ÌâÄ¿ÄÚÈÝ

½üÄêÀ´£¬ÓÉCO2´ß»¯¼ÓÇâºÏ³É¼×´¼µÄÏà¹ØÑо¿Êܵ½Ô½À´Ô½¶àµÄ¹Ø×¢¡£¸Ã·½·¨¼È¿É½â¾öCO2·ÏÆøµÄÀûÓÃÎÊÌ⣬ÓÖ¿É¿ª·¢Éú²ú¼×´¼µÄÐÂ;¾¶£¬¾ßÓÐÁ¼ºÃµÄÓ¦ÓÃǰ¾°¡£ÒÑÖª4.4 g CO2ÆøÌåÓëH2¾­´ß»¯¼ÓÇâÉú³ÉCH3OHÆøÌåºÍË®ÕôÆøÊ±·Å³ö4.95 kJµÄÄÜÁ¿¡£

(1)¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£º__________________________________________.

(2)ÔÚ270¡æ¡¢8 MPaºÍÊʵ±´ß»¯¼ÁµÄÌõ¼þÏ£¬CO2µÄת»¯ÂÊ´ïµ½22%£¬Ôò4.48 m3(ÒÑÕÛºÏΪ±ê×¼×´¿ö)µÄCO2ÔںϳÉCH3OHÆøÌå¹ý³ÌÖÐÄܷųöÈÈÁ¿________ kJ.

(3)ÓÖÒÑÖªH2O(g)===H2O(l)¡¡¦¤H£½£­44 kJ/mol£¬ÔòCO2ÆøÌåÓëH2ÆøÌå·´Ó¦Éú³ÉCH3OHÆøÌå16gºÍҺ̬ˮʱ·Å³öÈÈÁ¿Îª            kJ£®

 

(1) CO2(g)+3H2(g)=CH3OH(g)+H2O(g); ¦¤H=-49.5KJ/mol

(2) 2178kJ     (3)2.75 kJ

½âÎö:ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø