ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿¸ß´¿MnCO3ÊÇÖƱ¸¸ßÐÔÄÜ´ÅÐÔ²ÄÁϵÄÖ÷ÒªÔÁÏ¡£ÊµÑéÊÒÒÔMnO2ΪÔÁÏÖƱ¸ÉÙÁ¿¸ß´¿MnCO3µÄ²Ù×÷²½ÖèÈçÏÂËùʾ£º
¢ñ.ÖƱ¸MnSO4ÈÜÒº£º
ÔÚÈý¾±ÉÕÆ¿ÖмÓÈë4.35gMnSO4ºÍ×ãÁ¿µÄË®£¬½Á°è£¬Í¨ÈëSO2ºÍN2»ìºÏÆøÌ壬·´Ó¦3h£®Í£Ö¹Í¨ÈëSO2£¬¼ÌÐø·´Ó¦Æ¬¿Ì£¬¹ýÂË¡¢Ï´µÓ£¬µÃMnO2ÈÜÒº¡££¨N2²»²ÎÓë·´Ó¦£©
£¨1£©ÖÆÈ¡MnSO4µÄ»¯Ñ§·½³ÌʽΪ_______________£®
£¨2£©·´Ó¦¹ý³ÌÖУ¬ÎªÊ¹SO2¾¡¿ÉÄÜת»¯ÍêÈ«£¬ÔÚ²»¸Ä±ä¹ÌҺͶÁϵÄÌõ¼þÏ£¬¿É²ÉÈ¡µÄºÏÀí´ëÊ©ÓÐ________________£®
£¨3£©Ë®Ô¡¼ÓÈȵÄÓŵãÊÇ____________¡£
¢ò.ÖƱ¸¸ß´¿MnCO3¹ÌÌ壺
ÒÑÖªMnCO3ÄÑÈÜÓÚË®¡¢ÒÒ´¼£¬³±ÊªÊ±Ò×±»¿ÕÆøÑõ»¯£¬100¡æ¿ªÊ¼·Ö½â£»pH=7.7ʱMn(OH)2¿ªÊ¼³Áµí¡£
ʵÑé²½Öè: ¢Ù½«ËùµÃMnSO4ÈÜÒºÓë20.0mL 2.0mol/LµÄNa2CO3ÈÜÒº»ìºÏ£¬³ä·Ö·´Ó¦;¢Ú¹ýÂË£¬ÓÃÉÙÁ¿Ë®Ï´µÓ2~3´Î ;¢ÛÓÃÉÙÁ¿C2H5OHÏ´µÓ;¢ÜµÍΠ(µÍÓÚ100¡æ) ¸ÉÔµÃ¹ÌÌå3.45g¡£ (1) MnSO4ÈÜÒºÓëNa2CO3ÈÜÒº»ìºÏµÄÕýÈ·²Ù×÷Ϊ__________¡£(Ìî×Öĸ´úºÅ)
a. ½«Na2CO3ÈÜÒº»ºÂýµÎ¼Óµ½MnSO4ÈÜÒºÖУ¬±ß¼Ó±ß½Á°è
b. ½«MnSO4ÈÜÒº»ºÂýµÎ¼Óµ½Na2CO3ÈÜÒºÖУ¬±ß¼Ó±ß½Á°è
c. ½«Na2CO3ÈÜҺѸËÙµ¹Èëµ½MnSO4ÈÜÒºÖУ¬²¢³ä·Ö½Á°è
d. ½«MnSO4ÈÜҺѸËÙµ¹Èëµ½Na2CO3ÈÜÒºÖУ¬²¢³ä·Ö½Á°è
(2)¼ìÑéMnCO3¹ÌÌåÊÇ·ñÏ´µÓ¸É¾»µÄ·½·¨Îª______________¡£
(3) ÓÃÉÙÁ¿C2H5OHÏ´µÓµÄÄ¿µÄÊÇ_______________¡£
(4) MnCO3µÄ²úÂÊΪ_____________¡£
¡¾´ð°¸¡¿ MnO2+SO2=MnSO4£¨»òMnO2+H2SO4=MnSO4+H2O£© ¿ØÖÆÊʵ±µÄˮԡʪ¶È£¬¼õ»ºÍ¨ÈëÆøÌåµÄËٶȣ¬¼õСSO2ÔÚ»ìºÏÆø£¨ÓëN2£©ÖеıÈÀý ÊÜÈȾùÔÈ£¬ÈÝÒ׿ØÖÆʪ¶È a È¡×îºóÒ»´ÎÏ´µÓÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬µÎ¼ÓBaCl2ÈÝÒ×£¬Èç¹û²»²úÉú³Áµí£¬ËµÃ÷ÒÑÏ´µÓ¸É¾»£¬·ñÔò˵Ã÷δϴµÓ¸É¾» ÒÒ´¼·ÐµãµÍÓÚ100¡æ£¬±ãÓÚµÍθÉÔïÇÒMnCO3²»·Ö½â 75%
¡¾½âÎö¡¿±¾ÌâÖ÷Òª¿¼²é¶ÔÓÚʵÑéÊÒÒÔMnO2ΪÔÁÏÖƱ¸ÉÙÁ¿¸ß´¿MnCO3ʵÑéµÄÆÀ¼Û¡£
¢ñ.£¨1£©ÖÆÈ¡MnSO4µÄ»¯Ñ§·½³ÌʽΪMnO2+SO2=MnSO4¡£
£¨2£©·´Ó¦¹ý³ÌÖУ¬ÎªÊ¹SO2¾¡¿ÉÄÜת»¯ÍêÈ«£¬ÔÚ²»¸Ä±ä¹ÌҺͶÁϵÄÌõ¼þÏ£¬¿É²ÉÈ¡µÄºÏÀí´ëÊ©ÓпØÖÆÊʵ±µÄˮԡζȣ¬¼õ»ºÍ¨ÈëÆøÌåµÄËٶȣ¬¼õСSO2ÔÚ»ìºÏÆø£¨ÓëN2£©ÖеıÈÀý¡£
£¨3£©Ë®Ô¡¼ÓÈȵÄÓŵãÊÇÊÜÈȾùÔÈ£¬ÈÝÒ׿ØÖÆζȡ£
¢ò.(1) ÒòΪpH=7.7ʱMn(OH)2¿ªÊ¼³Áµí£¬¶øNa2CO3ÈÜÒºµÄpH>7.7£¬ËùÒÔÓ¦½«Na2CO3ÈÜÒº»ºÂýµÎ¼Óµ½MnSO4ÈÜÒºÖУ¬±ß¼Ó±ß½Á°è£¬¹ÊÑ¡a¡£
(2)¼ìÑéMnCO3¹ÌÌåÊÇ·ñÏ´µÓ¸É¾»¾ÍÊǼìÑéÊÇ·ñ²ÐÁôNa+¡¢£¬ÆäÖÐ
·½±ã¼ìÑ飬·½·¨Îª£ºÈ¡×îºóÒ»´ÎÏ´µÓÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬µÎ¼ÓBaCl2ÈÜÒº£¬Èç¹û²»²úÉú³Áµí£¬ËµÃ÷ÒÑÏ´µÓ¸É¾»£¬·ñÔò˵Ã÷δϴµÓ¸É¾»¡£
(3) ÓÃÉÙÁ¿C2H5OHÏ´µÓµÄÄ¿µÄÊÇÒÒ´¼·ÐµãµÍÓÚ100¡æ£¬±ãÓÚµÍθÉÔïÇÒMnCO3²»·Ö½â¡£
(4) 4.35g¼´0,05molMnO2ºÍ×ãÁ¿µÄË®ÓëSO2·´Ó¦µÃº¬ÓÐ0.05molMnSO4µÄÈÜÒº¡£20.0mL 2.0mol/LµÄNa2CO3ÈÜÒºº¬ÓÐ0.04molNa2CO3¡£½«ËùµÃMnSO4ÈÜÒºÓë20.0mL 2.0mol/LµÄNa2CO3ÈÜÒº»ìºÏ£¬³ä·Ö·´Ó¦£¬ÀíÂÛÉϵõ½0.04molMnCO3£¬ËùÒÔMnCO3µÄ²úÂÊΪ3.45g/[(0.04¡Á115)g]=75%¡£
![](http://thumb2018.1010pic.com/images/loading.gif)
¡¾ÌâÄ¿¡¿¡¾¼ÓÊÔÌâ¡¿³£ÎÂϵâËá¸Æ[Ca(IO3)2]ÊÇÒ»ÖÖ°×É«¾§Ì壬΢ÈÜÓÚË®£¬²»ÈÜÓÚÒÒ´¼£¬ÄÜÈÜÓÚÑÎËᣬÊÇÄ¿Ç°¹ã·ºÊ¹ÓõļÈÄܲ¹µâÓÖÄܲ¹¸ÆµÄÐÂÐÍʳƷºÍ´ÊÁÏÌí¼Ó¼Á¡£ÆäÖƱ¸ÔÀíÈçÏ£º
¢ñ.µâËá¸ÆµÄÖƱ¸
(1)ijͬѧÉè¼ÆµÄʵÑé×°ÖÃÈçÏÂͼËùʾ(¼Ð³Ö×°ÖÃÒÑÊ¡ÂÔ)£¬²½Öè¢ÚµÄ¼ÓÈÈ·½Ê½Îª_________¡£
(2)²½Öè¢ÝÀäÈ´Èȵı¥ºÍÈÜҺûÓÐÎö³ö¾§Ì壬ΪÁ˵õ½¾§Ìå¿ÉÒÔ²ÉÈ¡µÄÒ»ÖÖ´ëÊ©ÊÇ________¡£
(3)²½Öè¢ßÏ´µÓµÄ¾ßÌå²Ù×÷ÊÇ____________¡£
¢ò.²úÆ·´¿¶È²â¶¨
׼ȷ³ÆÈ¡0.5000 gÑùÆ·£¬ËữÈܽ⣬¶¨ÈÝÖÁ250 mL,´ÓÖÐÈ¡³ö25.00 mLÊÔÑùÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈë×ãÁ¿µÄKI³ä·Ö·´Ó¦£¬ÓÃ0.04000 mol¡¤L-1Áò´úÁòËáÄÆÈÜÒºµÎ¶¨ÖÁÖյ㣬Öظ´ÒÔÉϲ½Ö裬Ëù²âµÃʵÑéÊý¾ÝÈçÏ£º
1 | 2 | 3 | |
µÎ¶¨Æðʼ¶ÁÊý/mL | 1.52 | 1.16 | 0.84 |
µÎ¶¨ÖÕÖ¹¶ÁÊý/mL | 31.50 | 31.18 | 30.84 |
¼ºÖª£º2Na2S2O3 + I2= Na2S4O6 + 2NaI
(4)¼ÓÈëKIʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_____________¡£
(5)µÎ¶¨Ê±ËùÓõÄָʾ¼ÁÊÇ_________________.
(6)¾¼ÆËã²úÆ·µÄ´¿¶ÈΪ________________¡£