ÌâÄ¿ÄÚÈÝ

»ÆÌú¿ó(Ö÷Òª³É·ÖΪFeS2)Êǹ¤ÒµÖÆÈ¡ÁòËáµÄÖØÒªÔ­ÁÏ£¬ÆäìÑÉÕ²úÎïΪSO2ºÍFe3O4£®

(1)

½«0.050molSO2(g)ºÍ0.030molO2(g)·ÅÈëÈÝ»ýΪ1LµÄÃܱÕÈÝÆ÷ÖУ¬·´Ó¦£º2SO2(g)£«O2(g)2SO3(g)ÔÚÒ»¶¨Ìõ¼þÏ´ﵽƽºâ£¬²âµÃc(SO3)£½0.040mol/L£®¼ÆËã¸ÃÌõ¼þÏ·´Ó¦µÄƽºâ³£ÊýKºÍSO2µÄƽºâת»¯ÂÊ(д³ö¼ÆËã¹ý³Ì)£®

(2)

ÒÑÖªÉÏÊö·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬µ±¸Ã·´Ó¦´¦ÓÚƽºâ״̬ʱ£¬ÔÚÌå»ý²»±äµÄÌõ¼þÏ£¬ÏÂÁдëÊ©ÖÐÓÐÀûÓÚÌá¸ßSO2ƽºâת»¯ÂʵÄÓÐ________(Ìî×Öĸ)

A¡¢Éý¸ßζÈ
B¡¢½µµÍζÈ
C¡¢Ôö´óѹǿ
D¡¢¼õСѹǿ
E¡¢¼ÓÈë´ß»¯¼Á
F¡¢ÒƳöÑõÆø

(3)

SO2βÆøÓñ¥ºÍNa2SO3ÈÜÒºÎüÊտɵõ½¸üÒªµÄ»¯¹¤Ô­ÁÏ£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________£®

(4)

½«»ÆÌú¿óµÄìÑÉÕ²úÎïFe3O4ÈÜÓÚH2SO4ºó£¬¼ÓÈëÌú·Û£¬¿ÉÖƱ¸FeSO4£®ËáÈܹý³ÌÖÐÐè±£³ÖÈÜÒº×ã¹»ËáÐÔ£¬ÆäÔ­ÒòÊÇ________£®

´ð°¸£º2£®B¡¢C;
½âÎö£º

(1)

1.6¡Á103L/mol£¬80%(¼ÆËã¹ý³ÌÂÔ)

(3)

SO2£«H2O£«Na2SO3£½2NaHSO3

(4)

ÒÖÖÆFe2£«¡¢Fe3£«µÄË®½â£¬·ÀÖ¹Fe2£«±»Ñõ»¯³ÉFe3£«


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

»¯Ñ§¹¤ÒµÔÚ¹úÃñ¾­¼ÃÖÐÕ¼Óм«ÆäÖØÒªµÄµØ룬½­ËÕÊ¡ÊǹúÄÚ×îÔçµÄÁòËáÉú²ú»ùµØÖ®Ò»¡£Ö÷Òª·½³ÌʽºÍÖ÷ÒªÉ豸: £¨»ÆÌú¿óµÄÖ÷Òª³É·ÖΪFeS2£© 4FeS2+11O2 = 2Fe2O3+8SO2 £¨·ÐÌÚ¯£©  2SO2+O22SO£¨½Ó´¥ÊÒ£©    SO3+H2O=H2SO4£¨ÎüÊÕËþ£©

£¨1£©ÁòËáÉú²úÖУ¬¸ù¾Ý»¯Ñ§Æ½ºâÔ­ÀíÀ´È·¶¨µÄÌõ¼þ»ò´ëÊ©ÓР         £¨ÌîдÐòºÅ£©¡£

        A£®¿óʯ¼ÓÈë·ÐÌÚ¯֮ǰÏÈ·ÛËé      B£®½Ó´¥Êҵķ´Ó¦Ê¹ÓÃV2O5×÷´ß»¯¼Á

C£®½Ó´¥ÊÒÖв»Ê¹ÓúܸߵÄζȠ     D£®¾»»¯ºó¯ÆøÖÐÒªÓйýÁ¿µÄ¿ÕÆø

E£®½Ó´¥ÊÒÖеÄÑõ»¯ÔÚ³£Ñ¹Ï½øÐР   F£®ÎüÊÕËþÖÐÓÃ98.3%µÄŨÁòËáÎüÊÕSO3

£¨2£©0.1mol/LµÄNaHSO3ÈÜÒºÖУ¬ÓйØÁ£×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ£º

   c £¨Na+£©£¾c £¨HSO3£­£©£¾c £¨SO32£­£©£¾c £¨H2SO3£©

¢Ù¸ÃÈÜÒºÖÐc £¨H+£©        c £¨OH£­£©£¨Ìî¡°£¾¡±¡¢¡°£½¡± »ò¡°£¼¡± £©£¬ÆäÀíÓÉÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£º                                      ¡£

¢ÚÏÖÏòNaHSO3ÈÜÒºÖУ¬ÖðµÎ¼ÓÈëÉÙÁ¿º¬ÓзÓ̪µÄNaOHÈÜÒº£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ                    £¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º                     ¡£

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

¾«Ó¢¼Ò½ÌÍø