ÌâÄ¿ÄÚÈÝ
»ÆÌú¿ó£¨Ö÷Òª³É·ÖΪFeS2£©ÊÇÎÒ¹ú´ó¶àÊýÁòË᳧ÖÆÈ¡ÁòËáµÄÖ÷ÒªÔÁÏ¡£Ä³»¯Ñ§Ñ§Ï°Ð¡×é¶Ôij»ÆÌú¿óʯ½øÐÐÈçÏÂʵÑé̽¾¿¡£
[ʵÑéÒ»]²â¶¨ÁòÔªËصĺ¬Á¿¡£
¢ñ¡¢½«m1 g¸Ã»ÆÌú¿óÑùÆ··ÅÈëÈçÏÂͼËùʾװÖ㨼гֺͼÓÈÈ×°ÖÃÊ¡ÂÔ£©µÄʯӢ¹ÜÖУ¬´Óa´¦²»¶ÏµØ»º»ºÍ¨Èë¿ÕÆø£¬¸ßÎÂ×ÆÉÕʯӢ¹ÜÖеĻÆÌú¿óÑùÆ·ÖÁ·´Ó¦ÍêÈ«¡£Ê¯Ó¢¹ÜÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º4FeS2+11O22Fe2O3+8SO2
¢ò¡¢·´Ó¦½áÊøºó£¬½«ÒÒÆ¿ÖеÄÈÜÒº½øÐÐÈçÏ´¦Àí£º
|
ÎÊÌâÌÖÂÛ£º
£¨1£©IÖУ¬¼×Æ¿ÄÚËùÊ¢ÊÔ¼ÁÊÇ____ÈÜÒº¡£ÒÒÆ¿ÄÚ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____¡£
£¨2£©IIÖУ¬ÒÒÆ¿¼ÓÈëH2O2ÈÜҺʱ·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________¡£
£¨3£©¸Ã»ÆÌú¿óÖÐÁòÔªËصÄÖÊÁ¿·ÖÊýΪ_______________________¡£
£¨4£©IIIµÄ²½Öè¢ÛÖУ¬ÐèÒªÓõ½µÄÒÇÆ÷³ýÉÕ±¡¢²£Á§°ô¡¢½ºÍ·µÎ¹ÜÍ⣬»¹ÓÐ____________________________________________¡£
£¨5£©IIIµÄ²½Öè¢ÜÖУ¬±íʾµÎ¶¨ÒÑ´ïÖÕµãµÄÏÖÏóÊÇ
£¨6£©¢óµÄ²½Öè¢Ü½øÐÐÁËÈý´ÎƽÐÐʵÑ飬²âµÃÏûºÄKMnO4ÈÜÒºÌå»ý·Ö±ðΪ24.98mL¡¢24.80mL¡¢25.02mL£¨KMnO4±»»¹ÔΪMn2+£©¡£¸ù¾ÝÉÏÊöÊý¾Ý£¬¿É¼ÆËã³ö¸Ã»ÆÌú¿óÑùÆ·ÌúÔªËصÄÖÊÁ¿·ÖÊýΪ ¡£
£¨1£©ÇâÑõ»¯ÄÆ£¨»òÇâÑõ»¯¼Ø£©£¨2·Ö£© SO2+2OH¡ª=SO32¡ª+H2O£»£¨2SO32¡ª+O2=2SO42¡ªÎ´Ð´²»¿Û·Ö£©£¨2·Ö£©
£¨2£©SO32¡ª+H2O2=SO42¡ª+H2O £¨2·Ö£©
£¨3£©£¨2·Ö£©
£¨4£©250mlÈÝÁ¿Æ¿£¨2·Ö£©
£¨5£©×îºóÒ»µÎ¸ßÃÌËá¼ØÈÜÒºµÎÈëʱ£¬ÈÜÒºÑÕÉ«Í»±äΪ×ÏÉ«£¬ÇÒÔÚ30sÄÚ²»±äÉ«¡££¨2·Ö£©
£¨6£©»ò»òÆäËüºÏÀí´ð°¸£¨2·Ö£©
½âÎöÊÔÌâ·ÖÎö£º£¨1£©Îª·ÀÖ¹SO2½øÈëµ½¼××°ÖÃÖУ¬¿ÉÓÃÇâÑõ»¯ÄÆ£¨»òÇâÑõ»¯¼Ø£©ÎüÊÕ£¬ÒÒÆ¿ÖеÄÇâÑõ»¯ÄÆÎüÊÕÓ²Öʲ£Á§¹ÜÖвúÉúµÄ¶þÑõ»¯Áò£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºSO2+2OH¡ª=SO32¡ª+H2O£¬ÓÉÓÚ׶ÐÎÆ¿ÖÐÒ²´æÔÚ×ÅδÍêÈ«·´Ó¦µÄÑõÆø£¬¹ÊÒ²·¢Éú2SO32¡ª+O2=2SO42¡ª£¨2£©IIÖУ¬ÒÒÆ¿¼ÓÈëH2O2ÈÜҺʱ·´Ó¦µÄÀë×Ó·½³ÌʽΪSO32¡ª+H2O2=SO42¡ª+H2O£»£¨3£©ÒÒÆ¿ÖеÄÈÜÒº²úÉúSO42¡ª£¬SO42¡ªÓë¼ÓÈëµÄÂÈ»¯±µÖеÄBa2+³ÁµíÏà½áºÏ·¢Éú³Áµí·´Ó¦£ºSO42¡ª+Ba2+=BaSO4¡ý£¬¼´¹ÌÌåm2gΪBaSO4£¬½áºÏÌâÒâÁйØϵʽÈçÏ£º
FeS2¡ª2SO42¡ª¡ª2BaSO4
1mol 2mol
ÓÉ´Ë¿ÉÖª£¬ÁòÔªËØÔÚFeS2ÖеÄÎïÖʵÄÁ¿Îª£¬ÔòÆäÖÊÁ¿Îª£º£¬ËùÒԸûÆÌú¿óÖÐÁòÔªËصÄÖÊÁ¿·ÖÊýΪ
¡Á©‡£»£¨4£©Ó¦Îª250mlÈÝÁ¿Æ¿£»£¨5£©ÒòΪδ¼ÓÈëKMnO4ÈÜҺ֮ǰ£¬ÈÜÒºÖк¬ÓÐFe2+£¬ÈÜҺΪdzÂÌÉ«£¬
µ±¼ÓÈë×îºóÒ»µÎ¸ßÃÌËá¼ØÈÜÒºµÎÈëʱ£¬ÈÜÒºÑÕÉ«Í»±äΪ×ÏÉ«£¬ÇÒÔÚ30sÄÚ²»±äÉ«£¬ÔòÖ¤Ã÷µ½´ïµÎ¶¨Öյ㡣
£¨6£©·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ5Fe2++8H++MnO4-=Mn2++5Fe3++4H2O,È»ºó¸ù¾Ý5Fe2+¡ªMnO4-ÁйØϵʽ¿ÉÇóµÃ
¸Ã»ÆÌú¿óÑùÆ·ÌúÔªËصÄÖÊÁ¿·ÖÊýΪ»ò
¿¼µã£º¿¼²éÎïÖʵķÖÀëÓëÌá´¿£¬»¯Ñ§ÊµÑé»ù±¾²Ù×÷¡¢Ñõ»¹Ô·´Ó¦¡¢Àë×Ó·½³ÌʽµÄÊéд¡¢Ñõ»¯»¹Ô·´Ó¦µÎ¶¨ÔÀí
Óë¼ÆËã¡£
(14·Ö)½üÄêÀ´ÎÒ¹úµÄº½ÌìÊÂҵȡµÃÁ˾޴óµÄ³É¾Í£¬ÔÚº½Ìì·¢Éäʱ£¬ëÂ(N2H4)¼°ÆäÑÜÉúÎï³£ÓÃ×÷»ð¼ýÍƽø¼Á¡£
¢ÅҺ̬ëÂ×÷»ð¼ýȼÁÏʱ£¬ÓëҺ̬N2O4»ìºÏ·¢ÉúÑõ»¯»¹Ô·´Ó¦£¬ÒÑ֪ÿ1 gë³ä·Ö·´Ó¦ºóÉú³ÉÆø̬ˮ·Å³öÈÈÁ¿Îªa KJ£¬ÊÔд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ ¡£
¢ÆÔÚʵÑéÊÒÖУ¬ÓÃN2H4¡¤H2OÓëNaOH¿ÅÁ£Ò»ÆðÕôÁó£¬ÊÕ¼¯114¡«116 ¡æµÄÁó·Ö¼´ÎªÎÞˮ롣
¢ÙÔÚÕôÁó¹ý³ÌÖв»ÐèÒªµÄÒÇÆ÷ÊÇ (ÌîÐòºÅ×Öĸ)¡£
A£®¾Æ¾«µÆ | B£®³¤Ö±²£Á§µ¼¹Ü | C£®×¶ÐÎÆ¿ | D£®ÀäÄý¹Ü |
¢Ú³ýÉÏÊö±ØÐèµÄÒÇÆ÷Í⣬»¹È±ÉٵIJ£Á§ÒÇÆ÷ÊÇ ¡£
¢ÇëÂÄÜʹ¹ø¯ÄÚ±ÚµÄÌúÐâ(Ö÷Òª³É·ÖFe2O3)±ä³É´ÅÐÔÑõ»¯Ìú(Fe3O4)²ã£¬¿É¼õ»º¹ø¯ÐâÊ´¡£Èô·´Ó¦¹ý³ÌÖÐëÂת»¯ÎªµªÆø£¬ÔòÿÉú³É1 mol Fe3O4£¬ÐèÒªÏûºÄëµÄÖÊÁ¿Îª g¡£
¢È´ÅÐÔÑõ»¯Ìú(Fe3O4)µÄ×é³É¿Éд³ÉFeO¡¤Fe2O3¡£Ä³»¯Ñ§ÊµÑéС×éͨ¹ýʵÑéÀ´Ì½¾¿Ò»ºÚÉ«·ÛÄ©ÊÇ·ñÓÉFe3O4¡¢CuO×é³É(²»º¬ÓÐÆäËüºÚÉ«ÎïÖÊ)¡£Ì½¾¿¹ý³ÌÈçÏ£º
Ìá³ö¼ÙÉ裺¼ÙÉè1. ºÚÉ«·ÛÄ©ÊÇCuO£»¼ÙÉè2. ºÚÉ«·ÛÄ©ÊÇFe3O4£»
¼ÙÉè3. ¡£
Éè¼Æ̽¾¿ÊµÑ飺
È¡ÉÙÁ¿·ÛÄ©·ÅÈë×ãÁ¿Ï¡ÁòËáÖУ¬ÔÚËùµÃÈÜÒºÖеμÓKSCNÊÔ¼Á¡£
¢ÙÈô¼ÙÉè1³ÉÁ¢£¬ÔòʵÑéÏÖÏóÊÇ ¡£
¢ÚÈôËùµÃÈÜÒºÏÔѪºìÉ«£¬Ôò¼ÙÉè ³ÉÁ¢¡£
¢ÛΪ½øÒ»²½Ì½¾¿£¬¼ÌÐøÏòËùµÃÈÜÒº¼ÓÈë×ãÁ¿Ìú·Û£¬Èô²úÉú µÄÏÖÏó£¬Ôò¼ÙÉè3³ÉÁ¢¡£
ÓÐÁíһС×éͬѧÌá³ö£¬Èô»ìºÏÎïÖÐCuOº¬Á¿½ÏÉÙ£¬¿ÉÄܼÓÈëÌú·ÛºóʵÑéÏÖÏó²»Ã÷ÏÔ¡£
²éÔÄ×ÊÁÏ£ºCu2£«Óë×ãÁ¿°±Ë®·´Ó¦Éú³ÉÉîÀ¶É«ÈÜÒº£¬Cu2£«£«4NH3¡¤H2O£½Cu(NH3)42£«£«4H2O¡£
¢ÜΪ̽¾¿ÊǼÙÉè2»¹ÊǼÙÉè3³ÉÁ¢£¬ÁíÈ¡ÉÙÁ¿·ÛÄ©¼ÓÏ¡ÁòËá³ä·ÖÈܽâºó£¬ÔÙ¼ÓÈë×ãÁ¿°±Ë®£¬Èô¼ÙÉè2³ÉÁ¢£¬Ôò²úÉú ÏÖÏó£»Èô²úÉú ÏÖÏó£¬Ôò¼ÙÉè3³ÉÁ¢¡£
ijͬѧ½øÐÐÁòËá;§Ìå½á¾§Ë®º¬Á¿µÄ²â¶¨ÊµÑé¡£Íê³ÉÏÂÁÐÌî¿Õ£º
¡¾ÊµÑé²½Öè¡¿
£¨1£©ÓÃ_______£¨ÌîÒÇÆ÷Ãû³Æ£¬ÏÂͬ£©×¼È·³ÆÁ¿´ÉÛáÛöµÄÖÊÁ¿¡£
£¨2£©ÔÚ´ÉÛáÛöÖмÓÈëÔ¼2 gÑÐϸµÄÁòËá;§Ì壬²¢³ÆÁ¿¡£
£¨3£©°ÑÊ¢ÓÐÁòËá;§ÌåµÄ´ÉÛáÛö·ÅÔÚÄàÈý½ÇÉÏÂýÂý¼ÓÈÈ£¬Ö±µ½À¶É«ÍêÈ«±ä°×£¬È»ºó°ÑÛáÛöÒÆÖÁ____________ÖÐÀäÈ´µ½ÊÒΣ¬²¢³ÆÁ¿¡£
£¨4£©Öظ´(3)µÄʵÑé½øÐкãÖزÙ×÷£¬Ö±ÖÁÁ½´Î³ÆÁ¿½á¹ûÏà²î²»³¬¹ý0.001 g¡£
¡¾Êý¾Ý¼Ç¼Óë´¦Àí¡¿
| µÚÒ»´ÎʵÑé | µÚ¶þ´ÎʵÑé |
ÛáÛöµÄÖÊÁ¿£¨g£© | 29.563 | 30.064 |
ÛáÛö£«ÊÔÑùµÄÖÊÁ¿£¨g£© | 31.676 | 32.051 |
ºãÖغó£¬ÛáÛö£«ÁòËá͵ÄÖÊÁ¿£¨g£© | 30.911 | 31.324 |
xµÄÖµ | 5.05 | 5.13 |
¸ù¾ÝÉϱíÖеÄÊý¾Ý´¦Àí½á¹û£¬¼ÆËã±¾´ÎʵÑéµÄÏà¶ÔÎó²îΪ______%£¨ÒÑÖªxµÄÀíÂÛֵΪ5£©¡£
¡¾·ÖÎöÓëÌÖÂÛ¡¿
£¨1£©×öÒ»´ÎʵÑ飬ÖÁÉÙÐèÒª¼ÓÈÈ________´Î£¨ÌîÊý×Ö£¬ÏÂͬ£©£»ÖÁÉÙÐèÒª³ÆÁ¿_________´Î¡£
£¨2£©ºãÖزÙ×÷µÄÄ¿µÄÊÇ__________________¡£
£¨3£©Öظ´Á½´ÎʵÑéÇóxƽ¾ùÖµµÄÄ¿µÄÊÇ_____________________________¡£
£¨4£©ÊµÑéÖµ±ÈÀíÂÛֵƫ´óµÄÔÒò¿ÉÄÜÊÇ________£¨Ìî±àºÅ£©¡£
a£®¼ÓÈȹý³ÌÖÐÓо§Ì彦³ö b£®±»²âÑùÆ·Öк¬ÓмÓÈȲ»»Ó·¢µÄÔÓÖÊ
c£®ÊµÑéÇ°£¬¾§Ìå±íÃ泱ʪ d£®¾§Ìå×ÆÉÕºóÖ±½Ó·ÅÔÚ¿ÕÆøÖÐÀäÈ´
ʵÑéÊÒ³£ÀûÓü×È©£¨HCHO£©·¨²â¶¨(NH4)2SO4ÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊý£¬Æä·´Ó¦ÔÀíΪ£º4NH4£« £«6HCHO =3H£«£«6H2O£«(CH2)6N4H£« £ÛµÎ¶¨Ê±£¬1 mol (CH2)6N4H£«Óë l mol H£«Ï൱£¬È»ºóÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨·´Ó¦Éú³ÉµÄËᡣijÐËȤС×éÓü×È©·¨½øÐÐÁËÈçÏÂʵÑ飺
²½ÖèI ³ÆÈ¡ÑùÆ·1£®500 g¡£
²½ÖèII ½«ÑùÆ·Èܽâºó£¬ÍêȫתÒƵ½250 mLÈÝÁ¿Æ¿ÖУ¬¶¨ÈÝ£¬³ä·ÖÒ¡ÔÈ¡£
²½ÖèIII ÒÆÈ¡25£®00 mLÑùÆ·ÈÜÒºÓÚ250 mL׶ÐÎÆ¿ÖУ¬¼ÓÈë10 mL 20£¥µÄÖÐÐÔ¼×È©ÈÜÒº£¬Ò¡ÔÈ¡¢¾²ÖÃ5 minºó£¬¼ÓÈë1~2µÎ·Ó̪ÊÔÒº£¬ÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㡣°´ÉÏÊö²Ù×÷·½·¨ÔÙÖظ´2´Î¡£
£¨1£©¸ù¾Ý²½ÖèIIIÌî¿Õ£º
¢Ù¼îʽµÎ¶¨¹ÜÓÃÕôÁóˮϴµÓºó£¬Ö±½Ó¼ÓÈëNaOH±ê×¼ÈÜÒº½øÐе樣¬Ôò²âµÃÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊý________(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족)¡£
¢Ú׶ÐÎÆ¿ÓÃÕôÁóˮϴµÓºó£¬Ë®Î´µ¹¾¡£¬ÔòµÎ¶¨Ê±ÓÃÈ¥NaOH±ê×¼ÈÜÒºµÄÌå»ý_______(Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)
¢ÛµÎ¶¨Ê±±ßµÎ±ßÒ¡¶¯×¶ÐÎÆ¿£¬ÑÛ¾¦Ó¦¹Û²ì____________¡£
A£®µÎ¶¨¹ÜÄÚÒºÃæµÄ±ä»¯ B£®×¶ÐÎÆ¿ÄÚÈÜÒºÑÕÉ«µÄ±ä»¯
¢ÜµÎ¶¨´ïµ½ÖÕµãʱÏÖÏó£º__________________________________________________¡£
£¨2£©µÎ¶¨½á¹ûÈçϱíËùʾ£º
µÎ¶¨ ´ÎÊý | ´ý²âÈÜÒºµÄÌå»ý /mL | ±ê×¼ÈÜÒºµÄÌå»ý/mL | |
µÎ¶¨Ç°¿Ì¶È | µÎ¶¨ºó¿Ì¶È | ||
1 | 25£®00 | 1£®02 | 21£®03 |
2 | 25£®00 | 2£®00 | 21£®99 |
3 | 25£®00 | 0£®20 | 20£®20 |
ÈôNaOH±ê×¼ÈÜÒºµÄŨ¶ÈΪ0£®1010 mol¡¤L£1£¬Ôò¸ÃÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊýΪ___________¡£
ÌúÊÇÈËÌå²»¿ÉȱÉÙµÄ΢Á¿ÔªËØ£¬ÉãÈ뺬ÌúµÄ»¯ºÏÎï¿É²¹³äÌú¡£¡°ËÙÁ¦·Æ¡±ÊÇÊг¡ÉÏÒ»ÖÖ³£¼ûµÄ²¹ÌúÒ©Æ·£¬Ï±íÊÇ˵Ã÷ÊéµÄ²¿·ÖÄÚÈÝ¡£
[¹æ¸ñ]ÿƬº¬çúçêËáÑÇÌú [ÊÊÓ¦Ö¢]ÓÃÓÚȱÌúÐÔƶѪ֢£¬Ô¤·À¼°ÖÎÁÆÓᣠ[ÓÃÁ¿Ó÷¨]³ÉÈËÔ¤·ÀÁ¿/ÈÕ£¬³ÉÈËÖÎÁÆÁ¿¡ª/ÈÕ¡£ С¶ùÓÃÁ¿Ô¤·ÀÁ¿¡ª /ÈÕ£¬ÖÎÁÆÁ¿¡ª/ÈÕ [Öü²Ø]±Ü¹â¡¢ÃÜ·â¡¢ÔÚ¸ÉÔï´¦±£´æ¡£ |
£¨1£©¸ÃÒ©Æ·ÖÐFe2+»á»ºÂýÑõ»¯¡£¹ú¼Ò¹æ¶¨¸ÃÒ©ÎïÖÐFe2+µÄÑõ»¯ÂÊ£¨ÒѾ±»Ñõ»¯Fe2+µÄÖÊÁ¿ÓëFe2+×ÜÖÊÁ¿µÄ±ÈÖµ£©³¬¹ý10.00% ¼´²»ÄÜÔÙ·þÓá£
¢ÙʵÑéÊҿɲÉÓÃH2SO4ËữµÄKMnO4ÈÜÒº£¬¶Ô¡°ËÙÁ¦·Æ¡±ÖеÄFe2+½øÐеζ¨(¼ÙÉèÒ©Æ·ÖÐÆäËû³É·Ý²»ÓëKMnO4·´Ó¦)¡£Çëд³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º ¡£
¢ÚʵÑéÇ°£¬Ê×ÏÈÒª¾«È·ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄKMnO4ÈÜÒº250 mL£¬ÅäÖÆʱÐèÒªµÄ²£Á§ÒÇÆ÷³ý²£Á§°ô¡¢ÉÕ±¡¢½ºÍ·µÎ¹ÜÍ⣬»¹Ðè ¡£
¢ÛijͬѧÉè¼ÆÁËÏÂÁеζ¨·½Ê½£¨¼Ð³Ö²¿·ÖÂÔÈ¥£©£¬×îºÏÀíµÄÊÇ ¡££¨Ìî×ÖĸÐòºÅ£©
£¨2£©³ÆÁ¿ÉÏÊöº¬ÌúÔªËØÖÊÁ¿·ÖÊýΪ20.00%µÄ¡°ËÙÁ¦·Æ¡±10.00 g£¬½«ÆäÈ«²¿ÈÜÓÚÏ¡H2SO4ÖУ¬ÅäÖƳÉ1000 mlÈÜÒº£¬È¡³ö20.00 ml£¬ÓÃ0.01000 mol?L-1µÄKMnO4ÈÜÒºµÎ¶¨£¬ÓÃÈ¥KMnO4ÈÜÒº12.00 ml £¬¸ÃÒ©Æ·ÖÐFe2+µÄÑõ»¯ÂÊΪ ¡£
£¨3£©ÒÑÖªçúçêËáΪ¶þÔªÓлúôÈËᣬº¬23.6 gçúçêËáµÄÈÜÒºÓë4.0 mol?L-1 100.0 mlµÄÇâÑõ»¯ÄÆÈÜҺǡºÃÍêÈ«Öк͡£ºË´Å¹²ÕñÇâÆ×·ÖÎöÏÔʾ£¬çúçêËá·Ö×ÓÆ×ͼÉÏÖ»ÓÐÁ½×éÎüÊշ塣д³öçúçêËáÈÜÒºÓëÇâÑõ»¯ÄÆÈÜÒºÍêÈ«Öк͵Ļ¯Ñ§·½³Ìʽ(ÓлúÎïд½á¹¹¼òʽ) ¡£