¾ÛºÏÁòËáÌúÓֳƾÛÌú£¬»¯Ñ§Ê½Îª[Fe2(OH)n(SO4)3£­0.5n]m£¬ ¹ã·ºÓÃÓÚÎÛË®´¦Àí¡£ÊµÑéÊÒÀûÓÃÁòË᳧ÉÕÔü(Ö÷Òª³É·ÖΪÌúµÄÑõ»¯Îï¼°ÉÙÁ¿FeS¡¢SiO2µÈ)ÖƱ¸¾ÛÌúºÍÂÌ·¯(FeSO4¡¤7H2O)£¬¹ý³ÌÈçÏ£º

£¨1£©ÑéÖ¤¹ÌÌåW±ºÉÕºó²úÉúµÄÆøÌ庬ÓÐSO2µÄ·½·¨ÊÇ£º

___________________________________________________________________¡£

£¨2£©ÊµÑéÊÒÖƱ¸¡¢ÊÕ¼¯¸ÉÔïµÄSO2£¬ËùÐèÒÇÆ÷ÈçÏ¡£×°ÖÃA²úÉúSO2£¬°´ÆøÁ÷·½ÏòÁ¬½Ó¸÷ÒÇÆ÷½Ó¿Ú£¬Ë³ÐòΪa¡ú____¡ú____¡ú____¡ú____¡úf¡£

×°ÖÃAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________________________________________¡£

×°ÖÃDµÄ×÷ÓÃÊÇ_____________£¬×°ÖÃEÖÐNaOHÈÜÒºµÄ×÷ÓÃÊÇ__________________¡£

£¨3£©ÖƱ¸ÂÌ·¯Ê±£¬ÏòÈÜÒºXÖмÓÈë¹ýÁ¿__________£¬³ä·Ö·´Ó¦ºó£¬¾­¹ýÂ˲Ù×÷µÃµ½ÈÜÒºY£¬ÔÙ¾­Å¨Ëõ¡¢½á¾§µÈ²½ÖèµÃµ½ÂÌ·¯¡£¹ýÂËËùÐèµÄ²£Á§ÒÇÆ÷ÓÐ_______________________¡£

£¨4£©Óû²â¶¨ÈÜÒºYÖÐFe2+µÄŨ¶È£¬ÐèÒªÓÃÈÝÁ¿Æ¿ÅäÖÆKMnO4±ê×¼ÈÜÒº£¬¶¨ÈÝʱÊÓÏßÓ¦_____________________£¬Ö±µ½_____________________¡£ÓÃKMnO4±ê×¼ÈÜÒºµÎ¶¨Ê±Ó¦Ñ¡ÓÃ________µÎ¶¨¹Ü(Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±)¡£

£¨5£©ÈÜÒºZµÄpHÓ°Ïì¾ÛÌúÖÐÌúµÄÖÊÁ¿·ÖÊý¡£ÈôÈÜÒºZµÄpHƫС£¬½«µ¼Ö¾ÛÌúÖÐÌúµÄÖÊÁ¿·ÖÊý______(Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2010?ɽ¶«£©¾ÛºÏÁòËáÌúÓֳƾÛÌú£¬»¯Ñ§Ê½Îª[Fe2£¨OH£©n£¨SO4£©3-0.5n]m£¬¹ã·ºÓÃÓÚÎÛË®´¦Àí£®ÊµÑéÊÒÀûÓÃÁòË᳧ÉÕÔü£¨Ö÷Òª³É·ÖΪÌúµÄÑõ»¯Îï¼°ÉÙÁ¿FeS¡¢SiO2µÈ£©ÖƱ¸¾ÛÌúºÍÂÌ·¯£¨FeSO4?7H2O £©¹ý³ÌÈçÏ£º

£¨1£©ÑéÖ¤¹ÌÌåW±ºÉÕºó²úÉúµÄÆøÌ庬ÓÐSO2 µÄ·½·¨ÊÇ
½«ÆøÌåͨÈëÆ·ºìÈÜÒºÖУ¬Èç¹ûÆ·ºìÍÊÉ«£¬¼ÓÈȺóÓÖ»Ö¸´ºìÉ«£®Ö¤Ã÷ÓиÃÆøÌ壮
½«ÆøÌåͨÈëÆ·ºìÈÜÒºÖУ¬Èç¹ûÆ·ºìÍÊÉ«£¬¼ÓÈȺóÓÖ»Ö¸´ºìÉ«£®Ö¤Ã÷ÓиÃÆøÌ壮
£®
£¨2£©ÊµÑéÊÒÖƱ¸¡¢ÊÕ¼¯¸ÉÔïµÄSO2£¬ËùÐèÒÇÆ÷ÈçÏ£®×°ÖÃA²úÉúSO2£¬°´ÆøÁ÷·½ÏòÁ¬½Ó¸÷ÒÇÆ÷½Ó¿Ú£¬Ë³ÐòΪa¡ú
d
d
¡ú
e
e
¡ú
c
c
¡ú
b
b
¡úf×°ÖÃDµÄ×÷ÓÃÊÇ
°²È«Æ¿£¬·ÀÖ¹µ¹Îü
°²È«Æ¿£¬·ÀÖ¹µ¹Îü
£¬×°ÖÃEÖÐNaOHÈÜÒºµÄ×÷ÓÃÊÇ
βÆø´¦Àí£¬·ÀÖ¹ÎÛȾ
βÆø´¦Àí£¬·ÀÖ¹ÎÛȾ
£®


£¨3£©ÖƱ¸ÂÌ·¯Ê±£¬ÏòÈÜÒºXÖмÓÈë¹ýÁ¿
Ìú·Û
Ìú·Û
£¬³ä·Ö·´Ó¦ºó£¬¾­
¹ýÂË
¹ýÂË
²Ù×÷µÃµ½ÈÜÒºY£¬ÔÙ¾­Å¨Ëõ£¬½á¾§µÈ²½ÖèµÃµ½ÂÌ·¯£®
£¨4£©ÈÜÒºZµÄpHÓ°Ïì¾ÛÌúÖÐÌúµÄÖÊÁ¿·ÖÊý£¬ÓÃpHÊÔÖ½²â¶¨ÈÜÒºpHµÄ²Ù×÷·½·¨Îª
½«ÊÔÖ½·Åµ½±íÃæÃóÉÏ£¬ÓýྻµÄ²£Á§°ôպȡÉÙÐí´ý²âÒº£¬µÎÔÚÊÔÖ½µÄÖÐÑ룮ȻºóÓë±ê×¼±ÈÉ«¿¨¶Ô±È£®
½«ÊÔÖ½·Åµ½±íÃæÃóÉÏ£¬ÓýྻµÄ²£Á§°ôպȡÉÙÐí´ý²âÒº£¬µÎÔÚÊÔÖ½µÄÖÐÑ룮ȻºóÓë±ê×¼±ÈÉ«¿¨¶Ô±È£®
£®ÈôÈÜÒºZµÄpHƫС£¬½«µ¼Ö¾ÛÌúÖÐÌúµÄÖÊÁ¿·ÖÊýÆ«
Æ«µÍ
Æ«µÍ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2011?ÄÏ¿ªÇøһģ£©¾ÛºÏÁòËáÌúÓֳƾÛÌú£¬»¯Ñ§Ê½Îª[Fe2£¨OH£©n£¨SO4£©3-0.5n]m£¬¹ã·ºÓÃÓÚÎÛË®´¦Àí£®ÊµÑéÊÒÀûÓÃÁòË᳧ÉÕÔü£¨Ö÷Òª³É·ÖΪÌúµÄÑõ»¯Îï¼°ÉÙÁ¿FeS¡¢SiO2µÈ£©ÖƱ¸¾ÛÌúºÍÂÌ·¯£¨FeSO4?7H2O£©£¬¹ý³ÌÈçÏ£º

£¨1£©ÑéÖ¤¹ÌÌåW±ºÉÕºó²úÉúµÄÆøÌ庬ÓÐSO2µÄ·½·¨ÊÇ£º
½«ÆäÆøÌåͨÈëÆ·ºìÊÔÒº£¬ÈÜÒºÍÊÉ«£¬¼ÓÈȻָ´ºìÉ«£¬Ö¤Ã÷ÓжþÑõ»¯Áò
½«ÆäÆøÌåͨÈëÆ·ºìÊÔÒº£¬ÈÜÒºÍÊÉ«£¬¼ÓÈȻָ´ºìÉ«£¬Ö¤Ã÷ÓжþÑõ»¯Áò
£®
£¨2£©ÊµÑéÊÒÖƱ¸¡¢ÊÕ¼¯¸ÉÔïµÄSO2£¬ËùÐèÒÇÆ÷ÈçÏ£®×°ÖÃA²úÉúSO2£¬°´ÆøÁ÷·½ÏòÁ¬½Ó¸÷ÒÇÆ÷½Ó¿Ú£¬Ë³ÐòΪa¡ú
d
d
¡ú
e
e
¡ú
c
c
¡ú
b
b
¡úf£®
×°ÖÃAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Cu+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CuSO4+SO2¡ü+2H2O
Cu+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CuSO4+SO2¡ü+2H2O
£®
×°ÖÃDµÄ×÷ÓÃÊÇ
°²È«Æ¿¡¢·ÀÖ¹µ¹Îü
°²È«Æ¿¡¢·ÀÖ¹µ¹Îü
£¬×°ÖÃEÖÐNaOHÈÜÒºµÄ×÷ÓÃÊÇ
ÎüÊÕ¶àÓàµÄ¶þÑõ»¯Áò
ÎüÊÕ¶àÓàµÄ¶þÑõ»¯Áò
£®
£¨3£©ÖƱ¸ÂÌ·¯Ê±£¬ÏòÈÜÒºXÖмÓÈë¹ýÁ¿
Ìúм
Ìúм
£¬³ä·Ö·´Ó¦ºó£¬¾­¹ýÂ˲Ù×÷µÃµ½ÈÜÒºY£¬ÔÙ¾­Å¨Ëõ¡¢½á¾§µÈ²½ÖèµÃµ½ÂÌ·¯£®¹ýÂËËùÐèµÄ²£Á§ÒÇÆ÷ÓÐ
ÉÕ±­¡¢²£Á§°ô¡¢Â©¶·
ÉÕ±­¡¢²£Á§°ô¡¢Â©¶·
£®
£¨4£©Óû²â¶¨ÈÜÒºYÖÐFe2+µÄŨ¶È£¬ÐèÒªÓÃÈÝÁ¿Æ¿ÅäÖÆKMnO4±ê×¼ÈÜÒº£¬¶¨ÈÝʱÊÓÏßÓ¦
ƽÊӿ̶ÈÏß
ƽÊӿ̶ÈÏß
£¬Ö±µ½
°¼ÒºÃæ×îµÍ´¦ºÍ¿Ì¶ÈÏàÇÐ
°¼ÒºÃæ×îµÍ´¦ºÍ¿Ì¶ÈÏàÇÐ
£®ÓÃKMnO4±ê×¼ÈÜÒºµÎ¶¨Ê±Ó¦Ñ¡ÓÃ
Ëáʽ
Ëáʽ
µÎ¶¨¹Ü£¨Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±£©£®
£¨5£©ÈÜÒºZµÄpHÓ°Ïì¾ÛÌúÖÐÌúµÄÖÊÁ¿·ÖÊý£®ÈôÈÜÒºZµÄpHƫС£¬½«µ¼Ö¾ÛÌúÖÐÌúµÄÖÊÁ¿·ÖÊý
ƫС
ƫС
£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨15·Ö£©¾ÛºÏÁòËáÌúÓֳƾÛÌú£¬»¯Ñ§Ê½Îª£¬¹ã·ºÓÃÓÚÎÛË®´¦Àí¡£ÊµÑéÊÒÀûÓÃÁòË᳧ÉÕÔü£¨Ö÷Òª³É·ÖΪÌúµÄÑõ»¯Îï¼°ÉÙÁ¿FeS¡¢SiO2µÈ£©ÖƱ¸¾ÛÌúºÍÂÌ·¯£¨FeSO4¡¤7H2O £©¹ý³ÌÈçÏ£º

£¨1£©ÑéÖ¤¹ÌÌåW±ºÉÕºó²úÉúµÄÆøÌ庬ÓÐSO2 µÄ·½·¨ÊÇ___¡£

£¨2£©ÖƱ¸ÂÌ·¯Ê±£¬ÏòÈÜÒºXÖмÓÈë¹ýÁ¿___£¬³ä·Ö·´Ó¦ºó£¬¾­_____²Ù×÷µÃµ½ÈÜÒºY£¬ÔÙ¾­Å¨Ëõ£¬½á¾§µÈ²½ÖèµÃµ½ÂÌ·¯¡£

£¨3£©ÈÜÒºZµÄpHÓ°Ïì¾ÛÌúÖÐÌúµÄÖÊÁ¿·ÖÊý£¬ÓÃpHÊÔÖ½²â¶¨ÈÜÒºpHµÄ²Ù×÷·½·¨Îª____¡£ÈôÈÜÒºZµÄpHƫС£¬½«µ¼Ö¾ÛÌúÖÐÌúµÄÖÊÁ¿·ÖÊýÆ«_____¡£

£¨4£©¹Å´ú½«ÂÌ·¯ìÑÉÕ¿ÉÖÆÂÌ·¯ÓÍ£¨Ò²½ÐïêË®£¬¼´ÁòËᣩºÍºìÉ«ÑÕÁÏ£¨Fe2O3£©£¬Çëд³öÓйصĻ¯Ñ§·½³Ìʽ£º

                                                                         ¡£

£¨5£©ÂÌ·¯»¹¿ÉÓÃÒÔÏ·½·¨Éú²úºìÉ«ÑÕÁÏ£¨Fe2O3£©,Æä¾ßÌå²Ù×÷Á÷³ÌÊÇ£º½«5560kgÂÌ·¯£¨Ä¦¶ûÖÊÁ¿Îª278 g/mol£©ÈÜÓÚË®ÖУ¬¼ÓÈëÊÊÁ¿ÇâÑõ»¯ÄÆÈÜҺǡºÃÍêÈ«·´Ó¦£¬¹ÄÈë×ãÁ¿¿ÕÆø½Á°è£¬²úÉúºìºÖÉ«½ºÌ壻ÔÙÏòºìºÖÉ«½ºÌåÖмÓÈë16680 kg ÂÌ·¯ºÍ560 kgÌú·Û£¬¹ÄÈë×ãÁ¿¿ÕÆø½Á°è£¬·´Ó¦Íê³Éºó£¬ÓдóÁ¿Fe2O3¸½×ÅÔÚ½ºÌåÁ£×ÓÉÏÒÔ³ÁµíÐÎʽÎö³ö£»¹ýÂ˺󣬳Áµí¾­¸ßÎÂ×ÆÉյúìÉ«ÑÕÁÏ¡£ÈôËùµÃÂËÒºÖÐÈÜÖÊÖ»ÓÐÁòËáÄƺÍÁòËáÌú£¬ÔòÀíÂÛÉÏ¿ÉÉú²úºìÉ«ÑÕÁÏ____________________kg¡£

  

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ì¸£½¨Ê¡ÇåÁ÷Ò»ÖиßÈýµÚÈý½×¶Î¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

¾ÛºÏÁòËáÌúÓֳƾÛÌú, »¯Ñ§Ê½Îª[Fe(OH)(SO4)]m, ¹ã·ºÓÃÓÚÎÛË®´¦Àí¡£ÊµÑéÊÒÀûÓÃÁòË᳧ÉÕÔü(Ö÷Òª³É·ÖΪÌúµÄÑõ»¯Îï¼°ÉÙÁ¿FeS¡¢SiO2µÈ)ÖƱ¸¾ÛÌúºÍÂÌ·¯(FeSO4¡¤7H2O )¹ý³ÌÈçÏ£º

ÑéÖ¤¹ÌÌåW±ºÉÕºó²úÉúµÄÆøÌ庬ÓÐSO2µÄ·½·¨ÊÇ                                   
(2)ʵÑéÊÒÖƱ¸¡¢ÊÕ¼¯¸ÉÔïµÄSO2, ËùÐèÒÇÆ÷ÈçÏ¡£×°ÖÃA²úÉúSO2, °´ÆøÁ÷·½ÏòÁ¬½Ó¸÷ÒÇÆ÷½Ó¿Ú,˳ÐòΪa¡ú___ ¡ú___ ¡ú___ ¡ú___f¡£×°ÖÃDµÄ×÷ÓÃÊÇ          , ×°ÖÃEÖÐNaOHÈÜÒºµÄ×÷ÓÃÊÇ          ¡£

£¨3£©ÖƱ¸ÂÌ·¯Ê±£¬ÏòÈÜÒºXÖмÓÈë¹ýÁ¿_____£¬³ä·Ö·´Ó¦ºó£¬¾­_______²Ù×÷µÃµ½ÈÜÒºY£¬ÔÙ¾­Å¨Ëõ£¬½á¾§µÈ²½ÖèµÃµ½ÂÌ·¯¡£
(4)ÈÜÒºZµÄpHÓ°Ïì¾ÛÌúÖÐÌúµÄÖÊÁ¿·ÖÊý£¬ÓÃpHÊÔÖ½²â¶¨ÈÜÒºpHµÄ²Ù×÷·½·¨Îª                                  ¡£ÈôÈÜÒºZµÄpHƫС£¬½«µ¼Ö¾ÛÌúÖÐÌúµÄÖÊÁ¿·ÖÊýÆ«_______¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄ긣½¨Ê¡¸ßÈýµÚÈý½×¶Î¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

¾ÛºÏÁòËáÌúÓֳƾÛÌú, »¯Ñ§Ê½Îª[Fe(OH)(SO4)]m, ¹ã·ºÓÃÓÚÎÛË®´¦Àí¡£ÊµÑéÊÒÀûÓÃÁòË᳧ÉÕÔü(Ö÷Òª³É·ÖΪÌúµÄÑõ»¯Îï¼°ÉÙÁ¿FeS¡¢SiO2µÈ)ÖƱ¸¾ÛÌúºÍÂÌ·¯(FeSO4¡¤7H2O )¹ý³ÌÈçÏ£º

ÑéÖ¤¹ÌÌåW±ºÉÕºó²úÉúµÄÆøÌ庬ÓÐSO2µÄ·½·¨ÊÇ                                   

(2)ʵÑéÊÒÖƱ¸¡¢ÊÕ¼¯¸ÉÔïµÄSO2, ËùÐèÒÇÆ÷ÈçÏ¡£×°ÖÃA²úÉúSO2, °´ÆøÁ÷·½ÏòÁ¬½Ó¸÷ÒÇÆ÷½Ó¿Ú,˳ÐòΪa¡ú___ ¡ú___ ¡ú___ ¡ú___f¡£×°ÖÃDµÄ×÷ÓÃÊÇ          , ×°ÖÃEÖÐNaOHÈÜÒºµÄ×÷ÓÃÊÇ          ¡£

£¨3£©ÖƱ¸ÂÌ·¯Ê±£¬ÏòÈÜÒºXÖмÓÈë¹ýÁ¿_____£¬³ä·Ö·´Ó¦ºó£¬¾­_______²Ù×÷µÃµ½ÈÜÒºY£¬ÔÙ¾­Å¨Ëõ£¬½á¾§µÈ²½ÖèµÃµ½ÂÌ·¯¡£

(4)ÈÜÒºZµÄpHÓ°Ïì¾ÛÌúÖÐÌúµÄÖÊÁ¿·ÖÊý£¬ÓÃpHÊÔÖ½²â¶¨ÈÜÒºpHµÄ²Ù×÷·½·¨Îª                                  ¡£ÈôÈÜÒºZµÄpHƫС£¬½«µ¼Ö¾ÛÌúÖÐÌúµÄÖÊÁ¿·ÖÊýÆ«_______¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸