Ϊ²â¶¨Ä³Fe2O3ºÍFeO»ìºÏÎïÖÐFe2O3µÄº¬Á¿£¬ÍùÒ»¶¨Á¿ÑùÆ·ÖмÓÈë200 mL¡¡5.0 mol¡¤L£­1ÁòËᣬ³ä·Ö·´Ó¦ºó£¬²âµÃÈÜÒºÖÐc(H+)£½1.0 mol¡¤L£­1(ÉèÈÜÒºÌå»ýÈÔΪ200 mL)£¬ÔÙ¼ÓÈëÊÊÁ¿2.0 mol¡¤L£­1¡¡NaOHÈÜÒº£¬Ê¹Fe2+¡¢Fe3+ÍêÈ«³Áµí£¬´ý³ÁµíÈ«²¿±äΪºìºÖÉ«ºó£¬¹ýÂË£¬Ï´µÓ¡¢×ÆÉÕ³ÁµíÎ³ÆµÃ¹ÌÌåÖÊÁ¿±ÈÔ­ÑùÆ·Ôö¼Ó2.4 g£®

(1)ÓëÑùÆ··¢Éú·´Ó¦µÄÁòËáµÄÎïÖʵÄÁ¿Îª¶àÉÙ£¿

(2)Ðè¼ÓÈë¶àÉÙºÁÉýNaOHÈÜÒº£¬²ÅÄÜʹFe3+¡¢Fe2+Ç¡ºÃ³ÁµíÍêÈ«£¿

(3)¸Ã»ìºÏÎïÖÐFe2O3µÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¾«Ó¢¼Ò½ÌÍøÑо¿½ðÊôµÄÒ±Á¶¶ÔÈËÀàÓÐÖØÒªÒâÒ壮
£¨1£©ÈËÀàÀúÊ·ÉÏ´óÁ¿Éú²úºÍʹÓÃÂÁ¡¢Ìú¡¢îÑ¡¢Í­ËÄÖÖ½ðÊôµ¥ÖʵÄʱ¼ä˳ÐòÊÇ
 
£®
£¨2£©Í­µÄÒ±Á¶ÖÐÓÐÒ»²½ÊÇ2CuFeS2+4O2=Cu2S+3SO2+2FeO£¬·´Ó¦µÄÑõ»¯²úÎïÊÇ
 
£®
£¨3£©ÈçͼÊDz¿·Ö½ðÊôÑõ»¯Îï±»Ò»Ñõ»¯Ì¼»¹Ô­Ê±lg[p£¨CO£©/p£¨CO2£©]ÓëζȵĹØÏµÇúÏßͼ£®ÔòËÄÖÖ½ðÊôÑõ»¯ÎïÖÐ×îÒ×±»»¹Ô­µÄ½ðÊôÑõ»¯ÎïÊÇ£¬¸Ã·´Ó¦µÄ¡÷H
 
0£®
£¨4£©ÏÂÁÐ˵·¨ÕýÈ·ÊÇ
 

A£®Ñõ»¯Ã¾¸úÂÁ·Û¹²ÈÈÊÊÓÚÁ¶Ã¾
B£®²»Ðâ¸Öº¬ÓнðÊôºÍ·Ç½ðÊôÔªËØ
C£®ËùÓнðÊô¾ùÒÔ»¯ºÏÎï´æÔÚÓÚ×ÔÈ»½ç
D£®½ðÊôÒ±Á¶Öл¹Ô­¼Á¿ÉÒÔÊÇһЩ»îÆÃ½ðÊô
E£®»îÆÃ½ðÊôµÄÒ±Á¶¶¼ÊÇͨ¹ýµç½âÆäÑÎÈÜÒºÖÆµÃ
F£®½ðÊôÌáÁ¶Ò»°ãÒª¾­¹ý¿óʯµÄ¸»¼¯¡¢Ò±Á¶¡¢¾«Á¶Èý²½
£¨5£©Îª²â¶¨Ä³Ò»Ìú¿óʯÑùÆ·ÖеÄÌúÔªËØµÄÖÊÁ¿·ÖÊý£¬È¡3.702g¸ÃÌú¿óʯ£¨ÌúÒÔFe2O3ÐÎʽ´æÔÚ£¬ÆäÓàÎïÖʲ»º¬Ìú£©ÈÜÓÚŨÈÈÑÎËáÖУ¬Ï¡ÊÍÖÁ250mL£¬´ÓÖÐÈ¡³ö25mLÈÜÒº£¬×÷ÈçÏ´¦Àí£º¼ÓÈë¹ýÁ¿µÄ°±Ë®£¬Ê¹ÌúÒÔÇâÑõ»¯ÌúµÄÐÎʽ³Áµí£¬½«³Áµí¹ýÂË£¬Ï´¾»²¢×ÆÉÕ£¬Ê¹Ö®Íêȫת»¯ÎªÑõ»¯Ìú£®ÓйØÊµÑéÊý¾Ý£ºÛáÛöÖÊÁ¿£º15.2861g£¬Èý´ÎׯÉÕºóÛáÛöºÍ¹ÌÌåµÄÖÊÁ¿£º15.6209g¡¢15.6205g¡¢15.6205g£®ÔòÌú¿óʯÖеÄÌúÔªËØµÄÖÊÁ¿·ÖÊýΪ
 
£¨°Ù·ÖÊýÖеÄÊý×Ö±£Áôµ½Ð¡Êýµãºóһ룩£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ϊ²â¶¨Ä³ÌúÍ­ºÏ½ðµÄ×é³É£¬ÏÖ½«5.6g¸ÃºÏ½ð£¨±íÃæÑõ»¯Ä¤³É·ÖΪFe2O3ºÍCuO£©¼ÓÈëµ½¹ýÁ¿Ï¡ÁòËáÖУ¬ÍêÈ«·´Ó¦ºó²úÉúÆøÌå672mL£¬²¢µÃµ½Ç³ÂÌÉ«ÈÜÒºA£¨²»º¬Cu2+Àë×Ó£©ºÍ²»ÈÜÎïB£®¹ýÂË£¬½«B¼ÓÈëµ½ÊÊÁ¿µÄijŨ¶ÈµÄÏõËáÖУ¬ÍêÈ«Èܽâºó£¬µÃNO¡¢NO2µÄ»ìºÏÆø896mL£¬¾­²â¶¨¸Ã»ìºÏÆøÖÐV£¨NO£©£ºV£¨NO2£©=3£º1£®ÔÙ½«AÒ²¼ÓÈëµ½×ãÁ¿Í¬Å¨¶ÈµÄÏõËáÖгä·Ö·´Ó¦£¬ÓÃÅÅË®·¨ÊÕ¼¯²úÉúµÄÆøÌå½á¹ûµÃÒ»ÉÕÆ¿ÆøÌ壬£¨²»´ÓË®ÖÐÒÆ³öÉÕÆ¿£©½ô½Ó×ÅÏòÉÕÆ¿ÖÐͨÈë224mL O2£¬ÆøÌåÄÜÇ¡ºÃÍêÈ«ÈÜÓÚË®£¨ÒÔÉÏËùÓÐÆøÌåÌå»ý¶¼ÒÑ»»Ëã³É±ê×¼×´¿öʱµÄÊý¾Ý£©£®ÏÂÁнáÂÛÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢ºÏ½ðÖÐÑõÔªËØµÄÎïÖʵÄÁ¿ÊÇ0.01 molB¡¢ºÏ½ðÖÐÌúµ¥ÖʵÄÖÊÁ¿ÊÇ1.68 gC¡¢ÈÜÒºAÖдæÔÚµÄÑôÀë×ÓÖ»ÓÐFe2+Àë×ÓD¡¢²»ÈÜÎïBΪͭµ¥ÖÊÇÒÖÊÁ¿ÊÇ1.6 g

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄêÌì½òÊÐÄÏ¿ªÇø¸ßÈýµÚ¶þ´ÎÄ£Ä⿼ÊÔÀí×Û»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÊµÑéÌâ

I£®ÏÂͼ±íʾ´Ó¹ÌÌå»ìºÏÎïÖзÖÀëQµÄ2ÖÖ·½°¸£¬Çë»Ø´ðÓйØÎÊÌâ¡£

£¨1£©Ñ¡Ó÷½°¸(i)ʱ£¬QÓ¦¸Ã¾ßÓеÄÐÔÖÊÊÇ_____________£¬²ÐÁôÎïÓ¦¸Ã¾ß

ÓеÄÐÔÖÊÊÇ__________________________________¡£

£¨2£©Ñ¡Ó÷½°¸(ii)´Óij½ðÊô·ÛÄ©£¨º¬ÓÐAu¡¢AgºÍCu£©ÖзÖÀëAu£¬¼ÓÈëµÄÊÔ¼ÁΪ____________¡£

£¨3£©ÎªÌᴿijFe2O3ÑùÆ·£¨Ö÷ÒªÔÓÖÊÓÐSiO2.Al2O3£©£¬²ÎÕÕ·½°¸(i)ºÍ(ii)£¬ÇëÉè¼ÆÒ»ÖÖÒÔ¿òͼÐÎʽ±íʾµÄʵÑé·½°¸£¨×¢Ã÷ÎïÖʺͲÙ×÷£©£º

______________________________________________________________________________¡£

¢ò£®Ä³ÖÖº¬ÓÐÉÙÁ¿Ñõ»¯ÄƵĹýÑõ»¯ÄÆÑùÆ·£¨¼ºÖªÑùÆ·ÖÊÁ¿Îª1.560g¡¢×¶ÐÎÆ¿ºÍË®µÄÖÊÁ¿Îª

190.720g)£¬ÀûÓÃÓÒÏÂͼËùʾװÖòⶨ»ìºÏÎïÖÐNa2O2µÄÖÊÁ¿·ÖÊý£¬Ã¿¸ôÏàͬʱ¼ä¶ÁµÃµç

×ÓÌìÆ½µÄÊý¾ÝÈçÏÂ±í£º

£¨4£©Ð´³öNa2O2ÓëH2O·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________________________________£®

£¨5£©¼ÆËãNa2O2ÖÊÁ¿·ÖÊýʱ£¬±ØÐèµÄÊý¾ÝÊÇ_________________________________________.

²»±Ø×÷µÚ6´Î¶ÁÊýµÄÔ­ÒòÊÇ_____________________________________________________£®

£¨6£©²â¶¨ÉÏÊöÑùÆ·(1.560g)ÖÐNa2O2ÖÊÁ¿·ÖÊýµÄÁíÒ»ÖÖ·½°¸£¬Æä²Ù×÷Á÷³ÌÈçÏ£º

²Ù×÷¢ÚµÄÃû³ÆÊÇ____________£¬¸Ã·½°¸ÐèÖ±½Ó²â¶¨µÄÎïÀíÁ¿ÊÇ_____________ £¬²â¶¨¹ý³ÌÖÐ

ÐèÒªµÄÒÇÆ÷Óеç×ÓÌìÆ½¡¢Õô·¢Ã󡢾ƾ«µÆ£¬»¹ÐèÒª___________¡¢__________£¨¹Ì¶¨¡¢¼Ð

³ÖÒÇÆ÷³ýÍ⣩,ÔÚ×ªÒÆÈÜҺʱ£¬ÈçÈÜÒº×ªÒÆ²»ÍêÈ«£¬ÔòNa2O2ÖÊÁ¿·ÖÊýµÄ²â¶¨½á¹û_______£¨Ìî

¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°²»±ä¡±£©¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ϊ²â¶¨Ä³Fe2O3ºÍFeO»ìºÏÎïÖÐFe2O3µÄº¬Á¿£¬ÍùÒ»¶¨Á¿ÑùÆ·ÖмÓÈë200 mL 5.0 mol?L©¤1ÁòËᣬ³ä·Ö·´Ó¦ºó£¬²âµÃÈÜÒºÌå»ýÈÔΪ200 mL£¬ÈÜÒºÖÐc(H+)= 1.0 mol?L©¤1£¬ÔÙ¼ÓÈëÊÊÁ¿2.0 mol?L©¤1NaOHÈÜÒº£¬Ê¹Fe2+¡¢Fe3+Ç¡ºÃÍêÈ«³Áµí£¬³ä·Ö½Á°è²¢Î¢ÈÈ£¬´ý³ÁµíÈ«²¿±äΪºìºÖÉ«ºó£¬¹ýÂË¡¢Ï´µÓ¡¢×ÆÉÕ³ÁµíÎ³ÆµÃ¹ÌÌåÖÊÁ¿±ÈÔ­ÑùÆ·Ôö¼Ó2.4 g¡£

£¨1£©ÓëÑùÆ··¢Éú·´Ó¦µÄÁòËáµÄÎïÖʵÄÁ¿Îª¶àÉÙ£¿

£¨2£©Ðè¼ÓÈë¶àÉÙºÁÉýNaOHÈÜÒº£¬²ÅÄÜʹFe2+¡¢Fe3+Ç¡ºÃ³ÁµíÍêÈ«£¿

£¨3£©¸Ã»ìºÏÎïÖÐFe2O3µÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸