ÌâÄ¿ÄÚÈÝ

ijÈÜÒºÖнöº¬Ï±íÀë×ÓÖеÄ5ÖÖÀë×Ó£¨²»¿¼ÂÇË®µÄµçÀë¼°Àë×ÓµÄË®½â£©£¬ÇÒÀë×ÓµÄÎïÖʵÄÁ¿¾ùΪ1mol¡£
ÒõÀë×ÓSO42-¡¢NO3-¡¢Cl-
ÑôÀë×ÓFe3+¡¢Fe2+¡¢NH4+¡¢Cu2+¡¢Al3+
 
¢ÙÈôÏòÔ­ÈÜÒºÖмÓÈëKSCNÈÜÒº£¬ÎÞÃ÷ÏԱ仯¡£¢ÚÈôÏòÔ­ÈÜÒºÖмÓÈë¹ýÁ¿µÄÑÎËᣬÓÐÆøÌåÉú³É£¬ÈÜÒºÖÐÒõÀë×ÓÖÖÀ಻±ä¡£¢ÛÈôÏòÈÜÒºÖмÓÈëBaCl2ÈÜÒº£¬Óа×É«³ÁµíÉú³ÉÊԻشðÏÂÁÐÎÊÌâ¡£
£¨1£©ÈôÏÈÏòÔ­ÈÜÒºÖмÓÈë¹ýÁ¿µÄÑÎËᣬÔÙ¼ÓÈëKSCNÈÜÒº£¬ÏÖÏóÊÇ________¡£
£¨2£©Ô­ÈÜÒºÖк¬ÓеÄÑôÀë×ÓÊÇ________¡£
£¨3£©ÏòÔ­ÈÜÒºÖмÓÈë×ãÁ¿µÄÑÎËᣬ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ________¡£
£¨4£©ÏòÔ­ÈÜÒºÖмÓÈë×ãÁ¿µÄNaOHÈÜÒº£¬³ä·Ö·´Ó¦ºó£¬¹ýÂË¡¢Ï´µÓ¡¢×ÆÉÕ£¬×îÖÕËùµÃ¹ÌÌåÓÃÍÐÅÌÌìƽ³ÆÁ¿ÖÊÁ¿Îª________¡£

ÈÜÒº±äΪѪºìÉ«    Fe2+¡¢Cu2+    3Fe2++4H++NO3-£½3Fe3++NO¡ü+2H2O    160.0g
ÊÔÌâ·ÖÎö£ºÈôÏòÔ­ÈÜÒºÖмÓÈëKSCNÈÜÒº£¬ÎÞÃ÷ÏԱ仯£¬Õâ˵Ã÷ÈÜÒºÖÐûÓÐÌúÀë×Ó¡£ÈôÏòÔ­ÈÜÒºÖмÓÈë¹ýÁ¿µÄÑÎËᣬÓÐÆøÌåÉú³É£¬ÈÜÒºÖÐÒõÀë×ÓÖÖÀ಻±ä£¬Òò´ËÈÜÒºÖÐÒ»¶¨º¬ÓÐÂÈÀë×Ó¡£ÄܲúÉúÆøÅݵÄÖ»ÄÜÊÇNO3£­ÔÚËáÐÔÌõ¼þϱ»»¹Ô­Éú³ÉNO£¬Òò´ËÒ»¶¨»¹º¬Óл¹Ô­ÐÔÀë×ÓÑÇÌúÀë×Ó£¬Í¬Ê±»¹º¬ÓÐNO3£­¡£¢ÛÈôÏòÈÜÒºÖмÓÈëBaCl2ÈÜÒº£¬Óа×É«³ÁµíÉú³É£¬Õâ˵Ã÷»¹º¬ÓÐSO42-¡£ÓÉÓÚÀë×ÓµÄÎïÖʵÄÁ¿¾ùΪ1mol£¬Ôò¸ù¾ÝÈÜÒºµÄµçÖÐÐÔ¿ÉÖª£¬ÈÜÒºÖÐÒ»¶¨»¹º¬ÓÐÑôÀë×Ó¡£ÓÉÓÚÖ»Äܺ¬ÓÐ5ÖÖÀë×Ó£¬Ôò¸ù¾ÝÒõÀë×ӵĵçºÉÊýÊÇ4mol¿ÉÖª£¬ÁíÍâÒ»ÖÖÑôÀë×ÓÊÇÍ­Àë×Ó¡£
£¨1£©ÔÚËáÐÔÌõ¼þÏÂNO3£­ÄÜ°ÑFe2£«Ñõ»¯Éú³ÉFe3£«£¬ËùÒÔÈôÏÈÏòÔ­ÈÜÒºÖмÓÈë¹ýÁ¿µÄÑÎËᣬÔÙ¼ÓÈëKSCNÈÜÒº£¬ÏÖÏóÊÇÈÜÒº±äΪѪºìÉ«¡£
£¨2£©¸ù¾ÝÒÔÉÏ·ÖÎö¿ÉÖª£¬Ô­ÈÜÒºÖк¬ÓеÄÑôÀë×ÓÊÇFe2+¡¢Cu2+¡£
£¨3£©¼ÓÈëÑÎËᣬ¾ßÓÐÑõ»¯ÐÔµÄNO3-ºÍ»¹Ô­ÐÔµÄFe2+·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ3Fe2++4H++NO3-£½3Fe3++NO¡ü+2H2O¡£
£¨4£©Ô­ÈÜÒºÖÐËùº¬ÑôÀë×ÓÊÇFe2+¡¢Cu2+£¬ÈôÏòÔ­ÈÜÒºÖмÓÈë×ãÁ¿µÄNaOHÈÜÒº£¬³ä·Ö·´Ó¦ºó¹ýÂË£¬Ï´µÓ£¬×ÆÉÕÖÁºãÖØ£¬µÃµ½µÄ¹ÌÌåÊÇCuO¡¢Fe2O3£¬¸ù¾ÝÌâÒâ¸÷Àë×ÓµÄÎïÖʵÄÁ¿¾ùΪ1mol¿ÉÖª£¬m£¨CuO£©£½1mol¡Á80g/mol=80g£¬m£¨Fe2O3£©£½0.5mol¡Á160g/mol£½80g£¬ËùµÃ¹ÌÌåµÄÖÊÁ¿Îª80g+80g£½160.0g¡£
¿¼µã£º¿¼²éÀë×Ó¼ìÑé¡¢¼ø±ð¡¢Ñõ»¯»¹Ô­·´Ó¦µÄÊéдÒÔ¼°ÓйؼÆËã
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¢ñ£®Ä³ÈÜÒºÖнöº¬Ï±íÀë×ÓÖеÄ5ÖÖÀë×Ó£¨²»¿¼ÂÇË®µÄµçÀë¼°Àë×ÓµÄË®½â£©£¬ÇÒÀë×ÓµÄÎïÖʵÄÁ¿¾ùΪ1mol¡£

ÒõÀë×Ó

SO42-¡¢NO3-¡¢Cl-

ÑôÀë×Ó

Fe3+¡¢Fe2+¡¢NH4+¡¢Cu2+¡¢Al3+

 

¢ÙÈôÏòÔ­ÈÜÒºÖмÓÈëKSCNÈÜÒº£¬ÎÞÃ÷ÏԱ仯¡£¢ÚÈôÏòÔ­ÈÜÒºÖмÓÈë¹ýÁ¿µÄÑÎËᣬÓÐÆøÌåÉú³É£¬ÈÜÒºÖÐÒõÀë×ÓÖÖÀ಻±ä¡£¢ÛÈôÏòÔ­ÈÜÒºÖмÓÈëBaCl2ÈÜÒº£¬Óа×É«³ÁµíÉú³É¡£ÊԻشðÏÂÁÐÎÊÌâ

£¨1£©ÈôÏÈÏòÔ­ÈÜÒºÖмÓÈë¹ýÁ¿µÄÑÎËᣬÔÙ¼ÓÈëKSCNÈÜÒº£¬ÏÖÏóÊÇ             ¡£

£¨2£©Ô­ÈÜÒºÖк¬ÓеÄÑôÀë×ÓÊÇ                ¡£

£¨3£©ÏòÔ­ÈÜÒºÖмÓÈë×ãÁ¿µÄÑÎËᣬ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                   ¡£

£¨4£©ÏòÔ­ÈÜÒºÖмÓÈë×ãÁ¿µÄNaOHÈÜÒº£¬³ä·Ö·´Ó¦ºó£¬¹ýÂË¡¢Ï´µÓ¡¢×ÆÉÕ£¬×îÖÕËùµÃ¹ÌÌåÓÃÍÐÅÌÌìƽ³ÆÁ¿ÖÊÁ¿Îª             ¡£

¢ò. ²ÝËáÑÇÌú¾§Ì壨FeC2O4¡¤2H2O£©¡¢Ì¼Ëá﮺ͶþÑõ»¯¹èÔÚë²ÆøÖиßη´Ó¦¿ÉÖƱ¸ï®µç³ØµÄÕý¼«²ÄÁϹèËáÑÇÌúﮣ¨Li2FeSiO4£©¡£²ÝËáÑÇÌú¾§ÌåÔÚë²ÆøÆø·ÕÖнøÐÐÈÈÖØ·ÖÎö£¬½á¹ûÈçÓÒͼËùʾ£¨TG%±íʾ²ÐÁô¹ÌÌåÖÊÁ¿Õ¼Ô­ÑùÆ·×ÜÖÊÁ¿µÄ°Ù·ÖÊý£©,Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨5£©²ÝËáÑÇÌú¾§ÌåÖÐ̼ԪËصĻ¯ºÏ¼ÛΪ£º              

£¨6£©A¡úB·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                    ¡£

£¨7£©¾«È·Ñо¿±íÃ÷£¬B¡úCʵ¼ÊÊÇ·ÖÁ½²½½øÐеģ¬Ã¿Ò»²½Ö»ÊÍ·ÅÒ»ÖÖÆøÌ壬µÚ¶þ²½ÊͷŵÄÆøÌåµÄÏà¶Ô·Ö×ÓÖÊÁ¿½ÏµÚÒ»²½µÄ´ó£¬ÔòµÚÒ»²½ÊͷŵÄÆøÌ廯ѧʽΪ£º          £»ÊͷŵڶþÖÖÆøÌåʱ£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                ¡£

 

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

¾«Ó¢¼Ò½ÌÍø