ÌâÄ¿ÄÚÈÝ

ÖظõËá¼ØÊǹ¤ÒµÉú²úºÍʵÑéÊÒµÄÖØÒªÑõ»¯¼Á£¬¹¤ÒµÉϳ£ÓøõÌú¿ó£¨Ö÷Òª³É·ÖΪFeO¡¤Cr2O3£¬ÔÓÖÊΪSiO2¡¢Al2O3£©ÎªÔ­ÁϲúËü£¬ÊµÑéÊÒÄ£Ä⹤ҵ·¨ÓøõÌú¿óÖÆK2Cr2O7µÄÖ÷Òª¹¤ÒÕÈçÏÂͼ¡£Éæ¼°µÄÖ÷Òª·´Ó¦ÊÇ6FeO¡¤Cr2O3+24NaOH+7KClO312Na2CrO4+3Fe2O3+7KCl+12H2O¡£

£¨1£©¼î½þÇ°½«ÃúÌú¿ó·ÛËéµÄ×÷ÓÃÊÇ                              
£¨2£©²½Öè¢Ûµ÷½ÚpHºó¹ýÂ˵õ½µÄÂËÔüÊÇ                                   ¡£
£¨3£©²Ù×÷¢ÜÖУ¬Ëữʱ£¬CrO2- 4ת»¯ÎªCr2O2- 7£¬Ð´³öƽºâת»¯µÄÀë×Ó·½³Ìʽ
                                           £»
£¨4£©ÓüòÒªµÄÎÄ×Ö˵Ã÷²Ù×÷¢Ý¼ÓÈëKClµÄÔ­Òò                            ¡£
£¨5£©³ÆÈ¡ÖظõËá¼ØÊÔÑù2.500gÅä³É250mLÈÜÒº£¬È¡³ö25mLÓëµâÁ¿Æ¿ÖУ¬¼ÓÈë10mL2mol/ LH2SO4ºÍ×ãÁ¿µâ»¯¼Ø£¨¸õµÄ»¹Ô­²úÎïΪCr3+)£¬·ÅÓÚ°µ´¦5min¡£È»ºó¼ÓÈë100mLË®£¬¼ÓÈë3mLµí·Ûָʾ¼Á£¬ÓÃ0.1200mol/LNa2S2O3±ê×¼ÈÜÒºµÎ¶¨£¨I2+2S2O2- 3£½2I- +S4O2- 6£©
¢ÙÅжϴﵽµÎ¶¨ÖÕµãµÄÒÀ¾ÝÊÇ                                           £»
¢ÚÈôʵÑéÖй²ÓÃÈ¥Na2S2O3±ê×¼ÈÜÒº40.00mL£¬ÔòËùµÃ²úÆ·ÖÐÖظõËá¼ØµÄ´¿¶ÈΪ£¨ÉèÕû¸ö¹ý³ÌÖÐÆäËüÔÓÖʲ»²Î¼Ó·´Ó¦£©                              £¨±£Áô2λÓÐЧÊý×Ö£©¡£

(14·Ö)
£¨1£©Ôö´ó½Ó´¥Ãæ»ý£¬Ôö´ó·´Ó¦ËÙÂÊ£¨2·Ö£©
£¨2£©Al(OH)3¡¢H2SiO3£¨2·Ö£©
£¨3£©2 CrO2- 4+2H+ Cr2O2- 7+H2O£¨2·Ö£©
£¨4£©Î¶ȶÔÂÈ»¯ÄƵÄÈܽâ¶ÈÓ°ÏìС£¬µ«¶ÔÖظõËá¼ØµÄÈܽâ¶ÈÓ°Ïì½Ï´ó£¬ÀûÓø´·Ö½â·´Ó¦£¬¿ÉµÃµ½ÖظõËá¼Ø£¨2·Ö£©
£¨5£©¢Ùµ±µÎ¼Ó×îºóÒ»µÎÁò´úÁòËáÄÆÈÜÒº£¬ÈÜÒºÀ¶É«ÍÊÉ«£¨2·Ö£©¢Ú94.08 %£¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º
£¨1£©½«ÃúÌú¿ó·ÛËéµÄ×÷ÓÃÊÇÔö´ó½Ó´¥Ãæ»ý£¬Ôö´ó·´Ó¦ËÙÂÊ£»
£¨2£©¸ù¾Ý¹¤ÒÕÁ÷³ÌͼÖпÉÒÔ¿´³öÔÚ·´Ó¦Æ÷ÖÐSiO2¡¢Al2O3ÓëNaOH·´Ó¦·Ö±ðת»¯ÎªNa2SiO3ºÍNaAlO2, ²½Öè¢Ûµ÷½ÚpHºó¹ýÂ˵õ½µÄÂËÔüÊÇAl(OH)3¡¢H2SiO3£»
£¨3£©Ëữʱ£¬CrO2- 4ת»¯ÎªCr2O2- 7ƽºâת»¯µÄÀë×Ó·½³ÌʽΪ2CrO2- 4+2H+Cr2O2- 7+H2O£»
£¨4£©½áºÏÌâÖÐÐÅϢζȶÔÂÈ»¯ÄƵÄÈܽâ¶ÈÓ°ÏìС£¬µ«¶ÔÖظõËá¼ØµÄÈܽâ¶ÈÓ°Ïì½Ï´ó£¬ÀûÓø´·Ö½â·´Ó¦£¬½µÎ½ᾧʱµÃµ½µÄÊÇÖظõËá¼Ø¡£
£¨5£©¼ì²âµâµ¥ÖÊ´æÔÚ¿ÉÓõí·Û£¬Ö±µ½À¶É«ÍÊÈ¥£»·ÖÎö¿ÉµÃ¹Øϵʽ£ºCr2O2- 7~3I2~6S2O2- 3£¬n(S2O2- 3)=0.04L¡Á0.12mol/L£¬¶ø25mlÖÐ1/6n(S2O2- 3)= n(Cr2O2- 7)=0.0008mol£¬Ôò250mlÖк¬ÓеÄm(K2Cr2O7)=0.0008mol¡Á10¡Á294g/mol=2.352g£¬¿ÉÖªÖÊÁ¿·ÖÊýΪ2.352g/2.500g¡Á100%=94.08%¡£
¿¼µã£º±¾ÌâÒÔ¹¤ÒÕÁ÷³ÌΪ»ù´¡£¬¿¼²éÔªËؼ°»¯ºÏÎï¡¢»¯Ñ§ÊµÑé»ù±¾²Ù×÷¡¢»¯Ñ§¼ÆËãµÈÏà¹Ø֪ʶ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÖظõËá¼ØÊǹ¤ÒµÉú²úºÍʵÑéÊÒµÄÖØÒªÑõ»¯¼Á£¬¹¤ÒµÉϳ£ÓøõÌú¿ó£¨Ö÷Òª³É·ÝΪFeO ? Cr2O3)ΪԭÁÏÉú²ú£¬ÊµÑéÊÒÄ£Ä⹤ҵ·¨ÓøõÌú¿óÖÆK2Cr2O7µÄÖ÷Òª¹¤ÒÕÈçÏ£¬Éæ¼°µÄÖ÷Òª·´Ó¦ÊÇ£º
6FeO¡¤Cr2O3+24NaOH+7KClO312Na2CrO4+3Fe2O3+7KCl+12H2O

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÔÚ·´Ó¦Æ÷¢ÙÖУ¬ÓÐNa2CrO4Éú³É£¬Í¬Ê±Fe2O3ת±äΪNaFeO2£¬ÔÓÖÊSiO2¡¢Ai2O3Óë´¿¼î·´Ó¦×ª±äΪ¿ÉÈÜÐÔÑΣ¬Ð´³öÑõ»¯ÂÁÓë̼ËáÄÆ·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________¡£
£¨2) NaFeO2ÄÜÇ¿ÁÒË®½â£¬ÔÚ²Ù×÷¢ÚÉú³É³Áµí¶ø³ýÈ¥£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_______________¡£
£¨3£©¼òÒªÐðÊö²Ù×÷¢ÛµÄÄ¿µÄ£º________________________¡£
£¨4£©²Ù×÷¢ÜÖУ¬Ëữʱ£¬CrO42-ת»¯ÎªCr2O72-£¬Ð´³öƽºâת»¯µÄÀë×Ó·½³Ìʽ£º___________¡£
£¨5£©³ÆÈ¡ÖظõËá¼ØÊÔÑù2. 5000gÅä³É250mLÈÜÒº£¬È¡³ö25.00mLÓÚµâÁ¿Æ¿ÖУ¬¼ÓÈë10mL2mol/LH2SO4ºÍ×ãÁ¿µâ»¯¼Ø£¨¸õµÄ»¹Ô­²úÎïΪCr3+)£¬·ÅÓÚ°µ´¦5min£¬È»ºó¼ÓÈë100mLË®£¬¼ÓÈë3mLµí·Ûָʾ¼Á£¬ÓÃ0.1200mol/LNa2S2O3±ê×¼ÈÜÒºµÎ¶¨£¨I2+2S2O32£­=2I£­+S4O62£­£©£¬Èô¸Õ´ïµ½µÎ¶¨Öյ㹲ÓÃÈ¥Na2S2O3±ê×¼ÈÜÒº40.00mL£¬ÔòËùµÃ²úÆ·ÖظõËá¼ØµÄ´¿¶È________________ (ÉèÕû¸ö¹ý³ÌÖÐÆäËüÔÓÖʲ»²ÎÓë·´Ó¦)¡£

¹èÔåÍÁÊÇÓɹèÔåËÀÍöºóµÄÒź¡³Á»ýÐγɵģ¬Ö÷Òª³É·ÖÊÇ SiO2ºÍÓлúÖÊ£¬²¢º¬ÓÐÉÙÁ¿µÄAl2O3¡¢Fe2O3¡¢MgO µÈÔÓÖÊ¡£¾«ÖƹèÔåÍÁÒòΪÎü¸½ÐÔÇ¿¡¢»¯Ñ§ÐÔÖÊÎȶ¨µÈÌص㱻¹ã·ºÓ¦Óá£ÏÂͼÊÇÉú²ú¾«ÖƹèÔåÍÁ²¢»ñµÃAl(OH)3µÄ¹¤ÒÕÁ÷³Ì¡£

£¨1£©´Ö¹èÔåÍÁ¸ßÎÂìÑÉÕµÄÄ¿µÄÊÇ     ¡£
£¨2£©·´Ó¦¢óÖÐÉú³ÉAl(OH)3³ÁµíµÄ»¯Ñ§·½³ÌʽÊÇ     £»ÇâÑõ»¯ÂÁ³£ÓÃ×÷×èȼ¼Á£¬ÆäÔ­ÒòÊÇ     ¡£
£¨3£©ÊµÑéÊÒÓÃËá¼îµÎ¶¨·¨²â¶¨¹èÔåÍÁÖй躬Á¿µÄ²½ÖèÈçÏ£º
²½Öè1£º×¼È·³ÆÈ¡ÑùÆ·a g£¬¼ÓÈëÊÊÁ¿KOH¹ÌÌ壬ÔÚ¸ßÎÂϳä·Ö×ÆÉÕ£¬ÀäÈ´£¬¼ÓË®Èܽ⡣
²½Öè2£º½«ËùµÃÈÜÒºÍêȫתÒÆÖÁËÜÁÏÉÕ±­ÖУ¬¼ÓÈëÏõËáÖÁÇ¿ËáÐÔ£¬µÃ¹èËá×ÇÒº¡£
²½Öè3£ºÏò¹èËá×ÇÒºÖмÓÈëNH4FÈÜÒº¡¢±¥ºÍKClÈÜÒº£¬µÃK2SiF6³Áµí£¬ÓÃËÜÁÏ©¶·¹ýÂ˲¢Ï´µÓ¡£
²½Öè4£º½«K2SiF6תÒÆÖÁÁíÒ»ÉÕ±­ÖУ¬¼ÓÈëÒ»¶¨Á¿ÕôÁóË®£¬²ÉÓÃ70 ¡æˮԡ¼ÓÈÈʹÆä³ä·ÖË®½â£¨K2SiF6+3H2O=H2SiO3+4HF+2KF£©¡£
²½Öè5£ºÏòÉÏÊöË®½âÒºÖмÓÈëÊýµÎ·Ó̪£¬³ÃÈÈÓÃŨ¶ÈΪc mol¡¤L£­1 NaOHµÄ±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄNaOH±ê×¼ÈÜÒºVmL¡£
¢Ù²½Öè1ÖиßÎÂ×ÆÉÕʵÑéËùÐèµÄÒÇÆ÷³ýÈý½Ç¼Ü¡¢ÄàÈý½Ç¡¢¾Æ¾«ÅçµÆÍ⻹ÓР    ¡£
a£®Õô·¢Ãó  b£®±íÃæÃó  c£®´ÉÛáÛö   d£®ÌúÛáÛö
¢ÚʵÑéÖÐʹÓÃËÜÁÏÉÕ±­ºÍËÜÁÏ©¶·µÄÔ­ÒòÊÇ     ¡£
¢Û²½Öè3ÖвÉÓñ¥ºÍKClÈÜҺϴµÓ³Áµí£¬ÆäÄ¿µÄÊÇ     ¡£
¢Ü²½Öè4Öеζ¨ÖÕµãµÄÏÖÏóΪ     ¡£
¢ÝÑùÆ·ÖÐSiO2µÄÖÊÁ¿·ÖÊý¿ÉÓù«Ê½¡°¡Á100%¡±½øÐмÆËã¡£ÓÉ´Ë·ÖÎö²½Öè5Öеζ¨·´Ó¦µÄÀë×Ó·½³ÌʽΪ     ¡£

(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                        ¡£

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

¾«Ó¢¼Ò½ÌÍø