ÌâÄ¿ÄÚÈÝ

ij½á¾§Ë®ºÏÎﺬÓÐÁ½ÖÖÑôÀë×ÓºÍÒ»ÖÖÒõÀë×Ó¡£³ÆÈ¡ÖÊÁ¿Îª1.96 gµÄ¸Ã½á¾§Ë®ºÏÎÅä³ÉÈÜÒº¡£¼ÓÈë×ãÁ¿Ba(OH)2ÈÜÒº²¢¼ÓÈȸûìºÏÎÉú³É°×É«³Áµí£¬Ëæ¼´³Áµí±äΪ»ÒÂÌÉ«£¬×îºó±ä³ÉºìºÖÉ«£¬±ê¿öϲúÉú224 mLÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌ壻½«³Áµí¹ýÂË¡¢Ï´µÓ²¢×ÆÉÕÖÁºãÖØ£¬µÃµ½¹ÌÌå·ÛÄ©2.73 g£»ÓÃ×ãÁ¿Ï¡ÑÎËá´¦ÀíÉÏÊö·ÛÄ©£¬Ï´µÓºÍ¸ÉÔïºóµÃµ½°×É«¹ÌÌå2.33 g¡£
£¨1£©¸Ã½á¾§Ë®ºÏÎﺬÓеÄÁ½ÖÖÑôÀë×ÓÊÇ__________ºÍ__________£¬ÒõÀë×ÓÊÇ__________¡£Ð´³ö²úÉúÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌåµÄÀë×Ó·´Ó¦·½³Ìʽ£º            ¡£
£¨2£©ÊÔͨ¹ý¼ÆËãÈ·¶¨¸Ã½á¾§Ë®ºÏÎïµÄ»¯Ñ§Ê½         ¡£
(1)Fe2+   NH4+    SO42-     
NH4++OH-==NH3¡ü+H2
(2)(NH4)2Fe(SO4)2¡¤6H2O  
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2011?º¼ÖÝһ죩ij½á¾§Ë®ºÏÎﺬÓÐÁ½ÖÖÑôÀë×ÓºÍÒ»ÖÖÒõÀë×Ó£®³ÆÈ¡Á½·ÝÖÊÁ¿¾ùΪ45.3gµÄ¸Ã½á¾§Ë®ºÏÎ·Ö±ðÖÆ³ÉÈÜÒº£®ÏòÆäÖÐÒ»·ÝÖðµÎ¼ÓÈëNaOHÈÜÒº£¬¿ªÊ¼·¢ÏÖÈÜÒºÖгöÏÖ°×É«³Áµí²¢Öð½¥Ôö¶à£»Ò»¶Îʱ¼äºóÓÐÆøÌåÒݳö£¬¸ÃÆøÌåÓд̼¤ÐÔÆøÎ¶£¬ÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬¼ÓÈȺ󹲼ƿÉÊÕ¼¯µ½2.24L¸ÃÆøÌ壨±ê×¼×´¿ö£©£»×îºó°×É«³ÁµíÖð½¥¼õÉÙ²¢×îÖÕÏûʧ£®ÁíÒ»·ÝÖðµÎ¼ÓÈëBa£¨OH£©2ÈÜÒº£¬¿ªÊ¼ÏÖÏóÀàËÆ£¬µ«×îÖÕÈÔÓа×É«³Áµí£»¹ýÂË£¬ÓÃÏ¡ÏõËá´¦Àí³ÁµíÎ¾­Ï´µÓºÍ¸ÉÔµÃµ½°×É«¹ÌÌå46.6g£®
Çë»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©¸Ã½á¾§Ë®ºÏÎïÖк¬ÓеÄÁ½ÖÖÑôÀë×ÓÊÇ£º
NH4+
NH4+
ºÍ
Al3+
Al3+
£®
£¨2£©ÊÔͨ¹ý¼ÆËãÈ·¶¨¸Ã½á¾§Ë®ºÏÎïµÄ»¯Ñ§Ê½Îª
NH4Al£¨SO4£©2?12H2O[»ò£¨NH4£©2SO4?Al2£¨SO4£©3?24H2O]
NH4Al£¨SO4£©2?12H2O[»ò£¨NH4£©2SO4?Al2£¨SO4£©3?24H2O]
£®
£¨3£©¼ÙÉè¹ý³ÌÖÐÏò¸ÃÈÜÒºÖмÓÈëµÄBa£¨OH£©2ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ2.0mol?L-1£®
¢Ù¼ÓÈëBa£¨OH£©2ÈÜÒººó£¬ÈôËùµÃ³ÁµíµÄ×ÜÎïÖʵÄÁ¿×î´ó£¬Ôò·´Ó¦µÄÀë×Ó·½³ÌʽΪ
NH4++Al3++4OH-+2Ba2++2SO42-=2BaSO4¡ý+Al£¨OH£©3¡ý+NH3£®H2O
NH4++Al3++4OH-+2Ba2++2SO42-=2BaSO4¡ý+Al£¨OH£©3¡ý+NH3£®H2O
£®
¢ÚÈô¼ÓÈë75mLµÄBa£¨OH£©2ÈÜÒº£¬ÔòµÃµ½µÄ³ÁµíÖÊÁ¿Îª
42.75
42.75
g£®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø