ÌâÄ¿ÄÚÈÝ

ÄÆ¡¢Ã¾¡¢ÂÁ¡¢ÌúµÈ½ðÊôÔÚÈÕ³£Éú»î¡¢¹¤ÒµÉú²ú¡¢º½ÌìÊÂÒµÖÐÓÐ׏㷺µÄÓÃ;£®
£¨1£©4.8gAlÔÚ¿ÕÆøÖзÅÖÃÒ»¶Îʱ¼äÖ®ºó£¬ÖÊÁ¿±äΪ5.28g£¬Ôòδ·¢Éú±ä»¯µÄÂÁµÄÖÊÁ¿Îª
 
g£®
£¨2£©Ïò10mL 0.2mol/LµÄÂÈ»¯ÂÁÈÜÒºÖÐÖðµÎ¼ÓÈëδ֪Ũ¶ÈµÄÇâÑõ»¯±µÈÜÒº£¬²âµÃµÎ¼Ó10mLÓëµÎ¼Ó30mLʱËùµÃ³ÁµíÒ»Ñù¶à£®ÇóÇâÑõ»¯±µÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È£®
£¨3£©½«Ã¾¡¢ÂÁºÏ½ð6.3g»ìºÏÎï¼ÓÈ뵽ijŨ¶ÈµÄÏõËáÖУ¬²úÉú±ê¿öϵÄÆøÌå8.96L£¨¼ÙÉè·´Ó¦¹ý³ÌÖл¹Ô­²úÎïÊÇNOºÍNO2µÄ»ìºÏÆøÌ壩£¬ÏòËùµÃÈÜÒºÖмÓÈ백ˮÖÁ³Áµí´ïµ½×î´óÁ¿£¬³Áµí¸ÉÔïºó²âµÃµÄÖÊÁ¿±ÈÔ­ºÏ½ðµÄÖÊÁ¿Ôö¼Ó10.2g£¬¼ÆË㻹ԭ²úÎïÖÐNOºÍNO2µÄÎïÖʵÄÁ¿Ö®±È£®
£¨4£©ÏÖÓÐÒ»°üÂÁÈȼÁÊÇÂÁ·ÛºÍÑõ»¯Ìú·ÛÄ©µÄ»ìºÏÎÔÚ¸ßÎÂÏÂʹ֮³ä·Ö·´Ó¦£¬½«·´Ó¦ºóµÄ¹ÌÌå·ÖΪÁ½µÈ·Ý£¬½øÐÐÈçÏÂʵÑ飨¼ÆËãpHʱ¼Ù¶¨ÈÜÒºÌå»ýûÓб仯£©£º¢ÙÏòÆäÖÐÒ»·Ý¹ÌÌåÖмÓÈë50mL 3.0mol/LµÄNaOHÈÜÒº£¬¼ÓÈÈʹÆä³ä·Ö·´Ó¦ºó¹ýÂË£¬²âµÃÂËÒºµÄc£¨OH-£©Îª1.0mol/L£»¢ÚÏòÁíÒ»·Ý¹ÌÌåÖмÓÈë100mL 4.0mol/LµÄHClÈÜÒº£¬Ê¹¹ÌÌåÈ«²¿Èܽ⣬²âµÃ·´Ó¦ºóËùµÃÈÜÒºÖÐÖ»ÓÐH+¡¢Fe2+ºÍAl3+ÈýÖÖÑôÀë×ÓÇÒc£¨H+£©Îª0.2mol/L£¬¼ÆËãÕâ°üÂÁÈȼÁÖÐÂÁµÄÖÊÁ¿ºÍÑõ»¯ÌúµÄÖÊÁ¿£®
·ÖÎö£º£¨1£©¸ù¾ÝÂÁºÍÑõÆøµÄ¹ØϵÀûÓòîÁ¿·¨¼ÆË㣻
£¨2£©ÏòÂÈ»¯ÂÁÈÜÒºÖÐÖðµÎ¼ÓÈëÇâÑõ»¯±µÈÜҺʱ·¢ÉúµÄÀë×Ó·´Ó¦·½³ÌʽΪ£ºAl3++3OH-=Al£¨OH£©3¡ý¡¢Al£¨OH£©3+OH-=AlO2-+H2O£¬¸ù¾ÝÎïÖÊÖ®¼äµÄ¹Øϵʽ¼ÆË㣻
£¨3£©¸ù¾Ý½ðÊôµÄÖÊÁ¿¡¢×ª»¯Îª³Áµíʱ¹ÌÌåÔö¼ÓµÄÖÊÁ¿¼ÆËãþ¡¢ÂÁµÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾ÝÑõ»¯»¹Ô­·´Ó¦ÖеÃʧµç×ÓÏàËƼÆËãÒ»Ñõ»¯µªºÍ¶þÑõ»¯Ì¼µÄÎïÖʵÄÁ¿Ö®±È£»
£¨4£©¸ù¾ÝÌâÒâ¢ÙÖª£¬ÇâÑõ»¯ÄƹýÁ¿£¬¸ù¾Ý·´Ó¦µÄÇâÑõ»¯ÄƼÆËãÆ«ÂÁËáÄƵÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾ÝÂÁÔ­×ÓÊغã¼ÆËãÂÁµÄÖÊÁ¿£»
¢ÚÈÜÒº³ÊËáÐÔ£¬ËµÃ÷ÈÜÒºÖÐÑÎËá¹ýÁ¿£¬¸ù¾Ý¢ÙÖª£¬ÂÁÀë×ÓµÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾ÝµçºÉÊغã¼ÆËãÌúÀë×ÓµÄÎïÖʵÄÁ¿£¬¸ù¾ÝÖÊÁ¿¹«Ê½¼ÆËãÑõ»¯ÌúµÄÖÊÁ¿£®
½â´ð£º½â£º£¨1£©ÂÁºÍÑõÆøµÄ·´Ó¦·½³ÌʽΪ4Al+3O2=2Al2O3£¬¸ù¾Ý·½³Ìʽ֪£¬¹ÌÌåÔö¼ÓµÄÖÊÁ¿¾ÍÊÇÑõÆøµÄÖÊÁ¿£¬
Éèδ²Î¼Ó·´Ó¦µÄÂÁµÄÖÊÁ¿Îªag£¬Ôò²Î¼Ó·´Ó¦µÄÂÁµÄÖÊÁ¿Îª£¨4.8-a£©g
4Al+3O2=2Al2O3  ¹ÌÌåÖÊÁ¿Ôö¼Ó
108             96
£¨4.8-a£©g      £¨5.28-4.8£©g  
108£º96=£¨4.8-a£©g£º£¨5.28-4.8£©g  
9£º8=£¨4.8-a£©£º0.48£¬
a=4.26g£¬
¹Ê´ð°¸Îª£º4.26g£»
£¨2£©n£¨AlCl3£©=0.2mol/L¡Á0.01L=0.002mol£¬
ÉèÇâÑõ»¯±µµÄÎïÖʵÄÁ¿Å¨¶ÈΪx£¬
2AlCl3+3Ba£¨OH£©2=2Al£¨OH£©3¡ý+3BaCl2
2        3         2      
        0.01xmol 
0.01x¡Á2
3
mol
¸ù¾ÝÂÁÔ­×ÓÊغãÖª£¬µ±Íêȫת»¯ÎªÇâÑõ»¯ÂÁ³Áµíʱ£¬n£¨[Al£¨OH£©3]=n£¨AlCl3£©=0.002mol£¬
¸ù¾Ý2AlCl3+3Ba£¨OH£©2=2Al£¨OH£©3¡ý+3BaCl2Öª£¬ÍêÈ«³ÁµíʱÐèÒªn[Ba£¨OH£©2]=
0.002mol
2
¡Á3
=0.003mol£¬
¸ù¾ÝÌâÒâÖª£¬ÓëÇâÑõ»¯ÂÁ·´Ó¦µÄÇâÑõ»¯±µµÄÎïÖʵÄÁ¿=£¨0.03x-0.003£©mol£¬
¸ù¾Ý2Al£¨OH£©3+Ba£¨OH£©2=Ba£¨AlO2£©2+4H2O¿ÉÖª£¬Éú³ÉÆ«ÂÁËá±µÐèÒªn[Al£¨OH£©3]=2n[Ba£¨OH£©2]=2£¨0.03x-0.003£©mol£¬
¸ù¾ÝÂÁÔ­×ÓÊغãµÃÊ£Óàn[Al£¨OH£©3]=0.002mol-2£¨0.03x-0.003£©mol£¬
²âµÃµÎ¼Ó10mLÓëµÎ¼Ó30mLËùµÃ³ÁµíͬÑù¶à£¬ËùÒÔ
0.02x
3
mol=0.002mol-2£¨0.03x-0.003£©mol£¬
x=0.12mol/L£¬
´ð£ºÇâÑõ»¯±µµÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.12 mol/L£»
£¨3£©Ã¾¡¢ÂÁ×îºóת»¯Îª³ÁµíÇâÑõ»¯Ã¾ºÍÇâÑõ»¯ÂÁ£¬¹ÌÌåÔö´óµÄÖÊÁ¿ÊÇÇâÑõ¸ùÀë×ÓµÄÖÊÁ¿£¬ÔòÇâÑõ¸ùÀë×ÓµÄÎïÖʵÄÁ¿=
10.2g
17g/mol
=0.6mol£¬
ÉèþµÄÎïÖʵÄÁ¿Îªxmol£¬ÂÁµÄÎïÖʵÄÁ¿Îªymol£¬¸ù¾Ý½ðÊôµ¥ÖʵÄÖÊÁ¿¡¢×ª»¯Îª³ÁµíÔö¼ÓµÄÖÊÁ¿Áз½³ÌµÃ
24x+27y=6.3g
2x+3y=0.6
£¬
½âµÃ
x=0.15
y=0.1
£¬
þ¡¢ÂÁºÍÏõËá·´Ó¦Éú³ÉµªÑõ»¯ÎïʱתÒƵç×ÓÏàµÈ£¬µªÑõ»¯ÎïµÄÎïÖʵÄÁ¿=
8.96L
22.4L/mol
=0.4mol£¬Ã¿¸ö¶þÑõ»¯µª¡¢Ò»Ñõ»¯µª·Ö×ÓÖж¼Ö»º¬Ò»¸öµªÔ­×Ó£¬
ÉèÒ»Ñõ»¯µª¡¢¶þÑõ»¯µªµÄÎïÖʵÄÁ¿·Ö±ðΪn£¨NO£©¡¢n£¨NO2£©£¬
n(NO)+n(NO2)=0.4
n(NO)¡Á3+n(NO2)¡Á1=0.15¡Á2+0.1¡Á3
£¬
½âµÃ
n(NO)=0.1
n(NO2)=0.3
£¬
ÔòÒ»Ñõ»¯µªºÍ¶þÑõ»¯µªµÄÎïÖʵÄÁ¿Ö®±ÈΪ0.1mol£º0.3mol=1£º3£¬
´ð£ºÒ»Ñõ»¯µªºÍ¶þÑõ»¯µªµÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º3£»
£¨4£©¢ÙÖÐÇâÑõ»¯ÄƹýÁ¿£¬ÈÜÒºÖеÄÈÜÖÊÊÇÆ«ÂÁËáÄƺÍÇâÑõ»¯ÄÆ£¬ÈÜÒºÖÐn£¨NaOH£©=1.0mol/L¡Á0.05L=0.05mol£¬¸ù¾ÝÄÆÔ­×ÓÊغãµÃn£¨NaAlO2£©=£¨3.0-1.0£©mol/L¡Á0.05L=0.1mol£¬¸ù¾ÝÂÁÔ­×ÓÊغãÖª£¬ÂÁµÄÎïÖʵÄÁ¿Ò²ÊÇ0.1mol£¬Ôòm£¨Al£©=0.1mol¡Á27g/mol=2.7g£»Ã¿·ÝÖÐÂÁµÄÖÊÁ¿ÏàµÈ£¬ËùÒÔÂÁÈȼÁÖÐÂÁµÄÖÊÁ¿ÊÇ5.4g£»
¢Ú¸ÃÈÜÒº³ÊËáÐÔ£¬ÔòÑÎËá¹ýÁ¿£¬¸ù¾Ý¢ÚÖª£¬¸ÃÈÜÒºÖÐÂÁÀë×ÓµÄÎïÖʵÄÁ¿Îª0.1mol£¬ÂÁÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶È=
0.1mol
0.1L
=1mol/L£¬ÈÜÒº³ÊµçÖÐÐÔ£¬ÒõÑôÀë×ÓËù´øµçºÉÏàµÈ£¬ÉèÑÇÌúÀë×ÓµÄÎïÖʵÄÁ¿Îªn£¨Fe 2+ £©£¬¸ù¾ÝµçºÉÊغãµÃc£¨H+£©+2 c£¨Fe 2+ £©+3n£¨Al3+ £©=c£¨Cl-£©£¬
0.2mol/L+2c£¨Fe 2+£©+3¡Á1mol/L=4.0mol/L£¬
 c£¨Fe 2+£©=0.4mol/L£¬¸ù¾ÝÌúÔ­×ÓÊغãµÃÔòn£¨Fe2O3£©=
1
2
n£¨Fe 2+£©=
1
2
¡Á0.4mol/L¡Á0.1L=0.02mol£¬m£¨Fe2O3£©=0.02mol¡Á160g/mol=3.2g£¬Á½·ÝÖÐÑõ»¯ÌúÖÊÁ¿ÏàµÈ£¬ËùÒÔÂÁÈȼÁÖÐÑõ»¯ÌúµÄÖÊÁ¿Îª6.4g£¬
´ð£ºÂÁÈȼÁÖÐÂÁµÄÖÊÁ¿ÊÇ5.4g£¬Ñõ»¯ÌúµÄÖÊÁ¿ÊÇ6.4g£®
µãÆÀ£º±¾ÌâÒÔÂÁΪÔØÌ忼²éÁËÎïÖʵÄÁ¿µÄÓйؼÆË㣬Ã÷È·ÎïÖÊÖ®¼äµÄ·´Ó¦½áºÏÔ­×ÓÊغ㡢תÒƵç×ÓÊغãÀ´·ÖÎö½â´ð£¬×¢Ò⣨4£©ÖкܶàͬѧÍùÍùÖ»¼ÆËãÒ»·ÝÖÐÂÁºÍÑõ»¯ÌúµÄÖÊÁ¿¶øµ¼Ö´íÎó£¬ÎªÒ×´íµã£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨ 12·Ö£©îÑ£¨Ti£©±»³ÆΪ¼ÌÌú¡¢ÂÁÖ®ºóµÄµÚÈý½ðÊô£¬¶ÛîѺÍÒÔîÑΪÖ÷µÄºÏ½ðÊÇÐÂÐ͵Ľṹ²ÄÁÏ£¬Ö÷ÒªÓÃÓÚº½Ì칤ҵºÍº½º£¹¤Òµ£¬ÏÂÁÐÊÇÓйØîѵÄÒ±Á¶¼°Ó¦ÓõÄÎÊÌâ¡£
£¨1£©½ðÊôîÑÒ±Á¶¹ý³ÌÖÐÆäÖÐÒ»²½·´Ó¦Êǽ«Ô­ÁϽðºìʯת»¯£ºTiO2£¨½ðºìʯ£©+2C+2Cl2 TiCl4+2CO
ÒÑÖª£ºC£¨S£©+O2£¨g£©£½CO2£¨g£©        H£½£­393£®5 kJ¡¤mol-1
2CO£¨g£©£«O2£¨g£©£½2CO2£¨g£© H£½-566 kJ¡¤mol-1
TiO2£¨s£©+2Cl2£¨g£©£½TiCl4£¨s£©+O2£¨g£© H£½+141 kJ¡¤mol-1
ÔòTiO2£¨g£©+2Cl2£¨g£©+2C£¨s£©£½TiCl4£¨s£©+2CO£¨g£©µÄH£½                     £¬
£¨2£©ÄÆÈÈ»¹Ô­·¨ÊÇÒ±Á¶½ðÊôîѵķ½·¨Ö®Ò»£¬Ö÷Òª·´Ó¦Ô­ÀíΪ£º4Na+TiCl   4NaCl+Ti£¬¸Ã·´Ó¦²»ÄÜÔÚË®ÈÜÒºÖнøÐУ¬Ò»ÊÇÒòΪTiCl4»áÇ¿ÁÒË®½âÉú³ÉTiO2£¬ÁíÒ»Ô­Òò                                                             £¨ÓÃÊʵ±»¯Ñ§·½³Ìʽ¸¨ÒÔ±ØÒªµÄÎÄ×Ö˵Ã÷£©¡£
£¨3£©Ã¾»¹Ô­·¨Ò²ÊÇÒ±Á¶½ðÊôîѵij£Ó÷½·¨£¬ÆäÖ÷Òª·´Ó¦Ô­ÀíÈçÏ£º
MgCl2 Mg+Cl2               TiCl4+2Mg  2MgCl2+Ti
´Óº£Ë®ÖÐÌáÈ¡MgCl2ʱ£¬ÏÈÔÚº£Ë®ÖмÓÈëÊìʯ»Ò£¬³Áµí³öMg£¨OH£©2£¬Ð´³öMg£¨OH£©2ÈܶȻý±í´ïʽ£º                          
¿É¼ÓÈëÊʵ±¹ýÁ¿µÄÊìʯ»Ò£¬´ÓMg£¨OH£©2Èܽâƽºâ½Ç¶È½âÊÍÆäÔ­Òò               
£¨4£©TiCl4ÓëLiOHÔÚË®ÈÜÒºÖÐÒ»¶¨Ìõ¼þÏ¿ɷ´Ó¦Éú³ÉLi4Ti5O12£¨îÑËáﮣ©£¬Li4Ti5O12¿ÉÓëLiMn2O4£¨ÃÌËáﮣ©µÈÕý¼«²ÄÁÏ×é³ÉÀíÀë×Ó¶þ´Îµç³Ø£¬¹¤×÷ʱLi+ ÔÚµç³ØÄÚ¶¨ÏòÒƶ¯£¬Æäµç³Ø·´Ó¦Îª£º£¬Ê¹ÓÃʱÏȳäµç£¬Ð´³öÆä³äµçʽµÄÑô¼«·´Ó¦          £¬·ÅµçʱLi+µÄÒƶ¯·½Ïò          ¡£

£¨ 12·Ö£©îÑ£¨Ti£©±»³ÆΪ¼ÌÌú¡¢ÂÁÖ®ºóµÄµÚÈý½ðÊô£¬¶ÛîѺÍÒÔîÑΪÖ÷µÄºÏ½ðÊÇÐÂÐ͵Ľṹ²ÄÁÏ£¬Ö÷ÒªÓÃÓÚº½Ì칤ҵºÍº½º£¹¤Òµ£¬ÏÂÁÐÊÇÓйØîѵÄÒ±Á¶¼°Ó¦ÓõÄÎÊÌâ¡£

£¨1£©½ðÊôîÑÒ±Á¶¹ý³ÌÖÐÆäÖÐÒ»²½·´Ó¦Êǽ«Ô­ÁϽðºìʯת»¯£ºTiO2£¨½ðºìʯ£©+2C+2Cl2  TiCl4+2CO

ÒÑÖª£ºC£¨S£©+O2£¨g£©£½CO2£¨g£©         H£½£­393£®5 kJ¡¤mol-1

2CO£¨g£©£«O2£¨g£©£½2CO2£¨g£©  H£½-566 kJ¡¤mol-1

TiO2£¨s£©+2Cl2£¨g£©£½TiCl4£¨s£©+O2£¨g£©  H£½+141 kJ¡¤mol-1

ÔòTiO2£¨g£©+2Cl2£¨g£©+2C£¨s£©£½TiCl4£¨s£©+2CO£¨g£©µÄH£½                      £¬

£¨2£©ÄÆÈÈ»¹Ô­·¨ÊÇÒ±Á¶½ðÊôîѵķ½·¨Ö®Ò»£¬Ö÷Òª·´Ó¦Ô­ÀíΪ£º4Na+TiCl   4NaCl+Ti£¬¸Ã·´Ó¦²»ÄÜÔÚË®ÈÜÒºÖнøÐУ¬Ò»ÊÇÒòΪTiCl4»áÇ¿ÁÒË®½âÉú³ÉTiO2£¬ÁíÒ»Ô­Òò                                                              £¨ÓÃÊʵ±»¯Ñ§·½³Ìʽ¸¨ÒÔ±ØÒªµÄÎÄ×Ö˵Ã÷£©¡£

£¨3£©Ã¾»¹Ô­·¨Ò²ÊÇÒ±Á¶½ðÊôîѵij£Ó÷½·¨£¬ÆäÖ÷Òª·´Ó¦Ô­ÀíÈçÏ£º

         MgCl2  Mg+Cl2               TiCl4+2Mg  2MgCl2+Ti

´Óº£Ë®ÖÐÌáÈ¡MgCl2ʱ£¬ÏÈÔÚº£Ë®ÖмÓÈëÊìʯ»Ò£¬³Áµí³öMg£¨OH£©2£¬Ð´³öMg£¨OH£©2ÈܶȻý±í´ïʽ£º                           

¿É¼ÓÈëÊʵ±¹ýÁ¿µÄÊìʯ»Ò£¬´ÓMg£¨OH£©2Èܽâƽºâ½Ç¶È½âÊÍÆäÔ­Òò               

£¨4£©TiCl4ÓëLiOHÔÚË®ÈÜÒºÖÐÒ»¶¨Ìõ¼þÏ¿ɷ´Ó¦Éú³ÉLi4Ti5O12£¨îÑËáﮣ©£¬Li4Ti5O12¿ÉÓëLiMn2O4£¨ÃÌËáﮣ©µÈÕý¼«²ÄÁÏ×é³ÉÀíÀë×Ó¶þ´Îµç³Ø£¬¹¤×÷ʱLi+ ÔÚµç³ØÄÚ¶¨ÏòÒƶ¯£¬Æäµç³Ø·´Ó¦Îª£º£¬Ê¹ÓÃʱÏȳäµç£¬Ð´³öÆä³äµçʽµÄÑô¼«·´Ó¦           £¬·ÅµçʱLi+µÄÒƶ¯·½Ïò           ¡£

 

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

¾«Ó¢¼Ò½ÌÍø