀíÂÛÖµÉԶ࣬µ«²»Äܹý¶àµÄÔÒòÊÇ____________________________________________________________________¡£
(3)ÁòËáÌúÈÜҺˮ½â¿ÉÒԵõ½Ò»ÏµÁоßÓо»Ë®×÷ÓõļîʽÁòËáÌú(x Fe2O3¡¤y SO3¡¤z H2O)£¬ÏÖ²ÉÓÃÖØÁ¿·¨²â¶¨x¡¢y¡¢zµÄÖµ¡£
¢Ù²â¶¨Ê±ËùÐèµÄÊÔ¼Á____________¡£
(a) NaOH (b) Ba(OH)2 (c) BaCl2 (d) FeSO4
¢ÚÐèÒª²â¶¨____________ºÍ____________µÄÖÊÁ¿(Ìîд»¯ºÏÎïµÄ»¯Ñ§Ê½)¡£
(4)Ñ¡³ö²â¶¨¹ý³ÌÖÐËùÐèµÄ»ù±¾²Ù×÷____________(°´²Ù×÷ÏȺó˳ÐòÁгö)¡£
a.¹ýÂË¡¢Ï´µÓ
b.Õô·¢¡¢½á¾§
c.ÝÍÈ¡¡¢·ÖÒº
d.ÀäÈ´¡¢³ÆÁ¿
e.ºæ¸É»ò×ÆÉÕ
¡¾´ð°¸¡¿µç×ÓÌìƽ¡¢ÈÝÁ¿Æ¿ 4Fe2+ + O2 + 4H+ 4Fe3+ + 2H2O ÁòËáÉÔ¹ýÁ¿ÊÇΪÁË·ÀÖ¹Fe3+Ë®½â³ÉÇâÑõ»¯Ìú£¬µ±ÁòËá¹ý¶àʱ£¬ºóÐøÖкÍÐèÒªµÄ¼îµÄÓÃÁ¿´ó£¬Ôì³ÉÀË·Ñ ac Fe2O3 BaSO4 aed
¡¾½âÎö¡¿
(1)¢ÙÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒºÒª³ÆÁ¿ÈÜÖʵÄÖÊÁ¿£¬ÓÃÈÝÁ¿Æ¿¶¨ÈÝ£»
¢ÚÁòËáÑÇÌúÔÚËáÐÔÌõ¼þϱ»ÑõÆøÑõ»¯³ÉÁòËáÌú£»
(2)ÁòËá¹ý¶à£¬ÔÚºóÃæµ÷½ÚpHʱҪÏûºÄµÄÇâÑõ»¯ÄƵÄÁ¿¾Í¶à£¬Ôì³ÉÀË·Ñ£»
(3)¢Ù²ÉÓÃÖØÁ¿·¨²â¶¨¼îʽÁòËáÌú(x Fe2O3y SO3z H2O)ÖÐx¡¢y¡¢zµÄֵʱ£¬¿ÉÒÔ½«ÑùÆ·ÈÜÓÚÇâÑõ»¯ÄÆÈÜÒº£¬¸ù¾ÝµÃµ½µÄ¹ÌÌåÑõ»¯ÌúµÄÖÊÁ¿È·¶¨xµÄÖµ£¬¹ýÂË£¬ÔÚËùµÃÂËÒºÖмÓÂÈ»¯±µ£¬¸ù¾Ý²úÉúµÄÁòËá±µ³ÁµíµÄÖÊÁ¿¿ÉÈ·¶¨yµÄÖµ£¬¸ù¾ÝÑùÆ·µÄ×ÜÖÊÁ¿½áºÏFe2O3ºÍ¼ÆËãµÃµÄSO3µÄÖÊÁ¿¿ÉÈ·¶¨zµÄÖµ£»
¢Ú¸ù¾Ý¢ÙµÄ·ÖÎö¿ÉÖª£¬Òª²â¶¨Fe2O3¡¢BaSO4µÄÖÊÁ¿£»
(4)²â¶¨¹ý³ÌÖн«ÑùÆ·ÈÜÓÚÇâÑõ»¯ÄÆÈÜÒº£¬¾¹ý¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É»ò×ÆÉÕ¡¢ÀäÈ´¡¢³ÆÁ¿µÃÑõ»¯ÌúµÄÖÊÁ¿£¬ÔÚËùµÃÂËÒºÖмÓÂÈ»¯±µ£¬¾¹ý¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É»ò×ÆÉÕ¡¢ÀäÈ´¡¢³ÆÁ¿µÃÁòËá±µµÄÖÊÁ¿£¬¾Ý´Ë´ðÌâ¡£
(1)¢ÙÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒºÒª³ÆÁ¿ÈÜÖʵÄÖÊÁ¿£¬ÒªÓõç×ÓÌìƽ£¬ÓÃÈÝÁ¿Æ¿¶¨ÈÝ£¬¹Ê´ð°¸Îª£ºµç×ÓÌìƽ¡¢ÈÝÁ¿Æ¿£»
¢ÚÁòËáÑÇÌúÔÚËáÐÔÌõ¼þϱ»ÑõÆøÑõ»¯³ÉÁòËáÌú£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ4Fe2++O2+4H+=4Fe3++2H2O£¬¹Ê´ð°¸Îª£º4Fe2++O2+4H+=4Fe3++2H2O£»
(2)ÁòËáÉÔ¹ýÁ¿ÊÇΪÁË·ÀÖ¹Fe3+Ë®½â³ÉÇâÑõ»¯Ìú£¬µ±ÁòËá¹ý¶àʱ£¬ºóÐøÖкÍÐèÒªµÄ¼îµÄÓÃÁ¿Á¦´ó£¬Ôì³ÉÀË·Ñ£¬ËùÒÔ¼ÓÈëµÄÁòËá±ÈÀíÂÛÖµÉԶ࣬µ«²»Äܹý¶à£¬¹Ê´ð°¸Îª£ºÁòËáÉÔ¹ýÁ¿ÊÇΪÁË·ÀÖ¹Fe3+Ë®½â³ÉÇâÑõ»¯Ìú£¬µ±ÁòËá¹ý¶àʱ£¬ºóÐøÖкÍÐèÒªµÄ¼îµÄÓÃÁ¿Á¦´ó£¬Ôì³ÉÀË·Ñ£»
(3)¢Ù²ÉÓÃÖØÁ¿·¨²â¶¨¼îʽÁòËáÌú(x Fe2O3y SO3z H2O)ÖÐx¡¢y¡¢zµÄֵʱ£¬¿ÉÒÔ½«ÑùÆ·ÈÜÓÚÇâÑõ»¯ÄÆÈÜÒº£¬¸ù¾ÝµÃµ½µÄ¹ÌÌåÑõ»¯ÌúµÄÖÊÁ¿È·¶¨xµÄÖµ£¬¹ýÂË£¬ÔÚËùµÃÂËÒºÖмÓÂÈ»¯±µ£¬¸ù¾Ý²úÉúµÄÁòËá±µ³ÁµíµÄÖÊÁ¿¿ÉÈ·¶¨yµÄÖµ£¬¸ù¾ÝÑùÆ·µÄ×ÜÖÊÁ¿½áºÏFe2O3ºÍ¼ÆËãµÃµÄSO3µÄÖÊÁ¿¿ÉÈ·¶¨zµÄÖµ£¬¹ÊÑ¡ac£»
¢Ú¸ù¾Ý¢ÙµÄ·ÖÎö¿ÉÖª£¬Òª²â¶¨Fe2O3¡¢BaSO4µÄÖÊÁ¿£¬¹Ê´ð°¸Îª£ºFe2O3¡¢BaSO4£»
(4)²â¶¨¹ý³ÌÖн«ÑùÆ·ÈÜÓÚÇâÑõ»¯ÄÆÈÜÒº£¬¾¹ý¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É»ò×ÆÉÕ¡¢ÀäÈ´¡¢³ÆÁ¿µÃÑõ»¯ÌúµÄÖÊÁ¿£¬ÔÚËùµÃÂËÒºÖмÓÂÈ»¯±µ£¬¾¹ý¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É»ò×ÆÉÕ¡¢ÀäÈ´¡¢³ÆÁ¿µÃÁòËá±µµÄÖÊÁ¿£¬¹ÊÑ¡aed¡£