N2£¨g£©+3H2(g)2NH3(g)
£¨1£©Èô·´Ó¦½øÐе½Ä³Ê±¿Ìtʱ£¬n(N2)=13 mol£¬n(NH3)=6 mol£¬¼ÆËãaµÄÖµ¡£
£¨2£©·´Ó¦´ïµ½Æ½ºâʱ£¬»ìºÏÆøÌåµÄÌå»ýΪ716.8 L£¨±ê×¼×´¿ö£©£¬ÆäÖÐNH3µÄº¬Á¿£¨Ìå»ý·ÖÊý£©Îª25%¡£¼ÆËãƽºâʱNH3µÄÎïÖʵÄÁ¿¡£
£¨3£©Ô»ìºÏÆøÌåÓëƽºâ»ìºÏÆøÌåµÄ×ÜÎïÖʵÄÁ¿Ö®±È£¨Ð´³ö×î¼òÕûÊý±È£¬ÏÂͬ£©£¬nʼ¡Ãnƽ=________¡£
£¨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
½âÎö£º£¨1£©ÓÉ·´Ó¦·½³Ìʽ¿ÉÖª£¬tʱÉú³É6 mol NH3±ØÈ»·´Ó¦µô3 mol N2,¹Êa mol=n1(N2)+n(·´Ó¦µôµÄN2£©=13 mol+3 mol=16 mol¡£
£¨3£©¾Ý·´Ó¦µÄÌصã¿É¿´³ö£¬Éú³ÉNH3µÄÌå»ý¼´ÊÇÆøÌå¼õÉÙµÄ×ÜÌå»ý£¬Òò´Ën(ʼ£©¡Ãn(ƽ£©=V£¨Ê¼£©¡ÃV£¨Æ½£©=£¨1+25%£©¡Ã1=5¡Ã4¡£
£¨4£©n(ʼ)=n(ƽ)¡Á=¡Á=40 mol£¬b mol=n(ʼ£©-a mol=40 mol-16 mol=24 mol£¬¹Êa¡Ãb=16¡Ã24=2¡Ã3¡£
£¨5£©ÓÉ£¨2£©ÖªÆ½ºâʱn(NH3)=8 mol,Ôò·´Ó¦µôN2Ϊ8 mol¡Á=4 mol£¬·´Ó¦µôH2Ϊ8 mol¡Á=12mol,Ôò´ïµ½Æ½ºâʱ¦Á(N2)¡Ã¦Á(H2)==1¡Ã2¡£
£¨6£©Æ½ºâ»ìºÏÆøÌå×ÜÎïÖʵÄÁ¿Îª32 mol,ÆäÖÐNH3Ϊ8 mol,N2Ϊ16 mol-4 mol=12 mol,H2Ϊ24 mol-12 mol=12 mol¡£Ôòn(N2)¡Ãn(H2)¡Ãn(NH3)=12 mol¡Ã12 mol¡Ã8 mol=3¡Ã3¡Ã2¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¸ßΡ¢¸ßѹ | ´ß»¯¼Á |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ºãÎÂÏÂ,½«a mol N2Óëb mol H2µÄ»ìºÏÆøÌåͨÈëÒ»¸ö¹Ì¶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÈçÏ·´Ó¦£ºN2(g)+3H2(g) 2NH3(g)
£¨1£©Èô·´Ó¦½øÐе½Ä³Ê±¿Ìʱ£¬n£¨N2£©=13 mol, n£¨NH3£©=6 mol,¼ÆËãaµÄÖµ ¡£
£¨2£©·´Ó¦´ïƽºâʱ£¬»ìºÏÆøÌåµÄÌå»ýΪ716.8L£¨×´¿öÏ£©£¬ÆäÖÐNH3µÄº¬Á¿£¨Ìå»ý·ÖÊý£©Îª25%¡£¼ÆËãƽºâʱNH3µÄÎïÖʵÄÁ¿ ¡£
£¨3£©Ô»ìºÏÆøÌåÓëƽºâ»ìºÏÆøÌåµÄ×ÜÎïÖʵÄÁ¿Ö®±È£¨Ê¼£©£º£¨Æ½£©=_____________¡££¨Ð´³ö×î¼òÕûÊý±È£¬ÏÂͬ£©£¬
£¨4£©Ô»ìºÏÆøÌåÖУ¬a¡Ãb£½ ¡¡¡¡¡¡¡¡¡¡ ¡£
£¨5£©´ïµ½Æ½ºâʱ£¬N2ºÍH2µÄת»¯ÂÊÖ®±È£¬(N2)¡Ã (H2)£½ ¡¡¡¡¡¡¡¡¡¡ ¡£
£¨6£©Æ½ºâ»ìºÏÆøÌåÖУ¬n(N2)¡Ãn(H2)¡Ãn(NH3)£½ ¡¡¡¡¡¡¡¡ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ºãÎÂÏÂ,½«a mol N2Óëb mol H2µÄ»ìºÏÆøÌåͨÈëÒ»¸ö¹Ì¶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÈçÏ·´Ó¦£º N2(g)+3H2(g)2NH3(g)
(1)Èô·´Ó¦½øÐе½Ä³Ê±¿Ìʱ£¬£¨N2£©=13 mol, (NH3)=6 mol, ÇóaµÄÖµ
(2)·´Ó¦´ïƽºâʱ£¬»ìºÏÆøÌåµÄÌå»ýΪ716.8L£¨×´¿öÏ£©£¬ÆäÖÐNH3µÄÌå»ý·ÖÊýΪ25%£¬ÔòƽºâʱNH3µÄÎïÖʵÄÁ¿Îª¶àÉÙ£¿
(3)ÇóÔ»ìºÏÆøÌåÓëƽºâ»ìºÏÆøÌåµÄ×ÜÎïÖʵÄÁ¿Ö®±È£¨Ð´³ö×î¼òÕûÊý±È£©
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com