ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿»¯Ñ§ÖªÊ¶ÓëÉú²úÉú»îϢϢÏà¹Ø¡£
I.¹è¼°Æ仯ºÏÎï
£¨1£©¹è¼°Æ仯ºÏÎïÔÚÏÖ´úÐÅÏ¢¼¼ÊõÖÐÓ¦Óù㷺¡£ÆäÖÐÓÃÓÚÖÆÔì¼ÆËã»úоƬµÄ²ÄÁÏÊÇ__________£¨Ð´»¯Ñ§Ê½£©£¬ÓÃÓÚÖÆÔì¹âµ¼ÏËάµÄ²ÄÁÏÊÇ__________£¨Ð´»¯Ñ§Ê½£©¡£¹âµ¼ÏËάÔÚ¼îÐÔÍÁÈÀÖÐÒ×±»¸¯Ê´£¬Çëд³öÏà¹ØµÄÀë×Ó·´Ó¦·½³Ìʽ______________________
II.Âȼ°Æ仯ºÏÎï
£¨2£©Æ¯°×·ÛÊÇÉú»î³£±¸ÓÃÆ·£¬ÆäÖ÷Òª³É·ÝÊÇ__________________£¨Ð´»¯Ñ§Ê½£©£»Æ¯°×·Û±£´æ²»µ±Ò×±äÖÊ£¬Çëд³öƯ°×·ÛʧЧµÄ·½³Ìʽ________________£»È粻СÐÄ°ÑƯ°×·ÛÓë½à²ÞÁ飨Ö÷Òª³É·ÝΪÑÎËᣩ»ìºÏ£¬Ò×Éú³ÉÂÈÆøʹÈËÖж¾¡£Çëд³öÓйط´Ó¦µÄÀë×Ó·½³Ìʽ________________________¡£
£¨3£©Ä³ÊµÑéС×飬ÏòÒ»¶¨Á¿µÄʯ»ÒÈéÖлºÂýÔÈËÙµØͨÈë×ãÁ¿ÂÈÆø£¬·¢ÏÖ²úÎïÖл¹»ìÓÐÁËClO3££¬ËûÃÇÌÖÂÛºóÈÏΪ£¬ÊÇ·´Ó¦·ÅÈÈζÈÉý¸ßµÄÔÒò¡£²¢»³öÁËClO£¡¢ClO3£Á½ÖÖÀë×ÓµÄÎïÖʵÄÁ¿£¨n£©Ó뷴Ӧʱ¼ä(t)µÄ¹ØϵÇúÏߣ¬´ÖÂÔ±íʾΪÏÂͼ£¨²»¿¼ÂÇÂÈÆøºÍË®µÄ·´Ó¦£©¡£
¢ÙͼÖÐÇúÏßI±íʾ_____________Àë×ÓµÄÎïÖʵÄÁ¿Ë淴Ӧʱ¼ä±ä»¯µÄ¹Øϵ¡£
¢ÚËùȡʯ»ÒÈéÖк¬ÓÐCa(OH)2µÄÎïÖʵÄÁ¿Îª__________mol¡£
¢ÛÁíÈ¡Ò»·ÝÓë¢ÚµÈÎïÖʵÄÁ¿µÄʯ»ÒÈ飬ÒԽϴóµÄËÙÂÊͨÈë×ãÁ¿ÂÈÆø£¬·´Ó¦ºó²âµÃ²úÎïÖÐCl£µÄÎïÖʵÄÁ¿Îª0.37mol£¬Ôò²úÎïÖÐn(ClO£)/n(ClO3£)=______________¡£
III.µª¼°Æ仯ºÏÎï
£¨4£©ÏõËáÉú²ú¹ý³ÌÖÐÅŷųöÀ´µÄ·ÏÆø£¬º¬NO¡¢NO2µÈ´óÆøÎÛȾÎï¡£
¢Ù¹¤ÒµÉϳ£ÓÃÔÀí£ºNOx+NH3¡úN2+H2O£¬Ê¹Æäת»¯ÎªÎÞ¶¾µÄN2£¬ÏÖÓÐNO¡¢NO2µÄ»ìºÏ3.0L£¬¿ÉÓë3.5LÏàͬ״¿öµÄNH3ÍêÈ«·´Ó¦£¬È«²¿×ª»¯ÎªN2£¬ÔòÔÚÔ»ìºÏÆøÌåÖÐNOºÍNO2µÄÎïÖʵÄÁ¿Ö®±ÈÊÇ_____________£»
¢ÚÒÑÖª¼îÄܳýÈ¥ÏõËáβÆø£º2NO2+2NaOH£½NaNO2+NaNO3+H2O;NO+NO2+2NaOH£½2NaNO2+H2O
¸ù¾ÝÏõËáβÆø´¦ÀíµÄ·´Ó¦ÔÀí£¬ÏÂÁÐÆøÌåÖв»Äܱ»¹ýÁ¿NaOHÈÜÒºÎüÊÕµÄÊÇ_______________
A£®1molO2ºÍ4molNO2¡¡
B£®1molO2ºÍ4molNO¡¡
C£®1molNOºÍ5molNO2¡¡ ¡¡
D£®4molNOºÍ1molNO2
£¨5£©Æû³µÅŷŵÄβÆøÖÐÒòº¬ÓеªµÄÑõ»¯Îï¶øÎÛȾ´óÆø£¬Ôì³É²úÉúµªµÄÑõ»¯ÎïµÄÖ÷ÒªÔÒòÊÇ_____________
A£®È¼ÉÕº¬µª»¯ºÏÎïȼÁÏ¡¡¡¡¡¡¡¡¡¡¡¡¡¡
B£®È¼ÉÕº¬Ç¦ÆûÓÍ
C£®ÓÉÓÚȼÉÕ²»³ä·Ö¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
D£®ÄÚȼ»úÖпÕÆøÖеÄN2ÔÚ¸ßÎÂϱ»Ñõ»¯
¡¾´ð°¸¡¿ Si SiO2 SiO£²¡¡£«2OH£=SiO32-+H2O CaCl2¡¢Ca(ClO)2 Ca(ClO)2¡¡£«CO£²+H2O=2HClO+CaCO3¡ý 2HClO 2HCl£«O2¡ü ClO- £«Cl-£«2H+=Cl2 ¡ü+H2O ClO- 0.25 7£º6 1:3 D D
¡¾½âÎö¡¿£¨1£©¿¼²é¹èºÍ¶þÑõ»¯¹èµÄÓÃ;£¬ÀûÓþ§Ìå¹èÊÇÁ¼ºÃµÄ°ëµ¼Ì壬Òò´ËÖÆÔì¼ÆËã»úоƬµÄ²ÄÁÏÊǾ§Ìå¹è£¬»¯Ñ§Ê½ÎªSi£»¹âµ¼ÏËάµÄ³É·ÖÊÇSiO2£»SiO2ÊôÓÚËáÐÔÑõ»¯ÎÓë¼î·¢Éú¸´·Ö½â·´Ó¦£¬¼´SiO2£«2OH£=SiO32££«H2O£»£¨2£©¿¼²éƯ°×·ÛµÄ³É·ÖºÍʧЧµÄÔÒò£¬Æ¯°×·ÛÊÇCa(ClO)2ºÍCaCl2µÄ»ìºÏÎ̼ËáµÄËáÐÔÇ¿ÓÚ´ÎÂÈËᣬÒò´ËÓÐCa(ClO)2£«CO2£«H2O=CaCO3£«2HClO£¬HClO¼û¹âÊÜÈȷֽ⣬2HClO 2HCl£«O2¡ü£»Á½Õß·¢ÉúÑõ»¯»¹Ô·´Ó¦£¬Éú³ÉÂÈÆø£¬¼´Àë×Ó·´Ó¦·½³ÌʽΪClO££«Cl££«2H£«=H2O£«Cl2¡ü£»£¨3£©¿¼²éÑõ»¯»¹Ô·´Ó¦ÖеÃʧµç×ÓÊýÄ¿Êغ㣬¢Ù¸ù¾ÝÌâÒ⣬²úÉúClO3£µÄÔÒòÊÇ·´Ó¦ÊÇ·ÅÈÈ£¬Î¶ÈÉý¸ßÔì³ÉµÄ£¬ËµÃ÷·´Ó¦¿ªÊ¼Ê±Ã»ÓÐClO3£µÄÉú³É£¬¼´IµÄÇúÏßΪClO££¬IIµÄÇúÏßΪClO3££»¢Ú·´Ó¦Éú³ÉµÄ²úÎïÊÇCaCl2¡¢Ca(ClO)2¡¢Ca(ClO3)2ºÍH2O£¬t2ʱ¿Ìʱ£¬²úÉúClO3£ºÍClO£µÄÎïÖʵÄÁ¿Îª0.05mol¡¢0.10mol£¬¸ù¾ÝµÃʧµç×ÓÊýÄ¿Êغ㣬Çó³ön(Cl£)=n(ClO£)£«5n(ClO3£)£¬¼´n(Cl£)=( 0.10£«5¡Á0.05)mol=0.35mol£¬Òò´ËCa(OH)2µÄÎïÖʵÄÁ¿Îª[n(Cl£)£«n(ClO£)£«n(ClO3£)]/2=(0.05£«0.1£«0.35)/2mol=0.25mol£»¢Û¸ù¾ÝµÃʧµç×ÓÊýÄ¿Êغ㣬ÓÐn(Cl£)=n(ClO£)£«5n(ClO3£)=0.37mol£¬Ca(OH)2µÄÎïÖʵÄÁ¿Îª[n(Cl£)£«n(ClO£)£«n(ClO3£)]/2=0.25mol£¬½âµÃn(ClO£)=0.07mol£¬n(ClO3£)=0.06mol£¬Òò´ËÓÐn(ClO£)/n(ClO3£)=7/6£»£¨4£©¿¼²éÑõ»¯»¹Ô·´Ó¦ÖеÃʧµç×ÓÊýÄ¿Êغ㣬ÒÔ¼°»¯Ñ§¼ÆË㣬¢ÙÏàͬ״¿öÏ£¬ÆøÌåÌå»ý±ÈµÈÓÚÆäÎïÖʵÄÁ¿Ö®±È£¬ÁîNOµÄÌå»ýΪxL£¬ÔòNO2µÄÌå»ýΪ(3£x)L£¬¸ù¾ÝµÃʧµç×ÓÊýÄ¿Êغ㣬ÓÐ2x£«(3£x)¡Á4=3¡Á3.5£¬½âµÃx=0.75L£¬NO2µÄÌå»ýΪ2.25L£¬Á½Õß±ÈֵΪ0.75£º2.25=1£º3£»¢ÚA¡¢¸ù¾Ý4NO2£«O2£«2H2O=4HNO3£¬NO2ºÍO2Ç¡ºÃÍêÈ«·´Ó¦£¬Éú³ÉHNO3£¬È»ºóHNO3ÓëNaOH·´Ó¦£¬ÆøÌåÈ«²¿±»ÎüÊÕ£¬¹ÊA´íÎó£»B¡¢·¢2NO£«O2=2NO2£¬ÑõÆø²»×㣬°´ÕÕÑõÆø½øÐмÆË㣬Éú³É2molNO2£¬Ê£Óà2molNO£¬¸ù¾ÝÐÅÏ¢£¬2molNO2ºÍ2molNO£¬ÔÚÇâÑõ»¯ÄÆÈÜÒºÖÐÈ«²¿±»ÎüÊÕ£¬¹ÊB´íÎó£»C¡¢¸ù¾ÝÌâ¸ÉÖÐÁ½¸ö·´Ó¦·½³Ìʽ£¬ÆøÌåÈ«²¿ÎüÊÕ£¬¹ÊC´íÎó£»D¡¢¸ù¾ÝÉÏÊö·´Ó¦·½³Ìʽ£¬NO¹ýÁ¿£¬ÆøÌå²»Äܱ»ÍêÈ«ÎüÊÕ£¬¹ÊDÕýÈ·£»£¨5£©¿¼²é»·¾³ÎÛȾµÄÀ´Ô´£¬Æû³µÎ²ÆøÖеªµÄÑõ»¯ÎïÊÇÓÉ¿ÕÆøÖеĵªÆøÔÚ¸ßÎÂÏÂÓëÑõÆø·´Ó¦Éú³É£¬¹ÊÑ¡ÏîDÕýÈ·¡£