ÌâÄ¿ÄÚÈÝ

£¨8·Ö£©Cu¡ªFeºÏ½ðÓÉÓÚ³¤Ê±¼äÖÃÓÚ¿ÕÆøÖУ¬±íÃæ²úÉúÁËÒ»²ãÑõ»¯Ä¤£¨³É·ÝΪFe2O3ºÍCuO£©£¬ÏÖ½øÐÐÈçÏÂʵÑ飨ÒÔÏÂÆøÌåÌå»ý¾ùÔÚ±ê¿öϲâµÃ£©£º

¢Ù½«´ËºÏ½ð¿é5.76gÖÃÓÚÉÕ±­ÖУ¬È»ºó½«Ï¡H2SO4Öð½¥»ºÂý¼ÓÈëÖÁ¹ýÁ¿£¬ÊÕ¼¯²úÉúµÄÆøÌåΪ672mL£¬¹ýÂ˵ÃdzÂÌÉ«ÈÜÒºA£¬»¹ÓÐÂËÔüB¡£

¢Ú½«ÂËÔüBͶÈëµ½Ò»¶¨Å¨¶ÈµÄHNO3ÖУ¬ÍêÈ«Èܽ⣬µÃNO¡¢NO2µÄ»ìºÏÆøÌå896mL£¬¾­²â¶¨£¬Í¬ÎÂͬѹÏ´˻ìºÏÆøÌå¶ÔÇâÆøµÄÏà¶ÔÃܶÈΪ17¡£

¢Û½«¢ÙÖÐËùµÃÂËÒº¼ÓÈ뵽ͬŨ¶È×ãÁ¿µÄHNO3ÖУ¬ÓÃÅÅË®·¨ÊÕ¼¯Ò»ÉÕÆ¿ÆøÌ壬ÔÙÏòÉÕÆ¿ÖÐͨÈë224mL O2£¬ÆøÌåÇ¡ºÃÍêÈ«ÈÜÓÚË®¡£

£¨1£©AÖдæÔÚµÄÑôÀë×ÓÓР                £»

£¨2£©896mL»ìºÏÆøÌåÖÐNO¡¢NO2µÄÎïÖʵÄÁ¿Ö®±ÈΪ               £»

£¨3£©BµÄµ¥ÖÊΪ               £¬ÖÊÁ¿Îª                  g£»

£¨4£©¢ÛÖб»HNO3Ñõ»¯Á˵ÄÑôÀë×ÓµÄÎïÖʵÄÁ¿Îª               mol£»

£¨5£© ´ËºÏ½ð¿éÖÐÑõÔªËصÄÖÊÁ¿Îª                g¡£

 

¡¾´ð°¸¡¿

£¨¹²8·Ö£©

£¨1£©Fe2+¡¢H+  1·Ö)  

£¨2£©3£º1£¨1·Ö£© 

£¨3£©Cu £¨1·Ö£©  3.2 g£¨1·Ö£©

£¨4£©0.04 mol £¨2·Ö£© £¨5£©0.32£¨2·Ö£©

¡¾½âÎö¡¿ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
Cu-FeºÏ½ðÓÉÓÚ³¤Ê±¼äÖÃÓÚ¿ÕÆøÖбíÃæ²úÉúÁËÒ»²ãÑõ»¯Ä¤£¨³É·ÖΪFe2O3ºÍCuO£©£¬Ò»»¯Ñ§Ñо¿ÐÔѧϰС×é¶ÔÆä½øÐÐÈçÏÂ̽¾¿£¬Çë°´ÒªÇóÍê³ÉÏÂÁÐ̽¾¿±¨¸æ£®
[̽¾¿Ä¿µÄ]ʵÑé²â¶¨ÑùÆ·ÖÐFe£¬CuµÄÖÊÁ¿£®
[̽¾¿Ë¼Â·]Éè¼ÆʵÑé²âÁ¿Óйط´Ó¦ÎïºÍÉú³ÉÎïµÄÁ¿£¬²¢Í¨¹ý¼ÆËãÈ·¶¨ÑùÆ·ÖÐFe£¬CuµÄÖÊÁ¿£®
[ʵÑé̽¾¿]£¨±¾ÊµÑéÖеÄÆøÌåÌå»ý¾ùÒÑ»»Ëã³É±ê×¼×´¿ö£©
¢Ù½«´ËºÏ½ð5.76gÖÃÓÚÉÕ±­ÖУ¬È»ºó½«Ï¡H2SO4Öð½¥»ºÂý¼ÓÈëÖÁ¹ýÁ¿£¬ÊÕ¼¯²úÉúµÄÆøÌå²âµÃÆäÌå»ýΪVmL£®¹ýÂ˵ÃdzÂÌÉ«ÈÜÒºA£¨²»º¬Cu2+£©£¬»¹ÓÐÂËÔüB£®
¢Ú½«ÂËÔüBͶÈëµ½Ò»¶¨Å¨¶ÈµÄHNO3ÖУ¬ÍêÈ«Èܽ⣬ÊÕ¼¯²úÉúµÄÆøÌ壬¾­·ÖÎöÆøÌåÊÇNOºÍNO2µÄ»ìºÏÆøÌ壬×ÜÌå»ý896mL£¬ÆäÖÐNO2Ìå»ýΪ224mL£®
¢Û½«¢ÙÖÐËùµÃµÄÂËÒº¼ÓÈëµ½×ãÁ¿µÄÏ¡HNO3ÖУ¬³ä·Ö·´Ó¦ºóÔÙ¼ÓÈë×ãÁ¿µÄNaOHÈÜÒº£¬½«²úÉúµÄ³ÁµíÈ«²¿Â˳ö£¬³ä·Ö¼ÓÈÈ×ÆÉյúì×ØÉ«¹ÌÌ壬³ÆÁ¿¸Ã¹ÌÌåµÄÖÊÁ¿Îª3.2g£®
[½á¹û´¦Àí]
£¨1£©AÖдæÔÚµÄÑôÀë×ÓÓÐ
Fe2+£¬H+
Fe2+£¬H+
£¬ÂËÔüBΪ
Cu
Cu
£®
£¨2£©ÒÀ´Îд³ö²½Öè¢ÛÖмÓÈëHNO3ʱ·¢Éú·´Ó¦µÄÀë×Ó·´Ó¦·½³Ìʽ
3Fe2++4H++NO3-¨T3Fe3++NO¡ü+2H2O
3Fe2++4H++NO3-¨T3Fe3++NO¡ü+2H2O
£®
£¨3£©±»¸¯Ê´Ç°µÄºÏ½ðÖÐFeµÄÖÊÁ¿Îª
2.24
2.24
g£¬CuµÄÖÊÁ¿Îª
3.2
3.2
g£®
£¨4£©ÊµÑé̽¾¿¢ÙÖÐVΪ
448
448
mL£®

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

¾«Ó¢¼Ò½ÌÍø