ÌâÄ¿ÄÚÈÝ
£¨2010?·ą́Çøһģ£©£¨1£©»·¾³×¨¼ÒÈÏΪ¿ÉÒÔÓýðÊôÂÁ½«Ë®ÌåÖеÄNO3-ת»¯ÎªN2£¬´Ó¶øÇå³ýÎÛȾ£®¸Ã·´Ó¦ÖÐÉæ¼°µÄÁ£×ÓÓУºH2O¡¢Al¡¢OH-¡¢Al£¨OH£©3¡¢NO3-¡¢N2£¬Ç뽫¸÷Á£×Ó·Ö±ðÌîÈëÒÔÏ¿ոñÄÚ£¨²»ÓÃÅäƽ£©£®¸Ã·´Ó¦¹ý³ÌÖУ¬±»Ñõ»¯Óë±»»¹ÔµÄÔªËصÄÎïÖʵÄÁ¿Ö®±ÈΪ
£¨2£©ÎÒ¹úÊ×´´µÄº£Ñóµç³ØÒÔº£Ë®Îªµç½âÖÊÈÜÒº£¬µç³Ø×Ü·´Ó¦Îª£º4Al+3O2+6H2O=4Al£¨OH£©3£®µç³ØÕý¼«µÄµç¼«·´Ó¦Ê½Îª
£¨3£©ÒÑÖª£º4Al£¨s£©+3O2£¨g£©=2Al2O3£¨g£©¡÷H=-2834.9kJ/mol
Fe2O3£¨s£©+
C£¨s£©=
CO2£¨g£©+2Fe£¨s£©¡÷H=234.1kJ/mol
C£¨s£©+O2£¨g£©=CO2£¨g£©¡÷H=-393.5kJ/mol
д³öÂÁÓëÑõ»¯Ìú·¢ÉúÂÁÈÈ·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
£¨4£©½«Ò»¶¨ÖÊÁ¿µÄÄÆÂÁºÏ½ðÖÃÓÚË®ÖУ¬ºÏ½ðÈ«²¿Èܽ⣬µÃµ½20mol£¬pH=14µÄÈÜÒº£¬È»ºó2mol/LÑÎËáµÎ¶¨£¬¿ÉµÃ³ÁµíÖÊÁ¿ÓëÏûºÄµÄÑÎËáÌå»ý¹ØϵÈçÏÂͼ£ºÔò·´Ó¦¹ý³ÌÖвúÉúÇâÆøµÄ×ÜÌå»ýΪ
5£º3
5£º3
£®£¨2£©ÎÒ¹úÊ×´´µÄº£Ñóµç³ØÒÔº£Ë®Îªµç½âÖÊÈÜÒº£¬µç³Ø×Ü·´Ó¦Îª£º4Al+3O2+6H2O=4Al£¨OH£©3£®µç³ØÕý¼«µÄµç¼«·´Ó¦Ê½Îª
2H2O+O2+4e-=4OH-
2H2O+O2+4e-=4OH-
£»Õý¼«²ÄÁϲÉÓÃÁ˲¬Íø£¬ÀûÓò¬ÍøΪÕý¼«µÄÓŵãÊDz¬ÐÔÖÊÎȶ¨²»ÓëÑõÆø·´Ó¦£¬Íø×´½á¹¹¿ÉÒÔÔö´óÓëÑõÆøµÄ½Ó´¥Ãæ»ý
²¬ÐÔÖÊÎȶ¨²»ÓëÑõÆø·´Ó¦£¬Íø×´½á¹¹¿ÉÒÔÔö´óÓëÑõÆøµÄ½Ó´¥Ãæ»ý
£®£¨3£©ÒÑÖª£º4Al£¨s£©+3O2£¨g£©=2Al2O3£¨g£©¡÷H=-2834.9kJ/mol
Fe2O3£¨s£©+
3 |
2 |
3 |
2 |
C£¨s£©+O2£¨g£©=CO2£¨g£©¡÷H=-393.5kJ/mol
д³öÂÁÓëÑõ»¯Ìú·¢ÉúÂÁÈÈ·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
2Al£¨s£©+Fe2O3£¨s£©=2Fe£¨s£©=2Fe£¨s£©+Al2O3£¨s£©¡÷H=-593.1kJ/mol
2Al£¨s£©+Fe2O3£¨s£©=2Fe£¨s£©=2Fe£¨s£©+Al2O3£¨s£©¡÷H=-593.1kJ/mol
£®£¨4£©½«Ò»¶¨ÖÊÁ¿µÄÄÆÂÁºÏ½ðÖÃÓÚË®ÖУ¬ºÏ½ðÈ«²¿Èܽ⣬µÃµ½20mol£¬pH=14µÄÈÜÒº£¬È»ºó2mol/LÑÎËáµÎ¶¨£¬¿ÉµÃ³ÁµíÖÊÁ¿ÓëÏûºÄµÄÑÎËáÌå»ý¹ØϵÈçÏÂͼ£ºÔò·´Ó¦¹ý³ÌÖвúÉúÇâÆøµÄ×ÜÌå»ýΪ
2.016
2.016
L£¨±ê×¼×´¿ö£©£®·ÖÎö£º£¨1£©¸ù¾Ý½ðÊôÂÁ½«Ë®ÌåÖеÄNO3-ת»¯ÎªN2£¬¿ÉÖªAlÔªËصĻ¯ºÏ¼ÛÉý¸ß£¬ÔÚ·´Ó¦Öб»Ñõ»¯£¬NÔªËصĻ¯ºÏ¼Û½µµÍ£¬ÔÚ·´Ó¦Öб»»¹Ô£¬ÀûÓõç×ÓÊغãÀ´¼ÆË㣻
£¨2£©¸ù¾Ýµç³Ø×Ü·´Ó¦4Al+3O2+6H2O=4Al£¨OH£©3¿ÉÖª£¬Al×÷¸º¼«£¬ÑõÆøÔÚÕý¼«Éϵõç×Ó·¢Éú»¹Ô·´Ó¦£¬ÔÙÀûÓò¬µÄÐÔÖʼ°Íø×´½á¹¹·ÖÎöÆäÓŵ㣻
£¨3£©¸ù¾ÝÒÑÖªµÄ·´Ó¦ºÍ¸Ç˹¶¨ÂÉÀ´¼ÆËãÄ¿±ê·´Ó¦µÄ·´Ó¦ÈÈ£¬²¢ÊéдÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£»
£¨4£©¸ù¾Ý2Na+2H2O¨T2NaOH+H2¡ü¡¢2Al+2H2O+2NaOH¨T2NaAlO2+3H2¡ü¡¢NaOH+HCl¨TNaCl+H2O¡¢NaAlO2+HCl+H2O¨TAl£¨OH£©3¡ý+NaCl£¬½áºÏͼÏóÀ´¼ÆËã½ðÊôµÄÎïÖʵÄÁ¿£¬ÔÙÀûÓõç×ÓÊغãÀ´¼ÆËãÇâÆøµÄÎïÖʵÄÁ¿£¬×îºó¼ÆËãÆäÌå»ý£®
£¨2£©¸ù¾Ýµç³Ø×Ü·´Ó¦4Al+3O2+6H2O=4Al£¨OH£©3¿ÉÖª£¬Al×÷¸º¼«£¬ÑõÆøÔÚÕý¼«Éϵõç×Ó·¢Éú»¹Ô·´Ó¦£¬ÔÙÀûÓò¬µÄÐÔÖʼ°Íø×´½á¹¹·ÖÎöÆäÓŵ㣻
£¨3£©¸ù¾ÝÒÑÖªµÄ·´Ó¦ºÍ¸Ç˹¶¨ÂÉÀ´¼ÆËãÄ¿±ê·´Ó¦µÄ·´Ó¦ÈÈ£¬²¢ÊéдÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£»
£¨4£©¸ù¾Ý2Na+2H2O¨T2NaOH+H2¡ü¡¢2Al+2H2O+2NaOH¨T2NaAlO2+3H2¡ü¡¢NaOH+HCl¨TNaCl+H2O¡¢NaAlO2+HCl+H2O¨TAl£¨OH£©3¡ý+NaCl£¬½áºÏͼÏóÀ´¼ÆËã½ðÊôµÄÎïÖʵÄÁ¿£¬ÔÙÀûÓõç×ÓÊغãÀ´¼ÆËãÇâÆøµÄÎïÖʵÄÁ¿£¬×îºó¼ÆËãÆäÌå»ý£®
½â´ð£º½â£º£¨1£©ÓɽðÊôÂÁ½«Ë®ÌåÖеÄNO3-ת»¯ÎªN2¿ÉÖª£¬ÔÚNO3-+Al+H2O¡úAl£¨OH£©3+N2+OH-ÖУ¬AlÔªËر»Ñõ»¯£¬NÔªËر»»¹Ô£¬Ôò±»Ñõ»¯Óë±»»¹ÔµÄÔªËصÄÎïÖʵÄÁ¿·Ö±ðΪx¡¢y£¬Óɵç×ÓÊغã¿ÉÖª£¬
x¡Á£¨3-0£©=y¡Á£¨5-0£©£¬
½âµÃx£ºy=5£º3£¬
¼´±»Ñõ»¯Óë±»»¹ÔµÄÔªËصÄÎïÖʵÄÁ¿Ö®±ÈΪ5£º3£¬¹Ê´ð°¸Îª£º5£º3£»
£¨2£©Óɵç³Ø×Ü·´Ó¦4Al+3O2+6H2O=4Al£¨OH£©3¿ÉÖª£¬Al×÷¸º¼«£¬ÑõÆøÔÚÕý¼«Éϵõç×Ó·¢Éú»¹Ô·´Ó¦£¬
Õý¼«·´Ó¦Îª2H2O+O2+4e-=4OH-£¬
ÓÖÕý¼«²ÄÁÏʹÓò¬Íø£¬ÊÇÒò²¬ÐÔÖÊÎȶ¨²»ÓëÑõÆø·´Ó¦£¬Íø×´½á¹¹¿ÉÒÔÔö´óÓëÑõÆøµÄ½Ó´¥Ãæ»ý£¬
¹Ê´ð°¸Îª£º2H2O+O2+4e-=4OH-£»²¬ÐÔÖÊÎȶ¨²»ÓëÑõÆø·´Ó¦£¬Íø×´½á¹¹¿ÉÒÔÔö´óÓëÑõÆøµÄ½Ó´¥Ãæ»ý£»
£¨3£©ÓÉ¢Ù4Al£¨s£©+3O2£¨g£©=2Al2O3£¨g£©¡÷H=-2834.9kJ/mol£¬
¢ÚFe2O3£¨s£©+
C£¨s£©=
CO2£¨g£©+2Fe£¨s£©¡÷H=234.1kJ/mol£¬
¢ÛC£¨s£©+O2£¨g£©=CO2£¨g£©¡÷H=-393.5kJ/mol£¬
¸ù¾Ý¸Ç˹¶¨ÂÉ¿ÉÖª£¬¢Ù¡Á
+¢Ú-¢Û¡Á
¿ÉµÃ£¬
2Al£¨s£©+Fe2O3£¨s£©=2Fe£¨s£©=2Fe£¨s£©+Al2O3£¨s£©¡÷H=-593.1kJ/mol£¬
¹Ê´ð°¸Îª£º2Al£¨s£©+Fe2O3£¨s£©=2Fe£¨s£©=2Fe£¨s£©+Al2O3£¨s£©¡÷H=-593.1kJ/mol£»
£¨4£©ÓÉ2Na+2H2O¨T2NaOH+H2¡ü¡¢2Al+2H2O+2NaOH¨T2NaAlO2+3H2¡ü¡¢NaOH+HCl¨TNaCl+H2O¡¢
NaAlO2+HCl+H2O¨TAl£¨OH£©3¡ý+NaCl¿ÉÖª£¬
·´Ó¦ºón£¨Na£©=n£¨NaCl£©=n£¨NaOH£©+n£¨NaAlO2£©£¬
ÓÉͼÏó¿ÉÖª£¬30mLÑÎËáÉú³É³Áµí×î´ó£¬Ôòn£¨NaCl£©=n£¨HCl£©=0.03L¡Á2mol/L=0.06mol£¬
ÓɺϽðÈ«²¿ÈܽâºóµÃµ½20mol£¬pH=14µÄÈÜÒº£¬n£¨NaOH£©=0.02L¡Á1mol/L=0.02mol£¬
Ôòn£¨Al£©=n£¨NaAlO2£©=0.06mol-0.02mol=0.04mol£¬
¼´ºÏ½ðÖÐNaΪ0.06mol¡¢AlΪ0.04mol£¬Éè·´Ó¦¹ý³ÌÖвúÉúÇâÆøµÄÎïÖʵÄÁ¿Îªn£¬
Óɵç×ÓÊغã¿ÉÖª£¬0.06mol¡Á1+0.04mol¡Á3=n¡Á2¡Á1£¬
½âµÃn=0.09mol£¬
ËùÒÔÔÚ±ê¿öÏÂÆäÌå»ýΪ0.09mol¡Á22.4L/mol=2.016L£¬
¹Ê´ð°¸Îª£º2.016£®
x¡Á£¨3-0£©=y¡Á£¨5-0£©£¬
½âµÃx£ºy=5£º3£¬
¼´±»Ñõ»¯Óë±»»¹ÔµÄÔªËصÄÎïÖʵÄÁ¿Ö®±ÈΪ5£º3£¬¹Ê´ð°¸Îª£º5£º3£»
£¨2£©Óɵç³Ø×Ü·´Ó¦4Al+3O2+6H2O=4Al£¨OH£©3¿ÉÖª£¬Al×÷¸º¼«£¬ÑõÆøÔÚÕý¼«Éϵõç×Ó·¢Éú»¹Ô·´Ó¦£¬
Õý¼«·´Ó¦Îª2H2O+O2+4e-=4OH-£¬
ÓÖÕý¼«²ÄÁÏʹÓò¬Íø£¬ÊÇÒò²¬ÐÔÖÊÎȶ¨²»ÓëÑõÆø·´Ó¦£¬Íø×´½á¹¹¿ÉÒÔÔö´óÓëÑõÆøµÄ½Ó´¥Ãæ»ý£¬
¹Ê´ð°¸Îª£º2H2O+O2+4e-=4OH-£»²¬ÐÔÖÊÎȶ¨²»ÓëÑõÆø·´Ó¦£¬Íø×´½á¹¹¿ÉÒÔÔö´óÓëÑõÆøµÄ½Ó´¥Ãæ»ý£»
£¨3£©ÓÉ¢Ù4Al£¨s£©+3O2£¨g£©=2Al2O3£¨g£©¡÷H=-2834.9kJ/mol£¬
¢ÚFe2O3£¨s£©+
3 |
2 |
3 |
2 |
¢ÛC£¨s£©+O2£¨g£©=CO2£¨g£©¡÷H=-393.5kJ/mol£¬
¸ù¾Ý¸Ç˹¶¨ÂÉ¿ÉÖª£¬¢Ù¡Á
1 |
2 |
3 |
2 |
2Al£¨s£©+Fe2O3£¨s£©=2Fe£¨s£©=2Fe£¨s£©+Al2O3£¨s£©¡÷H=-593.1kJ/mol£¬
¹Ê´ð°¸Îª£º2Al£¨s£©+Fe2O3£¨s£©=2Fe£¨s£©=2Fe£¨s£©+Al2O3£¨s£©¡÷H=-593.1kJ/mol£»
£¨4£©ÓÉ2Na+2H2O¨T2NaOH+H2¡ü¡¢2Al+2H2O+2NaOH¨T2NaAlO2+3H2¡ü¡¢NaOH+HCl¨TNaCl+H2O¡¢
NaAlO2+HCl+H2O¨TAl£¨OH£©3¡ý+NaCl¿ÉÖª£¬
·´Ó¦ºón£¨Na£©=n£¨NaCl£©=n£¨NaOH£©+n£¨NaAlO2£©£¬
ÓÉͼÏó¿ÉÖª£¬30mLÑÎËáÉú³É³Áµí×î´ó£¬Ôòn£¨NaCl£©=n£¨HCl£©=0.03L¡Á2mol/L=0.06mol£¬
ÓɺϽðÈ«²¿ÈܽâºóµÃµ½20mol£¬pH=14µÄÈÜÒº£¬n£¨NaOH£©=0.02L¡Á1mol/L=0.02mol£¬
Ôòn£¨Al£©=n£¨NaAlO2£©=0.06mol-0.02mol=0.04mol£¬
¼´ºÏ½ðÖÐNaΪ0.06mol¡¢AlΪ0.04mol£¬Éè·´Ó¦¹ý³ÌÖвúÉúÇâÆøµÄÎïÖʵÄÁ¿Îªn£¬
Óɵç×ÓÊغã¿ÉÖª£¬0.06mol¡Á1+0.04mol¡Á3=n¡Á2¡Á1£¬
½âµÃn=0.09mol£¬
ËùÒÔÔÚ±ê¿öÏÂÆäÌå»ýΪ0.09mol¡Á22.4L/mol=2.016L£¬
¹Ê´ð°¸Îª£º2.016£®
µãÆÀ£º±¾Ì⿼²éÑõ»¯»¹Ô·´Ó¦¡¢Ôµç³Ø¡¢¸Ç˹¶¨ÂɼÆËã·´Ó¦ÈÈ¡¢»¯Ñ§·´Ó¦ÓëͼÏóµÄ¼ÆË㣬ÄѶȽϴó£¬ÖªÊ¶µÄ×ÛºÏÐԽϸߣ¬£¨4£©ÊÇѧÉú½â´ðÖеÄÄѵ㣬עÒâÊغ㷨ÔÚ½âÌâÖеÄÖØÒª×÷Óã®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿