ÌâÄ¿ÄÚÈÝ

£¨16·Ö£©ÒÑÖªFeSO4ÔÚ²»Í¬Ìõ¼þÏ·ֽâµÃµ½µÄ²úÎﲻͬ£¬¿ÉÄÜÊÇFeOºÍSO3£¬Ò²¿ÉÄÜÊÇFe2O3¡¢SO3ºÍSO2¡£Ä³Ñо¿Ð¡×é̽¾¿Ôھƾ«ÅçµÆ¼ÓÈÈÌõ¼þÏÂFeSO4·Ö½âµÄÆøÌå²úÎï¡£ÒÑÖªSO3µÄÈÛµãÊÇ16.8¡æ£¬·ÐµãÊÇ44.8¡æ¡£

£¨1£©×°ÖâòµÄÊÔ¹ÜÖв»×°ÈκÎÊÔ¼Á£¬Æä×÷ÓÃÊÇ______________________                       
ÊԹܽþÅÝÔÚ50¡æµÄˮԡÖУ¬Ä¿µÄÊÇ______________________________                         
£¨2£©×°ÖâóºÍ×°ÖâôµÄ×÷ÓÃÊÇ̽¾¿±¾ÊµÑ鯸Ìå²úÎï³É·Ö¡£ÇëÍê³ÉʵÑéÉè¼Æ£¬Ìîд¼ìÑéÊÔ¼Á¡¢Ô¤ÆÚÏÖÏóÓë½áÂÛ¡£
ÏÞÑ¡ÊÔ¼Á£º3 mol¡¤L£­1 H2SO4¡¢6 mol¡¤L£­1 NaOH¡¢0.5 mol¡¤L£­1 BaCl2¡¢0.5 mol¡¤L£­1 Ba(NO3)2¡¢
0.01 mol¡¤L£­1ËáÐÔKMnO4ÈÜÒº¡¢0.01 mol¡¤L£­1äåË®¡£

¼ìÑéÊÔ¼Á
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
×°ÖâóµÄÊÔ¹ÜÖмÓÈë__________       ___¡£
²úÉú´óÁ¿°×É«³Áµí£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO3¡£
×°ÖâôµÄÊÔ¹ÜÖмÓÈë_______     _________¡£
______________________________        
______________________________       
______________________________       
______________________________      
£¨3£©×°ÖâõµÄ×÷ÓÃÊÇ·ÀÖ¹Î²ÆøÎÛȾ»·¾³£¬ÉÕ±­ÖÐÓ¦¼ÓÈëµÄÊÔ¼ÁÊÇ              ¡£

£¨16·Ö£©
£¨1£©·ÀÖ¹ÈÜÒºµ¹ÎüÈë×°Öâñ£¨2·Ö£©  ·ÀÖ¹SO3Òº»¯»òÄý¹Ì£¨2·Ö£©
£¨2£©£¨10·Ö£©

¼ìÑéÊÔ¼Á
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
×°ÖâóµÄÊÔ¹ÜÖÐ×°ÓÐBaCl2ÈÜÒº¡££¨3·Ö£©
²úÉú´óÁ¿°×É«³Áµí£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO3¡£
×°ÖâôµÄÊÔ¹ÜÖÐ×°ÓÐËáÐÔKMnO4ÈÜÒº¡££¨3·Ö£©
ÈôÈÜÒº×ϺìÉ«ÍÊÈ¥£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO2£»£¨2·Ö£©
ÈôÈÜÒº×ϺìÉ«ÎÞÃ÷ÏԱ仯£¬Ö¤Ã÷ÆøÌå²úÎïÖв»º¬SO2£»£¨2·Ö£©
»ò£º
¼ìÑéÊÔ¼Á
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
×°ÖâóµÄÊÔ¹ÜÖÐ×°ÓÐBaCl2ÈÜÒº¡££¨3·Ö£©
²úÉú´óÁ¿°×É«³Áµí£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO3¡£
×°ÖâôµÄÊÔ¹ÜÖÐ×°ÓÐäåË®¡££¨3·Ö£©
ÈôäåË®³ÈÉ«ÍÊÈ¥£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO2£»£¨2·Ö£©
ÈôäåË®³ÈÉ«ÎÞÃ÷ÏԱ仯£¬Ö¤Ã÷ÆøÌå²úÎïÖв»º¬SO2£»£¨2·Ö£©
£¨3£©NaOHÈÜÒº£¨2·Ö£©

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÁòËáÑÇÌúÊDZȽÏÖØÒªµÄÑÇÌúÑΣ¬ÔÚũҵÉÏÓÃ×÷ũҩ£¬Ö÷ÖÎСÂóºÚË벡£¬ÔÚ¹¤ÒµÉÏÓÃÓÚȾɫ¡¢ÖÆÔìÀ¶ºÚīˮºÍľ²Ä·À¸¯¡¢³ý²Ý¼ÁµÈ£®
£¨1£©ÊµÑéÊÒÌõ¼þÏÂÓÃÌú·ÛºÍÏ¡ÁòËáÖÆÈ¡ÉÙÁ¿ÁòËáÑÇÌúÈÜÒº£¬Îª·ÀÖ¹Æä±äÖÊ£¬Ó¦ÔÚ¸ÃÈÜÒºÖÐÓ¦¼ÓÈëÊÊÁ¿
Ìú£¬ÁòËá
Ìú£¬ÁòËá
£®
£¨2£©ÐÂÖÆµÄÂÌ·¯¾§Ì壨FeSO4?7H2O£©ÊÇdzÂÌÉ«µÄ£¬µ«ÔÚ¿ÕÆøÖм«Ò×±ä³É»ÆÉ«»òÌúÐâÉ«µÄ¼îʽÁòËáÌú[Fe£¨OH£©SO4]£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
4FeSO4?7H2O+O2=4Fe£¨OH£©SO4+5H2O
4FeSO4?7H2O+O2=4Fe£¨OH£©SO4+5H2O
£®
£¨3£©ÒÑÖªFeSO4ÔÚ²»Í¬Ìõ¼þÏ·ֽâµÃµ½µÄ²úÎﲻͬ£¬¿ÉÄÜÊÇFeOºÍSO3£¬Ò²¿ÉÄÜÊÇFe2O3¡¢SO3ºÍSO2£®Ä³Ñо¿Ð¡×é̽¾¿Ôھƾ«ÅçµÆ¼ÓÈÈÌõ¼þÏÂFeSO4·Ö½âµÄÆøÌå²úÎÒÑÖªSO3µÄÈÛµãÊÇ16.8¡ãC£¬·ÐµãÊÇ44.8¡ãC£®
¢Ù×°ÖÃIIµÄÊÔ¹ÜÖв»×°ÈκÎÊÔ¼Á£¬Æä×÷ÓÃÊÇ
·ÀÖ¹ÈÜÒºµ¹ÎüÈë×°ÖâñÖУ¨»ò°²È«Æ¿£©
·ÀÖ¹ÈÜÒºµ¹ÎüÈë×°ÖâñÖУ¨»ò°²È«Æ¿£©
£¬ÊԹܽþÅÝÔÚ50¡æµÄÈÈˮԡÖУ¬Ä¿µÄÊÇ
·ÀÖ¹SO3Òº»¯»òÄý¹Ì
·ÀÖ¹SO3Òº»¯»òÄý¹Ì
£®
¢Ú×°ÖÃIIIºÍ×°ÖÃIVµÄ×÷ÓÃÊÇ̽¾¿±¾ÊµÑ鯸Ìå²úÎï³É·Ö£®ÇëÍê³ÉʵÑéÉè¼Æ£¬Ìîд¼ìÑéÊÔ¼Á¡¢Ô¤ÆÚÏÖÏóÓë½áÂÛ£®
ÏÞÑ¡ÊÔ¼Á£º3mol£®L-1£®H2S04¡¢6mol£®L-l NaOH¡¢0.5mol£®L -1 BaCl2¡¢O.5mol£®L-1Ba£¨NO3£©2¡¢0.01mol£®L-1 ËáÐÔ KMnO4 ÈÜÒº¡¢0.0l mol£®L-1 äåË®£®

¼ìÑéÊÔ¼Á Ô¤ÆÚÏÖÏóÓë½áÂÛ
×°ÖÃIIIµÄÊÔ¹ÜÖмÓÈë×ãÁ¿
0.5mol£®L-1BaCl2
0.5mol£®L-1BaCl2
£®
²úÉú´óÁ¿°×É«³Áµí£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO3£®
×°ÖÃIVµÄÊÔ¹ÜÖмÓÈë×ãÁ¿
0.01mol£®L-1ËáÐÔKMnO4ÈÜÒº£¨»ò0.0lmol£®L-1äåË®£©
0.01mol£®L-1ËáÐÔKMnO4ÈÜÒº£¨»ò0.0lmol£®L-1äåË®£©
£®
ÈôÈÜÒº×ÏÉ«£¨»ò³ÈÉ«£©ÍÊÈ¥£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO2£¬
ÈôÈÜÒº×ÏÉ«£¨»ò³ÈÉ«£©ÍÊÈ¥£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO2£¬

ÈôÈÜÒº×ÏÉ«£¨»ò³ÈÉ«£©ÎÞÃ÷ÏԱ仯£¬Ö¤Ã÷ÆøÌå²úÎïÖв»º¬SO2
ÈôÈÜÒº×ÏÉ«£¨»ò³ÈÉ«£©ÎÞÃ÷ÏԱ仯£¬Ö¤Ã÷ÆøÌå²úÎïÖв»º¬SO2
£®
¢Û×°ÖÃVµÄ×÷ÓÃÊÇ·ÀÖ¹Î²ÆøÎÛȾ»·¾³£¬ÉÕ±­ÖÐÓ¦¼ÓÈëµÄÊÔ¼ÁÊÇ
NaOHÈÜÒº
NaOHÈÜÒº
£®
¢ÜÈçºÎ¼ìÑé×°ÖâñÖйÌÌåÍêÈ«·Ö½âºóÉú³É¹ÌÌå²úÎïÓÐÎÞFeO£¿Ð´³ö²½Öè¡¢ÏÖÏó¼°½áÂÛ£º
È¡ÉÙÁ¿·Ö½âºóÊ£Óà¹ÌÌåÓÚÊÔ¹ÜÖУ¬ÍùÊÔ¹ÜÖмÓÈë×ãÁ¿Ï¡ÑÎËᣬ³ä·Ö·´Ó¦ºó£¬µÎÈëËáÐÔKMnO4ÈÜÒº£¬ÈôËáÐÔKMnO4ÈÜÒºÍÊÉ«£¬ËµÃ÷Éú³ÉµÄ¹ÌÌå²úÎﺬÓÐFeO
È¡ÉÙÁ¿·Ö½âºóÊ£Óà¹ÌÌåÓÚÊÔ¹ÜÖУ¬ÍùÊÔ¹ÜÖмÓÈë×ãÁ¿Ï¡ÑÎËᣬ³ä·Ö·´Ó¦ºó£¬µÎÈëËáÐÔKMnO4ÈÜÒº£¬ÈôËáÐÔKMnO4ÈÜÒºÍÊÉ«£¬ËµÃ÷Éú³ÉµÄ¹ÌÌå²úÎﺬÓÐFeO
£®
ÒÑÖªFeSO4ÔÚ²»Í¬Ìõ¼þÏ·ֽâµÃµ½µÄ²úÎﲻͬ£¬¿ÉÄÜÊÇFeOºÍSO3£¬Ò²¿ÉÄÜÊÇFe2O3¡¢SO3ºÍSO2£®Ä³Ñо¿Ð¡×é̽¾¿Ôھƾ«ÅçµÆ¼ÓÈÈÌõ¼þÏÂFeSO4·Ö½âµÄÆøÌå²úÎÒÑÖªSO3µÄÈÛµãÊÇ16.8¡æ£¬·ÐµãÊÇ44.8¡æ£®

£¨1£©×°ÖâòµÄÊÔ¹ÜÖв»×°ÈκÎÊÔ¼Á£¬Æä×÷ÓÃÊÇ
·ÀÖ¹ÈÜÒºµ¹ÎüÈë×°ÖâñÖУ¨»ò°²È«Æ¿£©
·ÀÖ¹ÈÜÒºµ¹ÎüÈë×°ÖâñÖУ¨»ò°²È«Æ¿£©
£¬ÊԹܽþÅÝÔÚ50¡æµÄˮԡÖУ¬Ä¿µÄÊÇ
·ÀÖ¹SO3Òº»¯»òÄý¹Ì
·ÀÖ¹SO3Òº»¯»òÄý¹Ì
£®
£¨2£©×°ÖâóºÍ×°ÖâôµÄ×÷ÓÃÊÇ̽¾¿±¾ÊµÑ鯸Ìå²úÎï³É·Ö£®ÇëÍê³ÉʵÑéÉè¼Æ£¬Ìîд¼ìÑéÊÔ¼Á¡¢Ô¤ÆÚÏÖÏóÓë½áÂÛ£®ÏÞÑ¡ÊÔ¼Á£º3mol?L-1 H2SO4¡¢6mol?L-1 NaOH¡¢0.5mol?L-1 BaCl2¡¢0.5mol?L-1 Ba£¨NO3£©2¡¢0.01mol?L-1ËáÐÔKMnO4ÈÜÒº¡¢0.01mol?L-1äåË®£®
¼ìÑéÊÔ¼Á Ô¤ÆÚÏÖÏóºÍ½áÂÛ
×°ÖâóµÄÊÔ¹ÜÖмÓÈë
0.5mol£®L-1BaCl2
0.5mol£®L-1BaCl2
£®
²úÉú´óÁ¿°×É«³Áµí£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO3£®
×°ÖâôµÄÊÔ¹ÜÖмÓÈë
0.01mol£®L-1ËáÐÔKMnO4ÈÜÒº£¨»ò0.0lmol£®L-1äåË®£©
0.01mol£®L-1ËáÐÔKMnO4ÈÜÒº£¨»ò0.0lmol£®L-1äåË®£©
£®
ÈôÈÜÒº×ÏÉ«£¨»ò³ÈÉ«£©ÍÊÈ¥£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO2£¬ÈôÈÜÒº×ÏÉ«£¨»ò³ÈÉ«£©ÎÞÃ÷ÏԱ仯£¬Ö¤Ã÷ÆøÌå²úÎïÖв»º¬SO2
ÈôÈÜÒº×ÏÉ«£¨»ò³ÈÉ«£©ÍÊÈ¥£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO2£¬ÈôÈÜÒº×ÏÉ«£¨»ò³ÈÉ«£©ÎÞÃ÷ÏԱ仯£¬Ö¤Ã÷ÆøÌå²úÎïÖв»º¬SO2


£¨3£©×°ÖâõµÄ×÷ÓÃÊÇ·ÀÖ¹Î²ÆøÎÛȾ»·¾³£¬ÉÕ±­ÖÐÓ¦¼ÓÈëµÄÊÔ¼ÁÊÇ
NaOHÈÜÒº
NaOHÈÜÒº
£®

£¨16·Ö£©ÒÑÖªFeSO4ÔÚ²»Í¬Ìõ¼þÏ·ֽâµÃµ½µÄ²úÎﲻͬ£¬¿ÉÄÜÊÇFeOºÍSO3£¬Ò²¿ÉÄÜÊÇFe2O3¡¢SO3ºÍSO2¡£Ä³Ñо¿Ð¡×é̽¾¿Ôھƾ«ÅçµÆ¼ÓÈÈÌõ¼þÏÂFeSO4·Ö½âµÄÆøÌå²úÎï¡£ÒÑÖªSO3µÄÈÛµãÊÇ16.8¡æ£¬·ÐµãÊÇ44.8¡æ¡£

£¨1£©×°ÖâòµÄÊÔ¹ÜÖв»×°ÈκÎÊÔ¼Á£¬Æä×÷ÓÃÊÇ______________________                       

ÊԹܽþÅÝÔÚ50¡æµÄˮԡÖУ¬Ä¿µÄÊÇ______________________________                         

£¨2£©×°ÖâóºÍ×°ÖâôµÄ×÷ÓÃÊÇ̽¾¿±¾ÊµÑ鯸Ìå²úÎï³É·Ö¡£ÇëÍê³ÉʵÑéÉè¼Æ£¬Ìîд¼ìÑéÊÔ¼Á¡¢Ô¤ÆÚÏÖÏóÓë½áÂÛ¡£

ÏÞÑ¡ÊÔ¼Á£º3 mol¡¤L£­1 H2SO4¡¢6 mol¡¤L£­1 NaOH¡¢0.5 mol¡¤L£­1 BaCl2¡¢0.5 mol¡¤L£­1 Ba(NO3)2¡¢

0.01 mol¡¤L£­1ËáÐÔKMnO4ÈÜÒº¡¢0.01 mol¡¤L£­1äåË®¡£

¼ìÑéÊÔ¼Á

Ô¤ÆÚÏÖÏóºÍ½áÂÛ

×°ÖâóµÄÊÔ¹ÜÖмÓÈë__________        ___¡£

²úÉú´óÁ¿°×É«³Áµí£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO3¡£

×°ÖâôµÄÊÔ¹ÜÖмÓÈë_______      _________¡£

______________________________        

______________________________       

______________________________       

______________________________      

£¨3£©×°ÖâõµÄ×÷ÓÃÊÇ·ÀÖ¹Î²ÆøÎÛȾ»·¾³£¬ÉÕ±­ÖÐÓ¦¼ÓÈëµÄÊÔ¼ÁÊÇ              ¡£

 

£¨9·Ö£©¡¢ÒÑÖªFeSO4ÔÚ²»Í¬Ìõ¼þÏ·ֽâµÃµ½µÄ²úÎﲻͬ£¬¿ÉÄÜÊÇFeOºÍSO3£¬Ò²¿ÉÄÜÊÇFe2O3¡¢SO3ºÍSO2¡£Ä³Ñо¿Ð¡×é̽¾¿Ôھƾ«ÅçµÆ¼ÓÈÈÌõ¼þÏÂFeSO4·Ö½âµÄÆøÌå²úÎï¡£ÒÑÖªSO3µÄÈÛµãÊÇ16.8¡æ£¬·ÐµãÊÇ44.8¡æ

¢Å×°ÖâòµÄÊÔ¹ÜÖв»×°ÈκÎÊÔ¼Á£¬Æä×÷ÓÃÊÇ                      ¡£ÊԹܽþÅÝÔÚ50¡æµÄˮԡÖУ¬Ä¿µÄÊÇ                                       ¡£

¼ìÑéÊÔ¼Á
Ô¤ÆÚÏÖÏóÓë½áÂÛ
×°ÖâóµÄÊÔ¹ÜÖмÓÈë                 ¡£
²úÉú´óÁ¿°×É«³Áµí£¬Ö¤Ã÷ÆøÌå²úÎïÖк¬ÓÐSO3
×°ÖâôµÄÊÔ¹ÜÖмÓÈë                  ¡£
                                           
                                            
¢Æ×°ÖâóºÍ×°ÖâôµÄ×÷ÓÃÊÇ̽¾¿±¾ÊµÑ鯸Ìå²úÎï³É·Ö¡£ÇëÍê³ÉʵÑéÉè¼Æ£¬Ìîд¼ìÑéÊÔ¼Á¡¢Ô¤ÆÚÏÖÏóÓë½áÂÛ¡£ÏÞÑ¡ÊÔ¼Á£º3mol¡¤L£­1H2SO4¡¢6 mol¡¤L£­1NaOH¡¢0.5 mol¡¤L£­1BaCl2¡¢0.5 mol¡¤L£­1Ba(NO3)2¡¢0.01 mol¡¤L£­1ËáÐÔKMnO4ÈÜÒº¡¢0.01 mol¡¤L£­1äåË®
¢Ç×°ÖâõµÄ×÷ÓÃÊÇ·ÀÖ¹Î²ÆøÎÛȾ»·¾³£¬ÉÕ±­ÖÐÓ¦¼ÓÈëµÄÊÔ¼ÁÊÇ           ¡£

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

¾«Ó¢¼Ò½ÌÍø