ijÌþµÄº¬ÑõÑÜÉúÎï0.05 mol£¬ÔÚ0.2mol O2 ÖÐȼÉÕ£¬ÑõÆø·´Ó¦ÍêÈ«ºó£¬¸ÃÓлúÎïÈ«²¿×ª»¯ÎªÆø̬ÎÞ»úÎÆäƽ¾ùĦ¶ûÖÊÁ¿Îª27 g/mol¡£Èô½«Æø̬»ìºÏÎïͨ¹ýÊ¢ÓÐ×ãÁ¿¼îʯ»ÒµÄ¸ÉÔï¹Ü£¬¸ÉÔï¹ÜÔöÖØ8 g£»Èô¸Ã½«Æø̬»ìºÏÆøÌåͨ¹ý×ãÁ¿µÄ³ÎÇåʯ»ÒË®£¬Ôò²úÉú³Áµí10 g¡£

£¨1£©È¼ÉÕºóµÄ²úÎïCO2¡¢COºÍH2O£¨g£©µÄÎïÖʵÄÁ¿·Ö±ðΪ________mol¡¢___________mol¡¢__________mol¡£

£¨2£©´ËÌþµÄº¬ÑõÑÜÉúÎïµÄ»¯Ñ§Ê½Îª__________¡£

£¨3£©Èô¸ÃÓлúÎïÄܺìÍâ¹âÆ×ÏÔʾÓÐÈ©»ùºÍôÇ»ù£¬ÆäºË´Å¹²ÕñÇâÆ×ÏÔʾÓÐÈýÖÖ»·¾³²»Í¬µÄÇ⣬ÔòÆä½á¹¹¼òʽΪ___________¡£

£¨1£© 0.1  0.1  0.2             £¨2£©C4H8O2                  £¨3£©(CH3)2C(OH)CHO

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑ֪ijÌþµÄº¬ÑõÑÜÉúÎïCxHyOz£¬ÆäÏà¶Ô·Ö×ÓÖÊÁ¿Îª60£¬Èô1mol¸ÃÓлúÎïÍêȫȼÉÕµÃ2molCO2ºÍ2mol H2O£¬Ôò£º
£¨1£©ÓлúÎïCxHyOzµÄ·Ö×ÓʽΪ
C2H4O2
C2H4O2
£»Ð´³ö¸ÃÓлúÎïÍêȫȼÉյĻ¯Ñ§·½³Ìʽ
C2H4O2+2O2
µãȼ
2CO2+2H2O
C2H4O2+2O2
µãȼ
2CO2+2H2O
£»
£¨2£©Èô1mol¸ÃÓлúÎïÓë×ãÁ¿Ì¼ËáÇâÄÆ·´Ó¦·Å³ö1molÓÐCO2£¬Ôò¸ÃÓлúÎïΪһԪÓлúôÈËᣬËüµÄÃû³ÆÊÇ
ÒÒËá
ÒÒËá
£»½á¹¹¼òʽΪ
CH3COOH
CH3COOH
£»
£¨3£©Èô¸ÃÓлúÎïΪõ¥À࣬ËüµÄÃû³ÆÊÇ
¼×Ëá¼×õ¥
¼×Ëá¼×õ¥
£»½á¹¹¼òʽΪ
HCOOCH3
HCOOCH3
£»
£¨4£©Èô¸ÃÓлúÎï²»Óë̼ËáÇâÄÆ·´Ó¦£¬µ«ÄÜÓë½ðÊôÄÆ·´Ó¦£®1mol¸ÃÓлúÎïÓë×ãÁ¿½ðÊôÄÆ·´Ó¦£¬Éú³É0.5molH2£¬Ôò¸ÃÓлúÎïΪôÇ»ùÈ©£¬ËüµÄ½á¹¹¼òʽΪ
HOCH2CHO
HOCH2CHO
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

L¡ª¶à°ÍÊÇÒ»ÖÖÓлúÎËü¿ÉÓÃÓÚÅÁ½ðÉ­×ÛºÏÕ÷ÖÎÁÆ£¬Æä·Ö×ÓʽΪ£ºC9H11O4N¡£ÕâÖÖÒ©ÎïµÄÑÐÖÆÊÇ»ùÓÚ»ñµÃ2000Äêŵ±´¶ûÉúÀíѧ»òҽѧ½±ºÍ»ñµÃ2001Äêŵ±´¶û»¯Ñ§½±µÄÑо¿³É¹û¡£ÒÑÖªº¬Ì¼¡¢Çâ¡¢Ñõ¡¢µª4ÖÖÔªËصÄÓлúÎïÔÚ¿ÕÆøÖÐÍêȫȼÉÕÉú³É¶þÑõ»¯Ì¼¡¢Ë®ºÍ°±Æø¡£

(1)L¡ª¶à°Í·Ö×ÓÔÚ¿ÕÆøÖÐȼÉյĻ¯Ñ§·½³ÌʽÊÇ______________________________£»

(2)ÈôÌþC7H12ºÍC5H12µÄ»ìºÏÎï1molÔÚ¿ÕÆøÖÐÍêȫȼÉÕʱÏûºÄµÄÑõÆøΪ9 mol£¬Ôò»ìºÏÎïÖÐC7H12ºÍC5H12µÄÖÊÁ¿±ÈΪ____________£»

(3)ÈôijÌþµÄº¬ÑõÑÜÉúÎï0.1 molÔÚ¿ÕÆøÖÐÍêȫȼÉÕʱÏûºÄµÄÑõÆøµÄÎïÖʵÄÁ¿ÓëµÈÎïÖʵÄÁ¿µÄL¡ª¶à°Í·Ö×ÓÍêȫȼÉÕʱÏûºÄµÄÑõÆøÏàͬ£¬ÇÒÉú³É¶þÑõ»¯Ì¼17.92L(±ê×¼×´¿ö)£¬Ôò¸ÃÌþµÄº¬ÑõÑÜÉúÎïͨʽΪ____________(Óú¬nµÄ»¯Ñ§Ê½±íʾ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

)LÒ»¶à°ÍÊÇÒ»ÖÖÓлúÎËü¿ÉÓÃÓÚÅÁ½ðÉ­×ÛºÏÖ¢µÄÖÎÁÆ£¬Æä·Ö×ÓʽΪ£ºC9H11N¡£ ÕâÖÖÒ©ÎïµÄÑÐÖÆÊÇ»ùÓÚ»ñµÃ2000Äêŵ±´¶ûÉúÀíѧ»òҽѧ½±ºÍ»ñµÃ2001Äêŵ±´¶û»¯Ñ§½±µÄÑо¿³É¹û¡£ÒÑÖªº¬Ì¼¡¢Çâ¡¢Ñõ¡¢µª4ÖÖÔªËصÄÓлúÎïÔÚ¿ÕÆøÖÐÍêȫȼÉÕÉú³É¶þÑõ»¯Ì¼¡¢Ë®ºÍ°±Æø¡£

    (1)LÒ»¶à°Í·Ö×ÓÔÚ¿ÕÆøÖÐȼÉյĻ¯Ñ§·½³ÌʽÊÇ                                ¡£

    (2)ÈôÌþC7H12ºÍC5H12µÄ»ìºÏÎï1 molÔÚ¿ÕÆøÖÐÍêȫȼÉÕʱÏûºÄµÄÑõÆøΪ9 mol£¬Ôò»ìºÏÎïÖÐC7H12ºÍC5H12µÄÖÊÁ¿±ÈΪ                                     ¡£

    (3)ÈôijÌþµÄº¬ÑõÑÜÉúÎï0.1 molÔÚ¿ÕÆøÖÐÍêȫȼÉÕʱÏûºÄµÄÑõÆøµÄÎïÖʵÄÁ¿ÓëµÈÎïÖʵÄÁ¿µÄLÒ»¶à°Í·Ö×ÓÍêȫȼÉÕʱÏûºÄµÄÑõÆøÏàͬ£¬ÇÒÉú³É¶þÑõ»¯Ì¼17.92L (±ê×¼×´¿ö)£¬Ôò¸ÃÌþµÄº¬ÑõÑÜÉúÎïͨʽΪ             (Óú¬µÄ»¯Ñ§Ê½±íʾ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄê±±¾©ÊÐÖصãÖÐѧ¸ß¶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§£¨Àí£©ÊÔ¾í ÌâÐÍ£º¼ÆËãÌâ

£¨7·Ö£©
ijÌþµÄº¬ÑõÑÜÉúÎï0.05 mol£¬ÔÚ0.2mol O2 ÖÐȼÉÕ£¬ÑõÆø·´Ó¦ÍêÈ«ºó£¬¸ÃÓлúÎïÈ«²¿×ª»¯ÎªÆø̬ÎÞ»úÎÆäƽ¾ùĦ¶ûÖÊÁ¿Îª27 g/mol¡£Èô½«Æø̬»ìºÏÎïͨ¹ýÊ¢ÓÐ×ãÁ¿¼îʯ»ÒµÄ¸ÉÔï¹Ü£¬¸ÉÔï¹ÜÔöÖØ8 g£»Èô¸Ã½«Æø̬»ìºÏÆøÌåͨ¹ý×ãÁ¿µÄ³ÎÇåʯ»ÒË®£¬Ôò²úÉú³Áµí10 g¡£
£¨1£©È¼ÉÕºóµÄ²úÎïCO2¡¢COºÍH2O£¨g£©µÄÎïÖʵÄÁ¿·Ö±ðΪ________mol¡¢___________mol¡¢__________mol¡£
£¨2£©´ËÌþµÄº¬ÑõÑÜÉúÎïµÄ»¯Ñ§Ê½Îª__________¡£
£¨3£©Èô¸ÃÓлúÎïÄܺìÍâ¹âÆ×ÏÔʾÓÐÈ©»ùºÍôÇ»ù£¬ÆäºË´Å¹²ÕñÇâÆ×ÏÔʾÓÐÈýÖÖ»·¾³²»Í¬µÄÇ⣬ÔòÆä½á¹¹¼òʽΪ___________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸