ÌâÄ¿ÄÚÈÝ

ºãÎÂÏ£¬½«a mol N2Óëb mol H2µÄ»ìºÏÆøÌåͨÈëÒ»¸ö¹Ì¶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷ÖУ¬·¢Éú·´Ó¦£º

N2(g) + 3H2(g)2NH3(g)

£¨1£©Èô·´Ó¦½øÐе½Ä³Ê±¿Ìtʱ£¬nt£¨N2£©=13 mol£¬nt£¨NH3£©=6 mol£¬¼ÆËãa=               ¡£

£¨2£©·´Ó¦´ïµ½Æ½ºâʱ£¬»ìºÏÆøÌåµÄÌå»ýΪ716.8 L£¨±ê×¼×´¿öÏ£©£¬ÆäÖÐNH3µÄº¬Á¿£¨Ìå»ý·ÖÊý£©Îª25%¡£¼ÆËãÆ½ºâʱNH3µÄÎïÖʵÄÁ¿=                ¡£

£¨3£©Ô­»ìºÏÆøÌåÓëÆ½ºâ»ìºÏÆøÌåµÄ×ÜÎïÖʵÄÁ¿Ö®±Èn£¨Ê¼£©¡Ãn£¨Æ½£©=              ¡£

£¨4£©Ô­»ìºÏÆøÌåÖУ¬a¡Ãb=              ¡£

£¨5£©´ïµ½Æ½ºâʱ£¬N2ºÍH2µÄת»¯ÂÊÖ®±È£¬¦Á£¨N2£©¡Ã¦Á£¨H2£©=                  ¡£

£¨6£©Æ½ºâ»ìºÏÆøÌåÖУ¬n£¨N2£©¡Ãn£¨H2£©¡Ãn£¨NH3£©=                 ¡£

 

£¨1£©16

£¨2£©8 mol

£¨3£©5¡Ã4

£¨4£©2¡Ã3

£¨5£©1¡Ã2

£¨6£©3¡Ã3¡Ã2

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø