ÌâÄ¿ÄÚÈÝ

25.¸ÖÌú¹¤ÒµÊǹú¼Ò¹¤ÒµµÄ»ù´¡¡£2006ÄêÎÒ¹ú´Ö¸Ö²úÁ¿Í»ÆÆ4ÒÚ¶Ö£¬¾ÓÊÀ½çÊ×λ¡£Ä³ÖÐѧÉç»áʵ¼ù»î¶¯Ð¡×éÀûÓÃ¼ÙÆÚ¶Ôµ±µØ¸ÖÌú³§½øÐÐÁ˵÷ÑУ¬¶Ô´Ó¿óʯ¿ªÊ¼µ½¸ÖÌú²ú³öµÄ¹¤ÒÕÁ÷³ÌÓÐÁËÈ«ÃæµÄ¸ÐÐÔÈÏʶ¡£ÇëÄú¶ÔÉç»áʵ¼ù»î¶¯Ð¡×é¸ÐÐËȤµÄÎÊÌâ½øÐмÆË㣺

£¨1£©½«6.62gÌú¿óʯÑùƷͶÈëÊÊÁ¿µÄÑÎËáÖУ¨³ä·Ö·´Ó¦£©£¬¹ýÂË£¬È»ºóÔÚÂËÒºÖмӹýÁ¿µÄNaOHÈÜÒº£¬³ä·Ö·´Ó¦ºó£¬¹ýÂË¡¢Ï´µÓ¡¢×ÆÉÕµÃ4.80gFe2O3¡£ÏÖÒÔ¸ÃÌú¿óʯΪԭÁÏÁ¶Ìú£¬ÈôÉú²ú¹ý³ÌÖÐÌúÔªËØËðʧ4%£¬¼ÆËãÿÉú²ú1.00tÉúÌú£¨º¬Ìú96%£©£¬ÖÁÉÙÐèÒªÕâÖÖÌú¿óʯ¶àÉÙ¶Ö£¿£¨±£ÁôÁ½Î»Ð¡Êý£©

£¨2£©È¡Ä³¸Ö·ÛÄ©28.12g£¨¼ÙÉèÖ»º¬FeºÍC£©£¬ÔÚÑõÆøÁ÷Öгä·Ö·´Ó¦£¬µÃµ½CO2ÆøÌå224mL£¨±ê×¼×´¿ö£©¡£

¢Ù¼ÆËã´Ë¸ÖÑù·ÛÄ©ÖÐÌúºÍ̼µÄÎïÖʵÄÁ¿Ö®±È¡£

¢ÚÔÙÈ¡Èý·Ý²»Í¬ÖÊÁ¿µÄ¸ÖÑù·ÛÄ©·Ö±ð¼Óµ½100mLÏàͬŨ¶ÈµÄH2SO4ÈÜÒºÖУ¬³ä·Ö·´Ó¦ºó£¬²âµÃµÄʵÑéÊý¾ÝÈçϱíËùʾ£º

ʵÑéÐòºÅ

¢ñ

¢ò

¢ó

¼ÓÈë¸ÖÑù·ÛÄ©µÄÖÊÁ¿/g

2.812

5.624

8.436

Éú³ÉÆøÌåµÄÌå»ý/L£¨±ê×¼×´¿ö£©

1.120

2.240

2.800

¼ÆËãÁòËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È¡£

¢ÛÈôÔÚʵÑé¢òÖмÌÐø¼ÓÈëmg¸ÖÑù·ÛÄ©£¬¼ÆËã·´Ó¦½áÊøÊ£ÓàµÄ¹ÌÌåÖÊÁ¿Îª¶àÉÙ£¿

£¨Óú¬mµÄ´úÊýʽ±íʾ£©

(1)½â£º 6.62gÌú¿óʯÖÐÌúµÄÖÊÁ¿Îªm(Fe)=4.80g¡Á=3.36gÉú²ú1.00tÉúÌúÐèÌú¿óʯµÄÖÊÁ¿Îª£º

m(Ìú¿óʯ=

´ð£ºÖÁÉÙÐèÒªÕâÖÖÌú¿óʯ1.97t¡£

£¨2£©¢Ùn(C)=

m(C)=0.010mol¡Á12g¡¤mol-1=0.12g

´ð£º´Ë¸ÖÑù·ÛÄ©ÖÐÌúºÍ̼µÄÎïÖʵÄÁ¿Ö®±ÈΪ50£º1.

¢Ú¸ù¾ÝʵÑé¢ó¿ÉµÃ£º

  Fe+H2SO4==FeSO4+H2

        1 mol                 22.4L

    n(H2SO4)             2.800L

n(H2SO4)=0.125mol

c(H2SO4)==1.25 mol¡¤L-1

´ð£ºÁòËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ1.25mol¡¤L-1.

¢Ûµ±¸ÖÑù·ÛÄ©ÖеÄÌúδȫ²¿ÈÜÒº½âʱ£¨m£¾1.406g£©£¬Ê£ÓàµÄ¹ÌÌåÖÊÁ¿Îª£º

£¨5.624g+mg£©-0.125mol¡Á56g¡¤mol-1=(m-1.376)g

µ±¸ÖÑù·ÛÄ©ÖеÄÌúÈ«²¿ÈÜÒº½âʱ£¨m1.406g£©£¬Ê£ÓàµÄ¹ÌÌåÖÊÁ¿Îª£º£¨5.624£«m£©g¡Á

´ð£ºµ±ÌúδÍêÈ«Èܽâʱ£¬Ê£Óà¹ÌÌåµÄÖÊÁ¿Îª£¨m-1.376£©g;µ±ÌúÍêÈ«Èܽâʱ£¬Ê£Óà¹ÌÌåµÄÖÊÁ¿Îª¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2011?µÂÖÝÄ£Ä⣩¸ÖÌú¹¤ÒµÊǹú¼Ò¹¤ÒµµÄ»ù´¡£¬¸ÖÌúÉúÐâÏÖÏóÈ´Ëæ´¦¿É¼û£¬Îª´ËÿÄê¹ú¼ÒËðʧ´óÁ¿×ʽð£®Çë»Ø´ð¸ÖÌú¸¯Ê´Óë·À»¤¹ý³ÌÖеÄÓйØÎÊÌ⣮

£¨1£©¸ÖÌúµÄµç»¯Ñ§¸¯Ê´Ô­ÀíÈçͼ1Ëùʾ£º
¢Ùд³öʯīµç¼«µÄµç¼«·´Ó¦Ê½
O2+4e-+2H2O¨T4OH-
O2+4e-+2H2O¨T4OH-
£»
¢Ú½«¸Ã×°ÖÃ×÷¼òµ¥Ð޸ļ´¿É³ÉΪ¸ÖÌúµç»¯Ñ§·À»¤µÄ×°Öã¬ÇëÔÚͼ1ÐéÏß¿òÄÚËùʾλÖÃ×÷³öÐ޸쬲¢ÓüýÍ·±ê³öµ¼ÏßÖеç×ÓÁ÷¶¯·½Ïò£®
¢Ûд³öÐ޸ĺóʯīµç¼«µÄµç¼«·´Ó¦Ê½
2Cl--2e-¨TCl2¡ü
2Cl--2e-¨TCl2¡ü
£®
£¨2£©Éú²úÖпÉÓÃÑÎËáÀ´³ýÌúÐ⣮ÏÖ½«Ò»ÉúÐâµÄÌúƬ·ÅÈëÑÎËáÖУ¬µ±ÌúÐâ±»³ý¾¡ºó£¬ÈÜÒºÖз¢ÉúµÄ»¯ºÏ·´Ó¦µÄ»¯Ñ§·½³Ìʽ
2FeCl3+Fe¨T3FeCl2
2FeCl3+Fe¨T3FeCl2
£®
£¨3£©ÔÚʵ¼ÊÉú²úÖУ¬¿ÉÔÚÌú¼þµÄ±íÃæ¶ÆÍ­·ÀÖ¹Ìú±»¸¯Ê´£®×°ÖÃʾÒâÈçͼ2£º
¢ÙAµç¼«¶ÔÓ¦µÄ½ðÊôÊÇ
Í­
Í­
£¨Ð´ÔªËØÃû³Æ£©£¬Bµç¼«µÄµç¼«·´Ó¦Ê½ÊÇ
Cu2++2e-¨TCu
Cu2++2e-¨TCu
£®
¢ÚÈôµç¶ÆÇ°Ìú¡¢Í­Á½Æ¬½ðÊôÖÊÁ¿Ïàͬ£¬µç¶ÆÍê³Éºó½«ËüÃÇÈ¡³öÏ´¾»¡¢ºæ¸É¡¢³ÆÁ¿£¬¶þÕßÖÊÁ¿²îΪ5.12g£¬Ôòµç¶ÆÊ±µç·ÖÐͨ¹ýµÄµç×ÓΪ
0.08
0.08
 mol£®
¢Û¶Æ²ãÆÆËðºó£¬¶ÆÍ­Ìú±È¶ÆÐ¿Ìú¸üÈÝÒ×±»¸¯Ê´£¬Çë¼òҪ˵Ã÷Ô­Òò
Ìú±ÈÍ­»îÆÃ£¬¶Æ²ãÆÆ»µºó£¬ÔÚ³±Êª»·¾³ÖÐÐγÉÔ­µç³Ø£¬ÌúΪ¸º¼«£¬¼ÓËÙÌúµÄ¸¯Ê´
Ìú±ÈÍ­»îÆÃ£¬¶Æ²ãÆÆ»µºó£¬ÔÚ³±Êª»·¾³ÖÐÐγÉÔ­µç³Ø£¬ÌúΪ¸º¼«£¬¼ÓËÙÌúµÄ¸¯Ê´
£®
¸ÖÌú¹¤ÒµÊǹú¼Ò¹¤ÒµµÄ»ù´¡£¬¸ÖÌúÉúÐâÏÖÏóÈ´Ëæ´¦¿É¼û£¬Îª´ËÿÄê¹ú¼ÒËðʧ´óÁ¿×ʽð£®Çë»Ø´ð¸ÖÌú¸¯Ê´Óë·À»¤¹ý³ÌÖеÄÓйØÎÊÌ⣮
¾«Ó¢¼Ò½ÌÍø
£¨1£©¸ÖÌúµÄµç»¯Ñ§¸¯Ê´Ô­Àí£¬ÔÚËáÐÔ»·¾³Öз¢ÉúÎöÇⸯʴ£¬ÔÚÖÐÐÔ»ò¼îÐÔ»·¾³Öз¢ÉúÎüÑõ¸¯Ê´£®
¢Ù·Ö±ðд³öͼ1ÖÐÌúµç¼«ºÍʯīµç¼«µÄµç¼«·´Ó¦Ê½£º
 
£®
¢Ú½«¸Ã×°ÖÃ×÷¼òµ¥Ð޸ļ´¿É³ÉΪ¸ÖÌúµç»¯Ñ§·À»¤µÄ×°Öã¬ÇëÔÚͼ1ÐéÏß¿òÄÚËùʾλÖÃ×÷³öÐ޸쬲¢ÓüýÍ·±ê³öµ¼ÏßÖеç×ÓÁ÷¶¯·½Ïò£®
¢Ûд³öÐ޸ĺóʯīµç¼«µÄµç¼«·´Ó¦Ê½
 
£®
£¨2£©¹¤ÒµÉϳ£ÓÃÑÎËá³ýÈ¥ÌúÐ⣮ÏÖ½«Ò»ÉúÐâµÄÌúƬ·ÅÈëÑÎËáÖУ¬ÈÜÒºÖпÉÄÜ·¢ÉúµÄ»¯Ñ§·´Ó¦µÄ»¯Ñ§·½³ÌʽÓÐ
 
£®
£¨3£©ÔÚʵ¼ÊÉú²úÖУ¬¿ÉÔÚÌúÖÆÆ·µÄ±íÃæ¶ÆÍ­·ÀÖ¹Ìú±»¸¯Ê´£®×°ÖÃʾÒâÈçͼ2£º
¢ÙAµç¼«¶ÔÓ¦µÄ½ðÊôÊÇ£¨Ð´ÔªËØÃû³Æ£©
 
£¬Bµç¼«µÄµç¼«·´Ó¦Ê½ÊÇ
 
£»
¢ÚÈôµç¶ÆÇ°Ìú¡¢Í­Á½Æ¬½ðÊôÖÊÁ¿Ïàͬ£¬µç¶ÆÍê³Éºó½«ËüÃÇÈ¡³öÏ´¾»¡¢ºæ¸É¡¢³ÆÁ¿£¬¶þÕßÖÊÁ¿²îΪ5.12g£¬Ôòµç¶ÆÊ±µç·ÖÐͨ¹ýµÄµç×ÓΪ
 
mol£®
¢Û¶Æ²ãÆÆËðºó£¬¶ÆÍ­Ìú±È¶ÆÐ¿Ìú¸üÈÝÒ×±»¸¯Ê´£¬Çë¼òҪ˵Ã÷Ô­Òò
 
£®

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

¾«Ó¢¼Ò½ÌÍø