2005Äê10ÔÂ12ÈÕ--17ÈÕ£¬ÎÒ¹ú¡°ÉñÖÛÁùºÅ¡±ÔØÈ˺½Ìì·É´¬£¬Ê¤Àû·¢ÉäºÍ·µ»Ø£¬±êÖ¾×ÅÎÒ¹úº½¿Õº½Ìì¼¼Êõ½øÈëÊÀ½çÁìÏÈˮƽ£®ÓîÖæ·É´¬ÄÚ£¬ÓԱËùºô³öµÄÆøÌåҪͨ¹ýÊ¢ÓÐÇâÑõ»¯ï®£¨LiOH£©µÄ¹ýÂËÍø£¬ÒÔ³ýÈ¥Ëùº¬µÄ¶þÑõ»¯Ì¼£¬ÈçÏÂʽËùʾ£º
2LiOH£¨¹Ì£©+CO2£¨Æø£©=Li2CO3£¨¹Ì£©+H2O£¨Òº£©
£¨1£©ÊÔ¼ÆËã1gÇâÑõ»¯ï®ËùÄÜÎüÊÕ¶þÑõ»¯Ì¼µÄÖÊÁ¿£®
£¨2£©ÈôÓÃÇâÑõ»¯¼Ø£¨KOH£©À´´úÌæÇâÑõ»¯ï®£¬ÊÔ¼ÆËã1gÇâÑõ»¯¼ØËùÄÜÎüÊÕ¶þÑõ»¯Ì¼µÄÖÊÁ¿£®
£¨3£©ÀûÓã¨1£©ºÍ£¨2£©ËùµÃ½á¹û£¬ÊÔ½âÊÍΪʲôÓîÖæ·É´¬Ñ¡ÓÃÇâÑõ»¯ï®À´ÎüÊÕ¶þÑõ»¯Ì¼½ÏÓÃÇâÑõ»¯¼ØÎª¼Ñ£®
£¨4£©ÈôÿλÓԱÿÌìËùºô³öµÄ¶þÑõ»¯Ì¼Æ½¾ùÊÇ502 L£¬±¾´ÎÓÐÁ½Î»ÓԱÔÚ½øÐÐÒ»ÏîΪÆÚ5ÌìµÄÌ«¿ÕÈÎÎñ£¬ÊÔ¼ÆËãÔÚÓîÖæ·É´¬ÉÏӦЯ´øÇâÑõ»¯ï®µÄÖÊÁ¿£¨ÔÚÓîÖæ·É´¬ÄÚÎÂ¶ÈºÍÆøÑ¹Ï£¬¶þÑõ»¯Ì¼ÆøÌåµÄÃܶÈΪ1.833g/L£©£®
½â£º£¨1£©Éè1gLiOHËùÄÜÎüÊÕCO
2µÄÖÊÁ¿Îªx
2LiOH+CO
2¨TLi
2CO
3+H
2O
48 44
1g x
¡àx=

=0.92g
´ð£º1gÇâÑõ»¯ï®ËùÄÜÎüÊÕ¶þÑõ»¯Ì¼µÄÖÊÁ¿Îª0.92g£»
£¨2£©Éè1gKOHËùÄÜÎüÊÕµÄCO
2µÄÖÊÁ¿Îªy
2KOH+CO
2¨TK
2CO
3+H
2O
112 44
1g y
¡ày=

=0.39g
´ð£º1gÇâÑõ»¯¼ØËùÄÜÎüÊÕ¶þÑõ»¯Ì¼µÄÖÊÁ¿Îª0.39g£»
£¨3£©0.92g£¾0.39gµÈÖÊÁ¿µÄLiOHÎüÊÕµÄCO
2±ÈKOHµÄ¶à£¬ÕâÑù¿ÉÒÔ¼õÉÙÓîÖæ·É´¬µÄ×ÔÖØ£»
£¨4£©ÉèÓîÖæ·É´¬Ó¦Ð¯´øµÄLiOHµÄÖÊÁ¿ÎªZ

¡àZ=10001.80g¡Ö10kg
´ð£ºÔÚÓîÖæ·É´¬ÉÏӦЯ´øÇâÑõ»¯ï®µÄÖÊÁ¿Ô¼Îª10kg£®
·ÖÎö£º±È½ÏÎüÊÕÏàͬÖÊÁ¿µÄ¶þÑõ»¯Ì¼ÆøÌåÐèÒªÇâÑõ»¯¼ØºÍÇâÑõ»¯ï®µÄÖÊÁ¿´óС£¬Ñ¡È¡ÖÊÁ¿Ð¡µÄÎïÖÊÎüÊÕ¶þÑõ»¯Ì¼¿É¼õÉÙ·É´¬×ÔÖØ£®
µãÆÀ£ºÁ½Î»ÓԱÔÚ5ÌìÀïºô³ö¶þÑõ»¯Ì¼µÄÖÊÁ¿=502 L¡Á2¡Á5¡Á1.833g/L=9201.66g£®