£¨2011?ËÄ´¨Ä£Ä⣩ʯÓÍÍÑÁò·ÏÆøÖк¬ÓÐÁò»¯Ç⣬ΪÁËÓÐЧ·ÀÖ¹ÎÛȾ£¬ÊµÏÖ·ÏÎïµÄ×ÛºÏÀûÓ㬹¤ÒµÉϲÉÓÃÏÂÁй¤ÒÕ¹ý³Ì½«Áò»¯Çâת»¯ÎªÁòËáºÍÇâÆøµÈ²úÆ·£®

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©·´Ó¦Æ÷ÖÐ×°ÓÐÁòËáÌúµÄËáÐÔÈÜÒº£¬¸ÃÈÜÒºÓëÁò»¯Çâ·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
2Fe3++H2S=2Fe2++2H++S¡ý
2Fe3++H2S=2Fe2++2H++S¡ý
£®
£¨2£©ÂËÒºµç½âºó£¬µÃµ½ÁòËáÌúµÄËáÐÔÈÜÒº£¬ÔÙ½«Æä·µ»Øµ½·´Ó¦Æ÷ÖеÄÄ¿µÄÊÇ
ʹÁòËáÌúµÄËáÐÔÈÜҺѭ»·Ê¹ÓÃ
ʹÁòËáÌúµÄËáÐÔÈÜҺѭ»·Ê¹ÓÃ
£®
£¨3£©ÔÚʵ¼ÊÉú²úÖУ¬¶þÑõ»¯Áòת»¯ÎªÈýÑõ»¯ÁòµÄÊÊÒËÌõ¼þÊÇ£ºV2O5×÷´ß»¯¼Á¡¢³£Ñ¹¡¢400¡æ¡«500¡æ£®ÈçͼÊDz»Í¬Ñ¹Ç¿Ï¸÷´Ó¦ÌåϵÖÐζÈÓë¶þÑõ»¯ÁòµÄƽºâת»¯Âʱ仯ÇúÏߣ®

¿¼Âǹ¤ÒµÉú²úЧÒæ²¢½áºÏͼʾ·ÖÎö£¬Ñ¡Ôñ400¡æ¡«500¡æµÄÔ­ÒòÊÇ
ÔÚ´ËζÈÏ£¬·´Ó¦ËÙÂʺÍƽºâת»¯Âʶ¼½Ï¸ß
ÔÚ´ËζÈÏ£¬·´Ó¦ËÙÂʺÍƽºâת»¯Âʶ¼½Ï¸ß
£»
Ñ¡Ôñ³£Ñ¹µÄÔ­ÒòÊÇ
³£Ñ¹Ï£¬Æ½ºâת»¯Âʽϸߣ¬ÇÒ²»»áÒò¼ÓѹÔì³ÉÉ豸Ôì¼ÛºÍºÄÄÜÔö¼Ó
³£Ñ¹Ï£¬Æ½ºâת»¯Âʽϸߣ¬ÇÒ²»»áÒò¼ÓѹÔì³ÉÉ豸Ôì¼ÛºÍºÄÄÜÔö¼Ó
£®
£¨4£©¹¤ÒµÉϳ£ÓùýÁ¿°±Ë®ÎüÊÕβÆøÖеĶþÑõ»¯Áò£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
SO2+2NH3?H2O=£¨NH4£©2SO3+H2O
SO2+2NH3?H2O=£¨NH4£©2SO3+H2O
£®
£¨5£©Éú²ú¹ý³ÌÖУ¬½Ó´¥ÊÒÉú³ÉµÄÈýÑõ»¯ÁòÓÃ98%µÄÁòËáÎüÊÕ£¬ÖƵÃÒ»ÖÖ·¢ÑÌÁòËᣨ9H2SO4?SO3£©£®Ä³ÁòËá³µ¼ä10СʱÏûºÄµÄÁòΪat£¬Ôòƽ¾ùÿСʱ´ÓÎüÊÕËþÁ÷³öµÄ¸Ã·¢ÑÌÁòËáΪ
481a
304
481a
304
t£¨ÔÚÉú²ú¹ý³ÌÖÐÁòµÄËðʧºöÂÔ²»¼Æ£©£®
·ÖÎö£º£¨1£©¸ù¾ÝFe3+µÄÑõ»¯ÐÔÓëÓëH2SµÄ»¹Ô­ÐÔÅж϶þÕßÖ®¼äµÄ·´Ó¦£¬ÒÔ´ËÊéд·´Ó¦µÄÀë×Ó·½³Ìʽ£»
£¨2£©ÁòËáÌúµÄËáÐÔÈÜÒº¾ßÓÐÑõ»¯ÐÔ£¬¿ÉÑ­»·Ê¹Óã»
£¨3£©¸ù¾ÝͼÏóÅжÏÔÚ400¡æ¡«500¡æ¡¢³£Ñ¹Ï·´Ó¦ÎïµÄת»¯ÂÊ£»
£¨4£©°±Ë®³Ê¼õС£¬¿ÉÓëËáÐÔÑõ»¯Îï·´Ó¦£»
£¨5£©98%µÄÁòËáÖÐn£¨H2SO4£©£ºn£¨H2O£©=
98g
98g/mol
£º
2g
18g/mol
=9£º1£¬¿ÉÉèŨÁòËáµÄ»¯Ñ§Ê½Îª9SO3?10H2O£¬
·¢ÑÌÁòËᣨ9H2SO4?SO3£©µÄ»¯Ñ§Ê½±äÐÎΪ10SO3?9H2O£¬¸ù¾Ý¹Øϵʽ£º9£¨9SO3?10H2O£©+19SO3=10£¨10SO3?9H2O£©¼ÆË㣮
½â´ð£º½â£º£¨1£©ÔÚËáÐÔÈÜÒºÖУ¬¾ßÓÐÑõ»¯ÐÔµÄFe3+Óë¾ßÓл¹Ô­ÐÔµÄH2S·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ2Fe3++H2S=2Fe2++2H++S¡ý£¬
¹Ê´ð°¸Îª£º2Fe3++H2S=2Fe2++2H++S¡ý£»
£¨2£©ÁòËáÌúµÄËáÐÔÈÜÒº¾ßÓÐÍÑÁò×÷Óã¬ÂËÒºµç½âºó£¬µÃµ½ÁòËáÌúµÄËáÐÔÈÜÒº¿ÉÑ­»·Ê¹Óã¬
¹Ê´ð°¸Îª£ºÊ¹ÁòËáÌúµÄËáÐÔÈÜҺѭ»·Ê¹Óã»
£¨3£©Ñ¡Ôñ400¡æ¡«500¡æ£¬·´Ó¦ËÙÂʺÍƽºâת»¯Âʶ¼½Ï¸ß£¬Èçζȹý¸ß£¬×ª»¯ÂÊ·´¶ø½µµÍ£¬³£Ñ¹Ï£¬Æ½ºâת»¯Âʽϸߣ¬ÈçÔö´óѹǿ£¬»áÔì³ÉÉ豸Ôì¼ÛºÍºÄÄÜÔö¼Ó£¬
¹Ê´ð°¸Îª£ºÔÚ´ËζÈÏ£¬·´Ó¦ËÙÂʺÍƽºâת»¯Âʶ¼½Ï¸ß£»³£Ñ¹Ï£¬Æ½ºâת»¯Âʽϸߣ¬ÇÒ²»»áÒò¼ÓѹÔì³ÉÉ豸Ôì¼ÛºÍºÄÄÜÔö¼Ó£»
£¨4£©°±Ë®³Ê¼îÐÔ£¬¿ÉÓë¾ßÓÐËáÐÔµÄSO2·¢Éú»¯ºÏ·´Ó¦£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪSO2+2NH3?H2O=£¨NH4£©2SO3+H2O£¬
¹Ê´ð°¸Îª£ºSO2+2NH3?H2O=£¨NH4£©2SO3+H2O£»
£¨5£©98%µÄÁòËáÖÐn£¨H2SO4£©£ºn£¨H2O£©=
98g
98g/mol
£º
2g
18g/mol
=9£º1£¬¿ÉÉèŨÁòËáµÄ»¯Ñ§Ê½Îª9SO3?10H2O£¬
·¢ÑÌÁòËᣨ9H2SO4?SO3£©µÄ»¯Ñ§Ê½±äÐÎΪ10SO3?9H2O£¬¿ÉµÃ·´Ó¦µÄ¹Øϵʽ£º9£¨9SO3?10H2O£©+19SO3=10£¨10SO3?9H2O£©£¬
Éè10СʱÉú³É·¢ÑÌÁòËáµÄÖÊÁ¿Îªx£¬
Ôò£º9£¨9SO3?10H2O£©+19SO3=10£¨10SO3?9H2O£©
                   19¡Á80   10¡Á962    
                   
at
32
¡Á80
  x
x=
10¡Á962¡Á
at
32
¡Á80
19¡Á80
=
4810a
304
t£¬
ËùÒÔƽ¾ùÿСʱÉú³É·¢ÑÌÁòËá
481a
304
t£¬
¹Ê´ð°¸Îª£º
481a
304
£®
µãÆÀ£º±¾Ì⿼²éº¬Áò»¯ºÏÎïµÄÐÔÖÊ×ÛºÏÓ¦ÓÃÒÔ¼°ÁòËáÉú²úµÄ¼ÆË㣬ÌâÄ¿ÄѶȽϴó£¬Ò×´íµãΪ£¨5£©£¬×¢Òâ¹ØϵʽµÄÀûÓã®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2011?ËÄ´¨Ä£Ä⣩̼ËáþºÍ̼Ëá¸ÆÓë´×ËáÇ¡ºÃÍêÈ«·´Ó¦µÃµ½µÄ»ìºÏÎï¿É×÷ÈÚÑ©¼Á£¬ÏÂÁÐÓйØ˵·¨´íÎóµÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2011?ËÄ´¨Ä£Ä⣩ÏÂÁз´Ó¦µÄÀë×Ó·½³ÌʽÊéдÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2011?ËÄ´¨Ä£Ä⣩X¡¢Y¡¢ZÊǶÌÖÜÆÚÔªËØ£¬ÔÚÖÜÆÚ±íÖеÄλÖùØϵÈçͼËùʾ£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2011?ËÄ´¨Ä£Ä⣩25¡æʱ£¬pH=aµÄHCOOHÈÜÒºÓëpH=bµÄNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºpH=7£¬Ôò£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2011?ËÄ´¨Ä£Ä⣩ÏòÁòËáÑÇÌúÓëÁòËáÂÁµÄ»ìºÏÈÜÒºÖмÓÈë×ãÁ¿µÄŨ°±Ë®£¬ÔÚ¿ÕÆøÖÐÂ˳öËùµÃ³Áµí£¬¾­Ï´¾»¡¢¸ÉÔï¡¢×ÆÉÕ£¬×îÖյõ½¹ÌÌåµÄ³É·ÖÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸