ÌâÄ¿ÄÚÈÝ

£¨11·Ö£©²ÝËáÑÇÌúÓÃ×÷·ÖÎöÊÔ¼Á¼°ÏÔÓ°¼ÁµÈ£¬ÆäÖƱ¸Á÷³ÌÈçÏ£º

£¨1£©ÅäÖÆ(NH4)2Fe(SO4)26H2OÈÜҺʱ£¬Ðè¼ÓÉÙÁ¿Ï¡ÁòËᣬĿµÄÊÇ________________________________________________________¡£
£¨2£©½«ÖƵõIJúÆ·ÔÚë²ÆøÆø·ÕÖнøÐÐÈÈÖØ·ÖÎö£¬½á¹ûÈçͼ10£¨TG%±íʾ²ÐÁô¹ÌÌåÖÊÁ¿Õ¼Ô­ÑùÆ·×ÜÖÊÁ¿µÄ°Ù·ÖÊý£©¡£

¢ÙAB·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
__________________________________________¡£
C´¦Ê±²ÐÁôÎïµÄ»¯Ñ§Ê½Îª______________¡£
¢ÚÓÐÑо¿Ñ§ÕßÔÚʵÑé¹ý³ÌÖÐÓÃÆøÏàÉ«Æ×»¹¼ì³ö
H2£¬×îÖÕ²úÎïÖÐÒ²ÓÐ΢Á¿µÄ´ÅÐÔÎïÖÊÉú³É£¬_____________________________
ÇëÄã²ÂÏëÓÃÒ»¸ö·½³Ìʽ½âÊÍÕâÖÖÊÂʵ£º______________________________ ¡£
¢Û ÏÖÈ¡ÔÚÕæ¿ÕÖÐ146¡æÍÑË®ºóµÄFeC2O41.44g·ÅÔÚijÕæ¿ÕµÄÃܱÕÈÝÆ÷ÖУ¬ÔÙ³äÈë0.04molCO¡£¼ÓÈÈÖÁ1100¡æ£¬ÆäÖÐFeO(s)+CO(g)Fe(s)+CO2(g)·´Ó¦Æ½ºâ³£ÊýK=0.4£¬Ôò¸Ã·´Ó¦´ïƽºâʱ£¬FeOµÄת»¯ÂÊΪ¶àÉÙ£¿____________________________¡£
£¨1£©ÒÖÖÆFe2£«Ë®½â£¨2·Ö£©
£¨2£©¢ÙFeC2O42H2O(s)FeC2O4(s)+2H2O(g) £¨2·Ö£©£»   FeO£¨2·Ö£©
¢Ú3FeO+H2OFe3O4+H2  £¨2·Ö£©  ¢Û71.4%£¨3·Ö£©
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø