ÌâÄ¿ÄÚÈÝ

ÂÈ»¯ï§¡¢¼×´¼¡¢Ñõ»¯ÂÁ¶¼ÊÇÖØÒª»¯ºÏÎï¡£

£¨1£©ÒÑÖª£º

I NH4Cl(s)=NH3(g)+HCl(g) ¡÷H=£«l63.9 kJ¡¤mol-1

II HCl(g)+CH3OH(g)CH3Cl(g)+H2O(g) ¡÷H=-31.9kJ¡¤mol-1

III NH4Cl(s)+CH3OH(g)NH3(g)+CH3Cl(g)+H2O(g)

¢Ù ·´Ó¦IIIÔÚ_________Ìõ¼þÏÂÄÜ×Ô·¢·´Ó¦£¨Ìî¡°½Ï¸ßζȡ±¡¢¡°½ÏµÍζȡ±»ò¡°ÈκÎζȡ±£©£¬ÀíÓÉÊÇ_______________¡£

¢Ú ͼ1ÊÇ·´Ó¦IIIʹÓÃÈýÖÖ²»Í¬´ß»¯¼ÁʱµÃµ½µÄCH3Cl²úÂÊÓëζȹØϵµÄ±ä»¯Í¼¡£

¼ºÖª£º´ß»¯¼ÁÓÃÁ¿¡¢´ß»¯¼ÁÁ£Êý¡¢n(¼×´¼)£ºn(ÂÈ»¯ï§)µÄÖµ¡¢¼×´¼½øÁÏËٶȡ¢·´Ó¦Ê±¼äµÈ²âÊÔÌõ¼þ¶¼Ïàͬ¡£

ͼ1 ÖÐaÇúÏßCH3Cl²úÂÊÏÈÔö´óºó¼õСµÄÔ­ÒòÊÇ___________¡£ÇëÔÚͼ2Öл­³öÆäËüÌõ¼þ¶¼Ïàͬʱ£¬ÔÚ370 ¡æÏÂʹÓÃÈýÖÖ²»Í¬´ß»¯¼ÁÖÁ·´Ó¦Æ½ºâʱ£¬CH3ClµÄ²úÂÊÓëʱ¼ä¹ØϵµÄ±ä»¯ÇúÏߣ¬²¢ÓÃa¡¢b¡¢c±ê³ö¶ÔÓ¦µÄÇúÏß¡£_______________

£¨2£©25 ¡æʱ£¬ÔÚijŨ¶ÈµÄNH4ClÈÜÒºÖеμÓÒ»¶¨Á¿µÄ°±Ë®ÖÁÖÐÐÔ£¬´Ëʱ²âµÃÈÜÒºÖÐc(Cl-)= 0.36mol ¡¤ L-1£¬Ôò»ìºÏÈÜÒºÖÐc (NH3 ¡¤ H2O)£½_______mol ¡¤ L-1¡££¨25¡æʱ£¬NH3¡¤H2OµÄKb=1.8¡Á10-5£©

£¨3£©¶à¿×Al2O3±¡Ä¤¿É×÷Ϊ´ß»¯¼ÁÔØÌ塢ģ°åºÏ³ÉÄÉÃײÄÁϵÈÓÃ;¡£ÏÖÒԸߴ¿ÂÁƬ×÷ΪÑô¼«£¬²»Ðâ¸Ö×÷ΪÒõ¼«£¬Ò»¶¨ÈܶȵÄÁ×ËáÈÜÒº×÷Ϊµç½âÖʽøÐеç½â£¬¼´¿É³õ²½ÖÆÈ¡¶à¿×Al2O3Ĥ¡£Çëд³ö¸ÃÖÆÈ¡¹ý³ÌµÄÑô¼«µç¼«·´Ó¦£º__________________¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

»¯Ñ§¶ÆÄøÊÇÖ¸²»Ê¹ÓÃÍâ¼ÓµçÁ÷£¬ÀûÓÃÑõ»¯»¹Ô­×÷ÓÃÔÚ½ðÊôÖƼþµÄ±íÃæÉϳÁ»ýÒ»²ãÄøµÄ·½·¨¡£´ÎÁ×ËáÄÆ(NaH2PO2)ÊÇ»¯Ñ§¶ÆÄøµÄÖØÒªÔ­ÁÏ£¬¹¤ÒµÉÏÖƱ¸NaH2PO2¡¤H2OµÄÁ÷³ÌÈçÏ£º

»Ø´ðÏÂÁÐÎÊÌâ:

¢Å´ÎÁ×ËáÄÆ(NaH2PO2)ÊÇ´ÎÁ×Ëá(H3PO2)Óë×ãÁ¿NaOHÈÜÒº·´Ó¦µÄ²úÎNaH2PO2ÊôÓÚ_______(Ìî¡°ÕýÑΡ±¡°ËáʽÑΡ±¡°¼îʽÑΡ±£©£¬ NaH2PO2ÖÐÁ×ÔªËصĻ¯ºÏ¼ÛΪ___________¡£

(2)ÔÚ·´Ó¦Æ÷ÖмÓÈëÈ黯¼Á²¢¸ßËÙ½Á°èµÄÄ¿µÄÊÇ________________

(3)ÔÚ·´Ó¦Æ÷Öз¢Éú¶à¸ö·´Ó¦£¬ÆäÖа×Á×(P4)ÓëCa(OH)2·´Ó¦Éú³É´ÎÁ×ËáÄƼ°Á×»¯ÇâµÄ»¯Ñ§·½³ÌʽΪ_________________________

(4)Á÷³ÌÖÐͨÈëCO2µÄÄ¿µÄÊÇ______________£¬ÂËÔüXµÄ»¯Ñ§Ê½Îª_____________

(5)Á÷³ÌÖÐĸҺÖеÄÈÜÖʳýNaH2PO2Í⣬»¹ÓеÄÒ»ÖÖÖ÷Òª³É·ÖΪ_____________

(6)º¬PH3µÄ·ÏÆø¿ÉÓÃNaClOºÍNaOHµÄ»ìºÏÈÜÒº´¦Àí½«Æäת»¯ÎªÁ×ËáÑΣ¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________

(7)ij´ÎÉú²úͶÈëµÄÔ­ÁÏ°×Á×Ϊ1240 kg£¬ÔÚ¸ßËÙÈ黯·´Ó¦Æ÷ÖÐÓÐ80%µÄ°×Á×ת»¯Îª´ÎÁ×ËáÄƼ°Á×»¯Ç⣬ºöÂÔÆäËü²½ÖèµÄËðʧ£¬ÀíÂÛÉÏ×îÖյõ½²úÆ·NaH2PO2¡¤H2OµÄÖÊÁ¿Ó¦Îª__________kg(NaH2PO2¡¤H2OµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª106)

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

¾«Ó¢¼Ò½ÌÍø