ÌâÄ¿ÄÚÈÝ

ÒÔÏÂÊÇ·ÖÎöÁòÌú¿óÖÐFeS2º¬Á¿µÄÈýÖÖ·½·¨£¬¸÷·½·¨µÄ²Ù×÷Á÷³ÌͼÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)Á÷³ÌͼÖвÙ×÷¢Ù¡¢¢Ú¡¢¢Û·Ö±ðÖ¸µÄÊÇ¢Ù_________¡¢¢Ú__________¡¢¢Û________¡£
²Ù×÷¢Ü¡¢¢ÝÓõ½µÄÖ÷ÒªÒÇÆ÷ÊÇ£º¢Ü__________¡¢¢Ý__________(ÿ¿ÕÌî1-2¸öÒÇÆ÷)¡£
(2)ÅжÏÈÜÒºÖÐSO42-Àë×ÓÒѳÁµíÍêÈ«µÄ·½·¨ÊÇ_________________________________¡£
(3)ijͬѧÓ÷½·¨¢ó²â¶¨ÊÔÑùÖÐFeÔªËصĺ¬Á¿£¬×¼È·³ÆÈ¡Ò»¶¨Á¿µÄ¿óʯÊÔÑù£¬ÊÔÑù¾­Èܽ⡢Ԥ´¦Àíºó£¬
A£®ÓôøÓп̶ȵÄÉÕ±­ÅäÖƳÉ100 mLÊÔÑùÈÜÒº¡£B£®ÓÃÁ¿Í²Á¿È¡25.00 mL´ý²âÈÜÒº£¬C£®²¢ÖÃÓÚ׶ÐÎÆ¿ÖС£D£®ÓÃÕôÁóˮϴµÓµÎ¶¨¹Üºó×°ÈëKMnO4±ê×¼ÈÜÒº£¬Óøñê×¼ÈÜÒºµÎ¶¨´ý²âÊÔÑù£¬(E)µ±ÈÜÒº±ä³Éµ­×Ϻìɫʱ£¬Í£Ö¹µÎ¶¨£¬Èç30ÃëÄÚ²»ÍÊÉ«£¬(F)¶ÁÈ¡²¢¼ÆËãµÎ¶¨¹ÜÖÐÏûºÄµÄKMnO4±ê×¼ÈÜÒºÌå»ý£¬¼ÆËãÊÔÑùÖеÄFeÔªËغ¬Á¿¡£ÇëÖ¸³ö¸ÃʵÑé¹ý³ÌÖдíÎó²Ù×÷²½ÖèµÄ±àºÅ________________________¡£
(4)ijͬѧ²ÉÓ÷½·¨¢ò·ÖÎö¿óʯÖеÄFeº¬Á¿£¬·¢Ïֲⶨ½á¹û×ÜÊÇÆ«¸ß£¬Ôò²úÉúÎó²îµÄ¿ÉÄÜÔ­ÒòÊÇ_____________________________________________¡£
(5)³ÆÈ¡¿óʯÊÔÑù1.60 g, °´·½·¨¢ñ·ÖÎö£¬³ÆµÃBaSO4µÄÖÊÁ¿Îª4.66 g£¬¼ÙÉè¿óʯÖеÄÁòÔªËØÈ«²¿À´×ÔÓÚFeS2£¬Ôò¸Ã¿óʯÖÐFeS2µÄÖÊÁ¿·ÖÊýÊÇ___________¡£

£¨1£©¹ýÂË £¨1·Ö£© Ï´µÓ £¨1·Ö£© ¸ÉÔï £¨1·Ö£© ÛáÛö¡¢¾Æ¾«µÆ £¨1·Ö£© Ììƽ £¨1·Ö£©
£¨2£©È¡ÉϲãÇåÒºµÎ¼ÓBaCl2ÈÜÒº£¬ÈôÎÞ°×É«³ÁµíÉú³É£¬ËµÃ÷SO42-³ÁµíÍêÈ« £¨2·Ö£©
£¨3£©A¡¢B¡¢D £¨3·Ö£©
£¨4£©£¨2·Ö£©ÓÐÈý¸ö¿ÉÄܵÄÔ­Òò£º£¨Ð´³öÆäÖÐÒ»ÖÖ£¬¼´¸ø2·Ö£©
¢Ù Fe(OH)3³Áµí±íÃæ»ý´ó£¬Ò×Îü¸½ÔÓÖÊ
¢Ú¹ýÂËÏ´µÓʱδ³ä·Ö½«Îü¸½µÄÔÓÖÊÏ´È¥
¢Û Fe(OH)3×ÆÉÕ²»³ä·Ö£¬Î´Íêȫת»¯ÎªFe2O3
£¨5£©75.0% £¨3·Ö£©

ÊÔÌâ·ÖÎö£º£¨1£©´ÓÈÜÒºÖеõ½´¿¾»µÄ¹ÌÌåÐè¾­¹ý¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔïÈý²½£¬µÚ¢Ü²½Îª×ÆÉÕʹÇâÑõ»¯Ìú·Ö½âΪÑõ»¯Ìú£¬È»ºó³ÆÁ¿Ñõ»¯ÌúµÄÖÊÁ¿È·¶¨FeS2º¬Á¿£¬ËùÒÔʹÓõÄÒÇÆ÷·Ö±ðΪÛáÛö¡¢¾Æ¾«µÆºÍÌìƽ£»£¨2£©È¡ÉϲãÇåÒºµÎ¼ÓBaCl2ÈÜÒº£¬ÈôÎÞ°×É«³ÁµíÉú³É£¬ËµÃ÷SO42-³ÁµíÍêÈ«£»£¨3£©A¡¢ÅäÖÆÈÜÒºÒªÓÃÈÝÁ¿Æ¿£¬´íÎó£»B¡¢Á¿Í²µÄ¶ÁÊýÖ»Äܾ«È·µ½Ê®·Ö룬´íÎó£»D¡¢µÎ¶¨¹ÜÒªÓôý×°ÒºÈóÏ´£¬´íÎó¡££¨4£©³ÆÁ¿µÄ¹ÌÌåÖÊÁ¿¸ß£¬µ¼Ö½á¹ûÆ«¸ß£¬¹Ê¿ÉÄܵÄÔ­ÒòÓУº¢Ù Fe(OH)3³Áµí±íÃæ»ý´ó£¬Ò×Îü¸½ÔÓÖÊ ¢Ú¹ýÂËÏ´µÓʱδ³ä·Ö½«Îü¸½µÄÔÓÖÊÏ´È¥ ¢Û Fe(OH)3×ÆÉÕ²»³ä·Ö£¬Î´Íêȫת»¯ÎªFe2O3£»
£¨5£©n(FeS2)=n(SO42-)/2=2n(BaSO4)/2=4.66¡Â233¡Â2=0.01mol
¸Ã¿óʯÖÐFeS2µÄÖÊÁ¿·ÖÊýÊÇ0.01¡Á120¡Â1.60=0.75
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
(15·Ö)¼îʽ̼ËáÍ­µÄ³É·ÖÓжàÖÖ£¬Æ仯ѧʽһ°ã¿É±íʾΪxCu(OH)2¡¤yCuCO3¡£
(1)¿×ȸʯ³ÊÂÌÉ«£¬ÊÇÒ»ÖÖÃû¹óµÄ±¦Ê¯£¬ÆäÖ÷Òª³É·ÖÊÇCu(OH)2¡¤CuCO3¡£Ä³ÐËȤС×éΪ̽¾¿ÖÆÈ¡¿×ȸʯµÄ×î¼Ñ·´Ó¦Ìõ¼þ£¬Éè¼ÆÁËÈçÏÂʵÑ飺
ʵÑé1£º½«2.0mL 0.50 mol¡¤L¨C1µÄCu(NO3)2ÈÜÒº¡¢2.0mL 0.50 mol¡¤L¨C1µÄNaOHÈÜÒººÍ0.25 mol¡¤L¨C1µÄNa2CO3ÈÜÒº°´±í¢ñËùʾÌå»ý»ìºÏ¡£
ʵÑé2£º½«ºÏÊʱÈÀýµÄ»ìºÏÎïÔÚ±í¢òËùʾζÈÏ·´Ó¦¡£
ʵÑé¼Ç¼ÈçÏ£º
±í¢ñ                                      ±í¢ò
񅧏
V (Na2CO3)/mL
³ÁµíÇé¿ö
 
񅧏
·´Ó¦Î¶È/¡æ
³ÁµíÇé¿ö
1
2.8
¶à¡¢À¶É«
 
1
40
¶à¡¢À¶É«
2
2.4
¶à¡¢À¶É«
 
2
60
ÉÙ¡¢Ç³ÂÌÉ«
3
2.0
½Ï¶à¡¢ÂÌÉ«
 
3
75
½Ï¶à¡¢ÂÌÉ«
4
1.6
½ÏÉÙ¡¢ÂÌÉ«
 
4
80
½Ï¶à¡¢ÂÌÉ«(ÉÙÁ¿ºÖÉ«)
 
¢ÙʵÑéÊÒÖÆÈ¡ÉÙÐí¿×ȸʯ£¬Ó¦¸Ã²ÉÓõÄ×î¼ÑÌõ¼þÊÇ                               ¡£
¢Ú80¡æʱ£¬ËùÖƵõĿ×ȸʯÓÐÉÙÁ¿ºÖÉ«ÎïÖʵĿÉÄÜÔ­ÒòÊÇ                          ¡£
(2)ʵÑéС×éΪ²â¶¨ÉÏÊöijÌõ¼þÏÂËùÖƵõļîʽ̼ËáÍ­ÑùÆ·×é³É£¬ÀûÓÃÏÂͼËùʾµÄ×°ÖÃ(¼Ð³ÖÒÇÆ÷Ê¡ÂÔ)½øÐÐʵÑ飺

²½Öè1£º¼ì²é×°ÖõÄÆøÃÜÐÔ£¬½«¹ýÂË¡¢Ï´µÓ²¢¸ÉÔï¹ýµÄÑùÆ·ÖÃÓÚƽֱ²£Á§¹ÜÖС£
²½Öè2£º´ò¿ª»îÈûK£¬¹ÄÈë¿ÕÆø£¬Ò»¶Îʱ¼äºó¹Ø±Õ£¬³ÆÁ¿Ïà¹Ø×°ÖõÄÖÊÁ¿¡£
²½Öè3£º¼ÓÈÈ×°ÖÃBÖ±ÖÁ×°ÖÃCÖÐÎÞÆøÅݲúÉú¡£
²½Öè4£º                                          (Çë²¹³ä¸Ã²½²Ù×÷ÄÚÈÝ)¡£
²½Öè5£º³ÆÁ¿Ïà¹Ø×°ÖõÄÖÊÁ¿¡£
¢Ù×°ÖÃAµÄ×÷ÓÃÊÇ               £»ÈôÎÞ×°ÖÃE£¬ÔòʵÑé²â¶¨µÄx/yµÄÖµ½«        ¡£(Ñ¡Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)¡£
¢ÚijͬѧÔÚʵÑé¹ý³ÌÖвɼ¯ÁËÈçÏÂÊý¾Ý£º
A£®·´Ó¦Ç°²£Á§¹ÜÓëÑùÆ·µÄÖÊÁ¿163.8g      B£®·´Ó¦ºó²£Á§¹ÜÖвÐÁô¹ÌÌåÖÊÁ¿56.0g
C£®×°ÖÃCʵÑéºóÔöÖØ9.0g                D£®×°ÖÃDʵÑéºóÔöÖØ8.8g
Ϊ²â¶¨x/yµÄÖµ£¬ÄãÈÏΪ¿ÉÒÔÑ¡ÓÃÉÏÊöËù²É¼¯Êý¾ÝÖеĠ          (д³öËùÓÐ×éºÏµÄ×Öĸ´úºÅ)ÈÎÒ»×é¼´¿É½øÐмÆË㣬²¢¸ù¾ÝÄãµÄ¼ÆËã½á¹û£¬Ð´³ö¸ÃÑùÆ·×é³ÉµÄ»¯Ñ§Ê½                        ¡£
£¨Ò»£©(4·Ö)À¨ºÅÖеÄÎïÖÊÊÇÔÓÖÊ£¬Ð´³ö³ýÈ¥ÕâЩÔÓÖʵÄÊÔ¼Á£º
£¨1£©MgO (Al2O3)                £¨2£©Cl2(HCl)           
£¨3£©FeCl3(FeCl2)               £¨4£©NaHCO3ÈÜÒº(Na2CO3)          
£¨¶þ£©(6·Ö)º£Ë®Öк¬ÓдóÁ¿µÄÂÈ»¯Ã¾£¬´Óº£Ë®ÖÐÌáȡþµÄÉú²úÁ÷³ÌÈçÏÂͼËùʾ£º

»Ø´ðÏÂÁÐÎÊÌ⣺
д³öÔÚº£Ë®ÖмÓÈëÑõ»¯¸ÆÉú³ÉÇâÑõ»¯Ã¾µÄ»¯Ñ§·½³Ìʽ                    £»
²Ù×÷¢ÙÖ÷ÒªÊÇÖ¸       £»ÊÔ¼Á¢Ù¿ÉÑ¡Óà        £»
²Ù×÷¢ÚÊÇÖ¸          £»¾­²Ù×÷¢Û×îÖտɵýðÊôþ¡£
£¨Èý£©£¨8·Ö£©ÊµÑéÊÒÅäÖÆ480ml 0£®1mol¡¤L-1µÄNa2CO3ÈÜÒº£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ó¦ÓÃÍÐÅÌÌìƽ³Æȡʮˮ̼ËáÄƾ§Ìå        g¡£
£¨2£©ÈçͼËùʾµÄÒÇÆ÷ÅäÖÆÈÜÒº¿Ï¶¨²»ÐèÒªµÄÊÇ      (ÌîÐòºÅ)£¬±¾ÊµÑéËùÐè²£Á§ÒÇÆ÷E¹æ¸ñΪ       mL¡£

£¨3£©ÈÝÁ¿Æ¿ÉϱêÓУº¢Ùζȡ¢¢ÚŨ¶È¡¢¢ÛÈÝÁ¿¡¢¢Üѹǿ¡¢¢Ý¿Ì¶ÈÏß¡¢¢ÞËáʽ»ò¼îʽÕâÁùÏîÖеĠ       ¡££¨ÌîÊý×Ö·ûºÅ£©
£¨4£©ÅäÖÆËùÐèµÄÖ÷ÒªÒÇÆ÷ÊÇ£ºaÈÝÁ¿Æ¿¡¢bÉÕ±­¡¢c½ºÍ·µÎ¹Ü¡¢dÍÐÅÌÌìƽ£¬ËüÃÇÔÚ²Ù×÷¹ý³ÌÖÐʹÓõÄÇ°ºó˳ÐòÊÇ       ¡££¨Ìîд×Öĸ£¬Ã¿ÖÖÒÇÆ÷Ö»ÄÜÑ¡ÔñÒ»´Î£©
£¨5£©²£Á§°ôÊÇ»¯Ñ§ÊµÑéÖг£ÓõÄÒ»ÖÖ²£Á§¹¤¾ß£¬ÔòÔÚÅäÖÆÈÜÒºµÄ¹ý³ÌÖв£Á§°ô¹²Æðµ½ÁË      ÖÖÓÃ;¡££¨ÌîдÊý×Ö£©
£¨6£©ÈôʵÑéʱÓöµ½ÏÂÁÐÇé¿ö£¬½«Ê¹ÈÜÒºµÄŨ¶ÈÆ«µÍµÄÊÇ       ¡£
A£®ÅäÖÆǰûÓн«ÈÝÁ¿Æ¿ÖеÄË®³ý¾¡£»
B£®Ì¼ËáÄÆʧȥÁ˲¿·Ö½á¾§Ë®£»
C£®Ì¼ËáÄƾ§Ìå²»´¿£¬ÆäÖлìÓÐÂÈ»¯ÄÆ£»
D£®³ÆÁ¿Ì¼ËáÄƾ§ÌåʱËùÓÃíÀÂëÉúÐ⣻
E£®¶¨ÈÝʱÑöÊӿ̶ÈÏß
ijʵÑéС×éΪÑо¿²ÝËáµÄÖÆÈ¡ºÍ²ÝËáµÄÐÔÖÊ£¬½øÐÐÈçÏÂʵÑé¡£
ʵÑéI:ÖƱ¸²ÝËá
ʵÑéÊÒÓÃÏõËáÑõ»¯µí·ÛË®½âÒºÖƱ¸²ÝËáµÄ×°ÖÃÈçͼËùʾ(¼ÓÈÈ¡¢½Á°èºÍÒÇÆ÷¹Ì¶¨×°ÖþùÒÑÂÔÈ¥£©£¬ÊµÑé¹ý³ÌÈçÏ£º

¢Ù½«Ò»¶¨Á¿µÄµí·ÛË®½âÒº¼ÓÈËÈý¾±Æ¿ÖÐ
¢Ú¿ØÖÆ·´Ó¦ÒºÎ¶ÈÔÚ55?600CÌõ¼þÏ£¬±ß½Á°è±ß»ºÂýµÎ¼ÓÒ»¶¨Á¿º¬ÓÐÊÊÁ¿´ß»¯¼ÁµÄ»ìËá (65%HNO3Óë98%H2S04µÄÖÊÁ¿±ÈΪ2 :1.5)ÈÜÒº
¢Û·´Ó¦3h×óÓÒ£¬ÀäÈ´£¬³éÂ˺óÔÙÖؽᾧµÃ²ÝËᾧÌå¡£
ÏõËáÑõ»¯µí·ÛË®½âÒº¹ý³ÌÖпɷ¢ÉúÏÂÁз´Ó¦£º
C6H12O6£«12HNO3¡ú3H2C2O4£«9NO2¡ü£«3NO¡ü£«9H2O
C6H12O6£«8HNO3¡ú6CO2¡ü£«8NO¡ü£«10H2O
3 H2C2O4£«2HNO3¡ú6CO2¡ü£«2NO¡ü£«4H2O
(1)¼ìÑéµí·ÛÊÇ·ñË®½âÍêÈ«ËùÐèÓõÄÊÔ¼ÁΪ________¡£
(2)ʵÑéÖÐÈô»ìËáµÎ¼Ó¹ý¿ì£¬½«µ¼Ö²ÝËá²úÂÊϽµ£¬ÆäÔ­ÒòÊÇ________¡£
ʵÑéII£º²ÝËᾧÌåÖнᾧˮ²â¶¨
²ÝËᾧÌåµÄ»¯Ñ§Ê½¿É±íʾΪH2C2O4? xH2O,Ϊ²â¶¨xµÄÖµ,½øÐÐÏÂÁÐʵÑ飺
¢Ù³ÆÈ¡6.3gij²ÝËᾧÌåÅä³É100.0mLµÄË®ÈÜÒº¡£
¢ÚÈ¡25.00mLËùÅäÈÜÒºÖÃÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈëÊÊÁ¿Ï¡H2SO4£¬ÓÃŨ¶ÈΪ0. 5mol/LµÄKMnO4ÈÜÒºµÎ¶¨£¬µÎ¶¨ÖÕµãʱÏûºÄKMnO4µÄÌå»ýΪ10.00mL¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨3£©Ð´³öÉÏÊö·´Ó¦µÄÀë×Ó·½³Ìʽ________________¡£
£¨4£©¼ÆËãx=________¡£
£¨5£©µÎ¶¨Ê±£¬³£·¢ÏÖ·´Ó¦ËÙÂÊ¿ªÊ¼ºÜÂý£¬ºóÀ´Ö𽥼ӿ죬¿ÉÄܵÄÔ­ÒòÊÇ________¡£
ʵÑéIII:²ÝËá²»Îȶ¨ÐÔ
²éÔÄ×ÊÁÏ£º²ÝËᾧÌå(H2C2O4 ?xH20),1000C¿ªÊ¼Ê§Ë®£¬100.5¡æ×óÓÒ·Ö½â²úÉúH2O¡¢COºÍCO2¡£ÇëÓÃÏÂͼÖÐÌṩµÄÒÇÆ÷¼°ÊÔ¼Á£¬Éè¼ÆÒ»¸öʵÑ飬֤Ã÷²ÝËᾧÌå·Ö½âµÃµ½µÄ»ìºÏÆøÖÐÓÐH2O¡¢COºÍCO2 (¼ÓÈÈ×°Öú͵¼¹ÜµÈÔÚͼÖÐÂÔÈ¥£¬²¿·Ö×°ÖÿÉÖظ´Ê¹Óã©¡£

»Ø´ðÏÂÁÐÎÊÌ⣺
(6)ÒÇÆ÷×°Öð´Á¬½Ó˳ÐòΪ________¡£
(7)ÒÇÆ÷BÖÐÎÞË®ÁòËáÍ­µÄ×÷ÓÃ________¡£
£¨8£©ÄÜÖ¤Ã÷»ìºÏÆøÖк¬ÓÐCOµÄʵÑéÒÀ¾ÝÊÇ________¡£

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

¾«Ó¢¼Ò½ÌÍø