ÌâÄ¿ÄÚÈÝ

3£®½«5.6g FeÈ«²¿ÈÜÓÚ200mLÒ»¶¨Å¨¶ÈµÄÏõËáÈÜÒºÖУ¬µÃµ½±ê×¼×´¿öÏÂµÄÆøÌå2.24L£¬ÓÖ²âµÃ·´Ó¦ºóÈÜÒºÖÐH+µÄŨ¶ÈΪ0.2mol•L-1£¨É跴ӦǰºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©£®ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®2.24 LÆøÌåÊÇNO
B£®·´Ó¦ºóFeÈ«²¿×ª»¯ÎªFe3+
C£®·´Ó¦ºóµÄÈÜÒºÖÐc£¨NO3-£©=1.7 mol•L-1
D£®·´Ó¦ºóµÄÈÜÒº×î¶à»¹ÄÜÔÙÈܽâ2.24 g Fe

·ÖÎö n£¨Fe£©=$\frac{5.6g}{56g/mol}$=0.1mol£¬n£¨ÆøÌ壩=$\frac{2.24L}{22.4L/mol}$=0.1mol£¬²âµÃ·´Ó¦ºóÈÜÒºÖÐH+µÄŨ¶ÈΪ0.2mol•L-1£¬ÔòÏõËá¹ýÁ¿£¬Óɵç×ÓÊØºã¿ÉÖª£¬Feʧȥ0.3molµç×Óʱ£¬NÔ­×ӵõ½0.3molµç×Ó£¬¼´Éú³ÉÆøÌåΪNO£¬ÒÔ´ËÀ´½â´ð£®

½â´ð ½â£ºn£¨Fe£©=$\frac{5.6g}{56g/mol}$=0.1mol£¬n£¨ÆøÌ壩=$\frac{2.24L}{22.4L/mol}$=0.1mol£¬²âµÃ·´Ó¦ºóÈÜÒºÖÐH+µÄŨ¶ÈΪ0.2mol•L-1£¬ÔòÏõËá¹ýÁ¿£¬
A£®Óɵç×ÓÊØºã¿ÉÖª£¬0.1mol¡Á£¨3-0£©=0.1mol¡Á£¨5-x£©£¬½âµÃx=2£¬¼´ÆøÌåΪNO£¬¹ÊAÕýÈ·£»
B£®ÏõËá¹ýÁ¿£¬·´Ó¦ºóÈÜÒºÖк¬ÓÐFe3+£¬¹ÊBÕýÈ·£»
C£®ÓÉNÔ­×ÓÊØºã¿ÉÖª£¬c£¨NO3-£©=$\frac{0.1mol¡Á3+0.2L¡Á0.2mol/L}{0.2L}$=1.7mol/L£¬¹ÊCÕýÈ·£»
D£®Ô­ÏõËáµÄÎïÖʵÄÁ¿Îª0.1mol¡Á3+0.2L¡Á0.2mol/L+0.1mol=0.44mol£¬ÓÉ3Fe+8HNO3=3Fe£¨NO3£©2+2NO¡ü+4H2O¿ÉÖª£¬×î¶àÏûºÄFeΪ0.44mol¡Á$\frac{3}{8}$¡Á56g/mol=9.24g£¬Ôò×î¶à»¹ÄÜÔÙÈܽâ9.24-5.6=3.64g£¬¹ÊD´íÎó£»
¹ÊÑ¡D£®

µãÆÀ ±¾Ì⿼²éÑõ»¯»¹Ô­·´Ó¦µÄ¼ÆË㣬Ϊ¸ßƵ¿¼µã£¬°ÑÎÕÏõËá¹ýÁ¿¼°µç×ÓÊØºã¡¢Ô­×ÓÊØºãµÄÓ¦ÓÃΪ½â´ðµÄ¹Ø¼ü£¬²àÖØ·ÖÎöÓë¼ÆËãÄÜÁ¦µÄ¿¼²é£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø