ÌâÄ¿ÄÚÈÝ

ij¹¤³§ÓÃÈíÃ̿󣨺¬MnO2Ô¼70%¼°Al2 O3£©ºÍÉÁп¿ó£¨º¬ZnSÔ¼80%¼°FeS£©¹²Í¬Éú²úMnO2£¬ºÍZn£¨¸Éµç³ØÔ­ÁÏ£©¡£Á÷³ÌÈçÏ£º

ÒÑÖª£º¢ÙAÊÇMnSO4¡¢ZnSO4¡¢Fe2(SO4)3£¬Al2(SO4)3µÄ»ìºÏÒº¡£
¢ÚIVÖеĵç½â·´Ó¦Ê½ÎªMnSO4+ZnSO4+2H2O    MnO2+Zn +2H2SO4¡£
£¨1£©AÖÐÊôÓÚ»¹Ô­²úÎïµÄÊÇ             ¡£
£¨2£©¼ÓÈëMnCO3¡¢Zn2(OH)2CO3µÄ×÷ÓÃÊÇ         £ºCµÄ»¯Ñ§Ê½ÊÇ           ¡£
£¨3£©¸ÃÉú²úÖгýµÃµ½Na2SO4¡¢SµÈ¸±²úÆ·Í⣬»¹¿ÉµÃµ½µÄ¸±²úÆ·ÊÇ            ¡£
£¨4£©¸±²úÆ·S¿ÉÓÃÓÚÖÆÁòËᣬת»¯¹ý³ÌÊÇ£ºS¡úSO2¡úSO3¡úH2SO4¡£Ð´³öµÚ¶þ²½×ª»¯µÄ»¯Ñ§·½³Ìʽ           ¡£
£¨5£©Òª´ÓNa2SO4ÈÜÒºÖеõ½Ã¢Ïõ£¨ Na2SO4£®10H2O£©£¬Ðè½øÐеIJÙ×÷ÓÐÕô·¢Å¨Ëõ¡¢          ¡¢
¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔïµÈ¡£
£¨6£©´ÓÉú²úMnO2ºÍZnµÄ½Ç¶È¼ÆË㣬ÈíÃÌ¿óºÍÉÁп¿óͶÁϵÄÖÊÁ¿±È´óÔ¼ÊÇ          ¡£

£¨1£©MnSO4
(2)Ôö´óÈÜÒºµÄp H,ʹFe3+ºÍAl3+Éú³É³Áµí  H2SO4
£¨3£©Fe2O3¡¢Al2O3
(4)   2SO2+O22SO3
(5) ÀäÈ´½á¾§
£¨6£©1£º1(»ò1.03:1)

½âÎöÊÔÌâ·ÖÎö£º±È½ÏÐÅÏ¢1¿ÉÖªMn»¯ºÏ¼Û½µµÍ£¬ÊôÓÚ»¹Ô­²úÎïµÄΪMnSO4£¬Óɹ¤ÒÕÁ÷³Ì·ÖÎö¼ÓÈëMnCO3¡¢Zn2(OH)2CO3µÄ×÷ÓÃÊÇÔö´óÈÜÒºµÄp H,ʹFe3+ºÍAl3+Éú³É³Áµí£»ÇÒCµÄ»¯Ñ§Ê½ÎªH2SO4£»²»ÄѵóöµÃµ½µÄ¸±²úÆ·»¹ÓÐFe2O3¡¢Al2O3£»¼ÙÉèÈíÃÌ¿óºÍÉÁп¿óµÄÖÊÁ¿·Ö±ðΪx ¡¢y£¬¸ù¾Ý¢Ú¿ÉÖªMnO2ºÍZnµÄÎïÖʵÄÁ¿Ö®±ÈΪ1:1£¬Ôò0.7x g/87g/mol:0.8y g/97g/mol=1:1,½âµÃx/y=1.03:1.
¿¼µã£ºÑõ»¯»¹Ô­·´Ó¦¡¢ÔªËØ»¯ºÏÎï֪ʶ¡¢ÎïÖʵķÖÀëºÍÌá´¿¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¶þÑõ»¯ÂÈ(ClO2)ΪһÖÖ»ÆÂÌÉ«ÆøÌ壬Êǹú¼ÊÉϹ«ÈϵĸßЧ¡¢¹ãÆס¢¿ìËÙ¡¢°²È«µÄɱ¾úÏû¶¾¼Á¡£
(1)¹¤ÒµÉÏÖƱ¸ClO2µÄ·´Ó¦Ô­ÀíΪ2NaClO3£«4HCl=2ClO2¡ü£«Cl2¡ü£«2H2O£«2NaCl¡£
¢ÙŨÑÎËáÔÚ·´Ó¦ÖÐÏÔʾ³öÀ´µÄÐÔÖÊÊÇ________¡£

A£®Ö»Óл¹Ô­ÐÔB£®»¹Ô­ÐÔºÍËáÐÔ
C£®Ö»ÓÐÑõ»¯ÐÔD£®Ñõ»¯ÐÔºÍËáÐÔ
¢ÚÈôÉÏÊö·´Ó¦ÖвúÉú0.1 mol ClO2£¬ÔòתÒƵç×ÓµÄÎïÖʵÄÁ¿Îª________mol¡£
(2)Ä¿Ç°ÒÑ¿ª·¢³öÓõç½â·¨ÖÆÈ¡ClO2µÄй¤ÒÕ¡£

¢ÙÉÏͼÊÇÓÃʯī×÷µç¼«£¬Ò»¶¨Ìõ¼þϵç½â±¥ºÍʳÑÎË®ÖÆÈ¡ClO2µÄʾÒâͼ¡£ÔòÑô¼«²úÉúClO2µÄµç¼«·´Ó¦Ê½Îª_____________________________________¡£
¢Úµç½âÒ»¶Îʱ¼ä£¬µ±Òõ¼«²úÉúµÄÆøÌåÌå»ýΪ112 mL(±ê×¼×´¿ö)ʱ£¬Í£Ö¹µç½â¡£Í¨¹ýÑôÀë×Ó½»»»Ä¤µÄÑôÀë×ÓµÄÎïÖʵÄÁ¿Îª________mol£¬ÓÃƽºâÒƶ¯Ô­Àí½âÊÍÒõ¼«ÇøpHÔö´óµÄÔ­Òò£º______________________________________¡£
(3)ClO2¶ÔÎÛË®ÖÐFe2£«¡¢Mn2£«¡¢S2£­ºÍCN£­µÈÓÐÃ÷ÏÔµÄÈ¥³ýЧ¹û¡£Ä³¹¤³§ÎÛË®Öк¬CN£­a mg/L£¬ÏÖÓÃClO2½«CN£­Ñõ»¯£¬²úÎïÖÐÓÐÁ½ÖÖÆøÌåÉú³É£¬ÆäÀë×Ó·´Ó¦·½³ÌʽΪ________________£»´¦Àí100 m3ÕâÖÖÎÛË®£¬ÖÁÉÙÐèÒªClO2________mol¡£

ʳÑÎÖк¬ÓÐÒ»¶¨Á¿µÄþ¡¢ÌúµÈÔÓÖÊ£¬¼ÓµâÑÎÖеâµÄËðʧÖ÷ÒªÊÇÓÉÓÚÔÓÖÊ¡¢Ë®·Ö¡¢¿ÕÆøÖеÄÑõÆøÒÔ¼°¹âÕÕ¡¢ÊÜÈȶøÒýÆðµÄ¡£ÒÑÖª£º
Ñõ»¯ÐÔ£ºIO3-£¾Fe3£«£¾I2£»»¹Ô­ÐÔ£ºS2O32-£¾I£­
3I2£«6OH£­=5I£­£«IO3-£«3H2O
KI£«I2KI3
(1)ijѧϰС×é¶Ô¼ÓµâÑνøÐÐÁËÈçÏÂʵÑ飺ȡһ¶¨Á¿Ä³¼ÓµâÑÎ(¿ÉÄܺ¬ÓÐKIO3¡¢KI¡¢Mg2£«¡¢Fe3£«)£¬ÓÃÊÊÁ¿ÕôÁóË®Èܽ⣬²¢¼ÓÏ¡ÑÎËáËữ£¬½«ËùµÃÊÔÒº·ÖΪ3·Ý¡£µÚÒ»·ÝÊÔÒºÖеμÓKSCNÈÜÒººóÏÔºìÉ«£»µÚ¶þ·ÝÊÔÒºÖмÓ×ãÁ¿KI¹ÌÌ壬ÈÜÒºÏÔµ­»ÆÉ«£¬ÓÃCCl4ÝÍÈ¡£¬Ï²ãÈÜÒºÏÔ×ϺìÉ«£»µÚÈý·ÝÊÔÒºÖмÓÈëÊÊÁ¿KIO3¹ÌÌåºó£¬µÎ¼Óµí·ÛÊÔ¼Á£¬ÈÜÒº²»±äÉ«¡£
¢Ù¼ÓKSCNÈÜÒºÏÔºìÉ«£¬¸ÃºìÉ«ÎïÖÊÊÇ       (Óû¯Ñ§Ê½±íʾ)£»CCl4ÖÐÏÔ×ϺìÉ«µÄÎïÖÊÊÇ      (Óõç×Óʽ±íʾ)¡£
¢ÚµÚ¶þ·ÝÊÔÒºÖмÓÈë×ãÁ¿KI¹ÌÌåºó£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ                               ¡¢                         ¡£
(2)KI×÷Ϊ¼Óµâ¼ÁµÄʳÑÎÔÚ±£´æ¹ý³ÌÖУ¬ÓÉÓÚ¿ÕÆøÖÐÑõÆøµÄ×÷Óã¬ÈÝÒ×ÒýÆðµâµÄËðʧ¡£Ð´³ö³±Êª»·¾³ÖÐKIÓëÑõÆø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                                          ¡£
½«I2ÈÜÓÚKIÈÜÒº£¬ÔÚµÍÎÂÌõ¼þÏ£¬¿ÉÖƵÃKI3¡¤H2O¡£¸ÃÎïÖÊ×÷ΪʳÑμӵâ¼ÁÊÇ·ñºÏÊÊ£¿       (Ìî¡°ÊÇ¡±»ò¡°·ñ¡±)£¬²¢ËµÃ÷ÀíÓÉ£º               ¡£
(3)ΪÁËÌá¸ß¼ÓµâÑÎ(Ìí¼ÓKI)µÄÎȶ¨ÐÔ£¬¿É¼ÓÎȶ¨¼Á¼õÉÙµâµÄËðʧ¡£ÏÂÁÐÎïÖÊÖÐÓпÉÄÜ×÷ΪÎȶ¨¼ÁµÄÊÇ       ¡£

A£®Na2S2O3B£®AlCl3
C£®Na2CO3D£®NaNO2
(4)¶Ôº¬Fe2£«½Ï¶àµÄʳÑÎ(¼ÙÉè²»º¬Fe3£«)£¬¿ÉÑ¡ÓÃKI×÷Ϊ¼Óµâ¼Á¡£ÇëÉè¼ÆʵÑé·½°¸£¬¼ìÑé¸Ã¼ÓµâÑÎÖеÄFe2£«£º                                                   

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

¾«Ó¢¼Ò½ÌÍø