ÌâÄ¿ÄÚÈÝ

ijʵÑéС×éͬѧΪÁË̽¾¿Í­ÓëŨÁòËáµÄ·´Ó¦£¬½øÐÐÁËÈçÏÂʵÑ飬ʵÑé×°ÖÃÈçͼËùʾ¡£

ʵÑé²½Ö裺
¢ÙÏÈÁ¬½ÓÈçͼËùʾµÄ×°Ö㬼ì²éºÃÆøÃÜÐÔ£¬ÔÙ¼ÓÈëÊÔ¼Á£»
¢Ú¼ÓÈÈAÊԹܣ¬´ýBÊÔ¹ÜÖÐÆ·ºìÈÜÒºÍËÉ«ºó£¬Ï¨Ãð¾Æ¾«µÆ£»
¢Û½«CuË¿ÏòÉϳ鶯À뿪ҺÃæ¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)AÊÔ¹ÜÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                  ¡£
(2)Äܹ»Ö¤Ã÷Í­ÓëŨÁòËá·´Ó¦Éú³ÉÆøÌåµÄʵÑéÏÖÏóÊÇ                           ¡£
(3)ÔÚÊ¢ÓÐBaCl2ÈÜÒºµÄCÊÔ¹ÜÖУ¬³ýÁ˵¼¹Ü¿ÚÓÐÆøÅÝÍ⣬ÎÞÆäËûÃ÷ÏÔÏÖÏó£¬Èô½«ÆäÖеÄÈÜÒº·Ö³ÉÁ½·Ý£¬·Ö±ðµÎ¼ÓÏÂÁÐÈÜÒº£¬½«²úÉú³ÁµíµÄ»¯Ñ§Ê½ÌîÈë±íÖжÔÓ¦µÄλÖá£
µÎ¼ÓµÄÈÜÒº
ÂÈË®
°±Ë®
³ÁµíµÄ»¯Ñ§Ê½
 
 
 
д³öÆäÖÐSO2±íÏÖ»¹Ô­ÐÔµÄÀë×Ó·´Ó¦·½³Ìʽ£º                                  ¡£
(4)ʵÑéÍê±Ïºó£¬ÏÈϨÃð¾Æ¾«µÆ£¬ÓÉÓÚµ¼¹ÜEµÄ´æÔÚ£¬ÊÔ¹ÜBÖеÄÒºÌå²»»áµ¹ÎüÈëÊÔ¹ÜAÖУ¬ÆäÔ­ÒòÊÇ                                                        ¡£
(5)ʵÑéÍê±Ïºó£¬×°ÖÃÖвÐÁôµÄÆøÌåÓж¾£¬²»ÄÜ´ò¿ªµ¼¹ÜÉϵĽºÈû¡£ÎªÁË·ÀÖ¹¸ÃÆøÌåÅÅÈë¿ÕÆøÖÐÎÛȾ»·¾³£¬²ð³ý×°ÖÃÇ°£¬Ó¦µ±²ÉÈ¡µÄ²Ù×÷ÊÇ                                ¡£
(6)½«SO2ÆøÌåͨÈ뺬ÓÐn mol Na2SµÄÈÜÒºÖУ¬³ä·Ö·´Ó¦ºó£¬ÈÜÒºÖгöÏÖ»ÆÉ«»ë×Ç£¬ÊÔ·ÖÎö¸ÃÈÜÒº×î¶àÄÜÎüÊÕSO2ÆøÌå        mol(²»¿¼ÂÇÈܽâµÄSO2)¡£
(1)Cu£«2H2SO4(Ũ) CuSO4£«SO2¡ü£«2H2O
(2)BÊÔ¹ÜÖÐÆ·ºìÈÜÒºÍËÉ«
(3)BaSO4¡¡BaSO3¡¡SO2£«Cl2£«2H2O=4H£«£«SO42-£«2Cl£­(»òBa2£«£«SO2£«Cl2£«2H2O=BaSO4¡ý£«4H£«£«2Cl£­)
(4)µ±AÊÔ¹ÜÄÚÆøÌåѹǿ¼õСʱ£¬¿ÕÆø´ÓEµ¼¹Ü½øÈëAÊÔ¹ÜÖУ¬Î¬³ÖAÊÔ¹ÜÖÐѹǿƽºâ
(5)´ÓEµ¼¹Ü¿ÚÏòAÊÔ¹ÜÖлºÂýµØ¹ÄÈë×ãÁ¿µÄ¿ÕÆø£¬½«²ÐÁôµÄSO2ÆøÌå¸ÏÈëNaOHÈÜÒºÖУ¬Ê¹Ö®±»ÍêÈ«ÎüÊÕ
(6)2.5n
(1)AÊÔ¹ÜÖз¢ÉúµÄÊÇCuÓëŨÁòËáÉú³ÉSO2µÄ·´Ó¦£ºCu£«2H2SO4(Ũ) CuSO4£«SO2¡ü£«2H2O¡£(2)ÀûÓÃSO2ÄÜʹƷºìÈÜÒºÍËÉ«µÄÐÔÖÊÀ´Ö¤Ã÷CuÓëŨÁòËá·´Ó¦²úÉúÁËSO2ÆøÌå¡£(3)ÂÈË®ÖеÄCl2¾ßÓÐÑõ»¯ÐÔ£¬Äܽ«SO2Ñõ»¯ÎªSO42-£¬SO42-ÓëBa2£«·´Ó¦Éú³ÉBaSO4³Áµí£»°±Ë®¾ßÓмîÐÔ£¬ÎüÊÕSO2Éú³ÉSO32-£¬SO32-ÓëBa2£«·´Ó¦Éú³ÉBaSO3³Áµí¡£(4)µ¼¹ÜEÓë´óÆøÏàͨ£¬ÄÜʹװÖÃÄÚµÄѹǿºã¶¨£¬·ÀÖ¹³öÏÖµ¹ÎüÏÖÏó¡£(5)²ÐÁôµÄÓж¾ÆøÌåΪSO2£¬Ö»ÒªÀûÓÿÕÆø½«SO2»ºÂý¸ÏÈëNaOHÈÜÒºÖУ¬Ê¹SO2±»ÍêÈ«ÎüÊÕ¼´¿É¡£(6)SO2ÓëNa2S·´Ó¦£¬Ê×ÏÈÊÇSO2ÓëH2O·´Ó¦Éú³ÉH2SO3£ºSO2£«H2O??H2SO3£¬H2SO3ÓëNa2S·´Ó¦Éú³ÉH2S£ºNa2S£«H2SO3=H2S¡ü£«Na2SO3£¬SO2ÓëH2S·´Ó¦Éú³ÉS£º2H2S£«SO2=3S¡ý£«2H2O£¬¸Ã¹ý³Ì¿É±íʾΪ3SO2£«2Na2S=3S¡ý£«2Na2SO3£¬n mol Na2SÎüÊÕ1.5n mol SO2£¬Í¬Ê±Éú³Én mol Na2SO3£¬Na2SO3ÎüÊÕSO2Éú³ÉNaHSO3£ºNa2SO3£«SO2£«H2O=2NaHSO3£¬¸Ã²½·´Ó¦ÓÖ¿ÉÎüÊÕn mol SO2£¬¹Ê×î¶àÄÜÎüÊÕ2.5n mol SO2¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
(15·Ö)½¹ÑÇÁòËáÄÆ(Na2S2O5)Êdz£ÓõÄʳƷ¿¹Ñõ»¯¼ÁÖ®Ò»¡£Ä³Ñо¿Ð¡×é½øÐÐÈçÏÂʵÑ飺
ʵÑéÒ»  ½¹ÑÇÁòËáÄƵÄÖÆÈ¡
²ÉÓÃÓÒͼװÖÃ(ʵÑéÇ°Òѳý¾¡×°ÖÃÄڵĿÕÆø)ÖÆÈ¡Na2S2O5¡£×°ÖâòÖÐÓÐNa2S2O5¾§ÌåÎö³ö£¬·¢ÉúµÄ·´Ó¦Îª£ºNa2SO3£«SO2£½Na2S2O5

£¨1£©×°ÖÃIÖвúÉúÆøÌåµÄ»¯Ñ§·½³ÌʽΪ                 ¡£
£¨2£©Òª´Ó×°ÖâòÖлñµÃÒÑÎö³öµÄ¾§Ì壬¿É²ÉÈ¡µÄ·ÖÀë·½·¨ÊÇ                   ¡£
£¨3£©×°ÖâóÓÃÓÚ´¦ÀíβÆø£¬¿ÉÑ¡ÓõÄ×îºÏÀí×°ÖÃ(¼Ð³ÖÒÇÆ÷ÒÑÂÔÈ¥)Ϊ              (ÌîÐòºÅ)¡£

ʵÑé¶þ    ½¹ÑÇÁòËáÄƵÄÐÔÖÊ
Na2S2O5ÈÜÓÚË®¼´Éú³ÉNaHSO3¡£
£¨4£©Ö¤Ã÷NaHSO3ÈÜÒºÖÐHSO3£­     µÄµçÀë³Ì¶È´óÓÚË®½â³Ì¶È£¬¿É²ÉÓõÄʵÑé·½·¨ÊÇ       (ÌîÐòºÅ)¡£
a£®²â¶¨ÈÜÒºµÄpH     b£®¼ÓÈëBa(OH)2ÈÜÒº   c£®¼ÓÈëÑÎËá  
d£®¼ÓÈëÆ·ºìÈÜÒº  e£®ÓÃÀ¶É«Ê¯ÈïÊÔÖ½¼ì²â
£¨5£©¼ìÑéNa2S2O5¾§ÌåÔÚ¿ÕÆøÖÐÒѱ»Ñõ»¯µÄʵÑé·½°¸ÊÇ                         ¡£
ʵÑéÈý ÆÏÌѾÆÖп¹Ñõ»¯¼Á²ÐÁôÁ¿µÄ²â¶¨
£¨6£©ÆÏÌѾƳ£ÓÃNa2S2O5×÷¿¹Ñõ»¯¼Á¡£²â¶¨Ä³ÆÏÌѾÆÖп¹Ñõ»¯¼ÁµÄ²ÐÁôÁ¿(ÒÔÓÎÀëSO2¼ÆËã)µÄ·½°¸ÈçÏ£º

(ÒÑÖª£ºµÎ¶¨Ê±·´Ó¦µÄ»¯Ñ§·½³ÌʽΪSO2£«I2£«2H2O£½H2SO4£«2HI)
¢Ù°´ÉÏÊö·½°¸ÊµÑ飬ÏûºÄ±ê×¼I2ÈÜÒº25.00 mL£¬¸Ã´ÎʵÑé²âµÃÑùÆ·Öп¹Ñõ»¯¼ÁµÄ²ÐÁôÁ¿(ÒÔÓÎÀëSO2¼ÆËã)Ϊ               g¡¤L£­1¡£
¢ÚÔÚÉÏÊöʵÑé¹ý³ÌÖУ¬ÈôÓв¿·ÖHI±»¿ÕÆøÑõ»¯£¬Ôò²âµÃ½á¹û     (Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°²»±ä¡±)¡£
£¨17·Ö£©I.Ϊ̽¾¿SO2µÄÐÔÖÊ£¬ÐèÒª±ê×¼×´¿öÏÂ11£®2 LSO2ÆøÌå¡£»¯Ñ§Ð¡×éͬѧÒÀ¾Ý»¯Ñ§·½³ÌʽZn+ 2H2SO4£¨Å¨£©=ZnSO4+SO2¡ü+2H2O¼ÆËãºó£¬È¡32.5gпÁ£ÓëÖÊÁ¿·ÖÊýΪ98%µÄŨÁòËᣨÃܶȣ©  60mL³ä·Ö·´Ó¦£¬Ð¿È«²¿Èܽ⣬¶ÔÓÚÖƵõÄÆøÌ壬ÓÐͬѧÈÏΪ¿ÉÄÜ»ìÓÐÔÓÖÊ¡£
£¨1£©»¯Ñ§Ð¡×éËùÖƵõÄÆøÌåÖлìÓеÄÖ÷ÒªÔÓÖÊÆøÌå¿ÉÄÜÊÇ            £¨Ìî·Ö×Óʽ£©¡£²úÉúÕâÖÖ½á¹ûµÄÖ÷ÒªÔ­ÒòÊÇ                             £¨Óû¯Ñ§·½³ÌʽºÍ±ØÒªµÄÎÄ×Ö¼ÓÒÔ˵Ã÷£©
£¨2£©ÎªÖ¤ÊµÏà¹Ø·ÖÎö£¬»¯Ñ§Ð¡×éµÄͬѧÉè¼ÆÁËʵÑ飬×é×°ÁËÈçÏÂ×°Ö㬶ÔËùÖÆÈ¡µÄÆøÌå½øÐÐ̽¾¿¡£

¢ÙÖÃBÖмÓÈëµÄÊÔ¼Á         £¬×°ÖÃCÖÐÆ·ºìÈÜÒºµÄ×÷ÓÃÊÇ                     ¡£
¢Ú×°ÖÃD¼ÓÈëµÄÊÔ¼Á         £¬×°ÖÃF¼ÓÈëµÄÊÔ¼Á                     ¡£
¢Û¿É֤ʵһ¶¨Á¿µÄпÁ£ºÍÒ»¶¨Á¿µÄŨÁòËá·´Ó¦ºóÉú³ÉµÄÆøÌåÖлìÓÐijÔÓÖÊÆøÌåµÄʵÑéÏÖÏóÊÇ                                                                              ¡£
¢ÜUÐ͹ÜG¼ÓÈëµÄÊÔ¼Á         £¬×÷ÓÃΪ                                       .
II£®¹¤ÒµÉϿɲÉÓõ绯ѧ·¨ÀûÓÃH2S·ÏÆøÖÆÈ¡ÇâÆø£¬¸Ã·¨ÖÆÇâ¹ý³ÌµÄʾÒâͼËùʾ£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©²úÉú·´Ó¦³ØÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                  ¡£
£¨2£©·´Ó¦ºóµÄÈÜÒº½øÈëµç½â³Ø£¬µç½âʱÑô¼«·´Ó¦Ê½Îª                         ¡£
£¨3£©Èôµç½â³ØÖÐÉú³É5. 6 L H2(±ê×¼×´¿ö)£¬ÔòÀíÂÛÉÏÔÚ·´Ó¦³ØÖпÉÉú³ÉS³ÁµíµÄÎïÖʵÄÁ¿Îª___________ mol¡£
ij¿ÎÍâÐËȤС×éΪ̽¾¿Í­¸úŨÁòËá·´Ó¦Çé¿ö£¬ÓÃÏÂͼËùʾװÖýøÐÐʵÑé¡£ÒÑÖª£º¢ÙSO2ÄÑÈÜÓÚ±¥ºÍÑÇÁòËáÇâÄÆÈÜÒº£»¢ÚSO2ÄÜÓëËáÐÔ¸ßÃÌËá¼ØÈÜÒº·¢ÉúÑõ»¯»¹Ô­·´Ó¦Ê¹Ö®ÍÊÉ«£¨»¯Ñ§·½³ÌʽΪ5SO2+2KMnO4+2H2O=K2SO4+2MnSO4+2H2SO4£©¡£

»Ø´ðÏÂÁÐÎÊÌ⣨ע£ºEΪֹˮ¼Ð£¬FΪÐýÈû£©£º
£¨1£©¼ì²éA×°ÖõÄÆøÃÜÐԵķ½·¨            ¡£
£¨2£©×°ÖÃAÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ           ¡£
£¨3£©×°ÖÃDÖÐÊԹܿڷÅÖõÄÃÞ»¨Ó¦ÕºÓÐNaOHÈÜÒº£¬
Æä×÷ÓÃÊÇ              ¡£
£¨4£©×°ÖÃB¾ßÓÐÖü´æÆøÌåµÄ×÷Óᣵ±D´¦ÓÐÃ÷ÏÔµÄÏÖÏóºó£¬¹Ø±ÕÐýÈûF²¢ÒÆÈ¥¾Æ¾«µÆ£¬ÓÉÓÚÓàÈȵÄ×÷Óã¬A´¦ÈÔÓÐÆøÌå²úÉú£¬´ËʱBÖеÄÏÖÏóÊÇ               £¬BÖÐÓ¦·ÅÖõÄÒºÌåÊÇ£¨Ìî×Öĸ£©           ¡£
A£®Ë®B£®±¥ºÍNaHSO3ÈÜÒºC£®ËáÐÔKMnO4ÈÜÒºD£®NaOHÈÜÒº
£¨5£©¸ÃС×éѧÉú×öÁËÈçÏÂʵÑ飺ȡһ¶¨ÖÊÁ¿µÄͭƬºÍÒ»¶¨Ìå»ý18.4 mol¡¤L£­1µÄŨÁòËá·ÅÔÚÔ²µ×ÉÕÆ¿Öй²ÈÈ£¬Ö±µ½·´Ó¦Íê±Ï£¬·¢ÏÖÉÕÆ¿ÖÐûÓÐͭƬʣÓà¡£Íù·´Ó¦ºóµÄÈÜÒºÖмÓÈë×ãÁ¿µÄBaCl2ÈÜÒº£¬»ñµÃ³Áµí3.495 g£»²úÉúµÄÆøÌåÇ¡ºÃʹ200ml 0.01mol¡¤L-1µÄËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£¬ÔòʵÑéÖÐÈ¡ÓõÄÁòËáµÄÎïÖʵÄÁ¿Îª           ¡£

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

¾«Ó¢¼Ò½ÌÍø