ÏõËṤҵβÆøÖеĵªµÄÑõ»¯Îï(NOºÍ)ÊÇÖ÷ÒªµÄ´óÆøÎÛÎÆä³£ÓõÄÖÎÀí·½·¨ÓÐÒÔÏÂÁ½ÖÖ(ÒѾ¼ò»¯)£®
¢ÙNaOHÎüÊÕ·¨£¬·´Ó¦ÔÀíÈçÏ£º
¢Ú°±´ß»¯»¹Ô·¨£¬·´Ó¦ÔÀí¿É±íʾΪ
ÏÖÓÐÒ»¶¨Á¿µÄº¬ºÍNOµÄÏõË᳧βÆø(²»º¬ÆäËûÆøÌå)£¬ÈôÓùýÁ¿µÄNaOHÈÜÒºÎüÊÕºó£¬ÈÜÒºÖÐÓëµÄÎïÖʵÄÁ¿Ö®±ÈÇ¡ºÃÓëβÆøÖÐNOºÍµÄÎïÖʵÄÁ¿Ö®±ÈÏàµÈ£®
(1)ÈôÓñíʾ¸ÃβÆøÖеªµÄÑõ»¯ÎïµÄƽ¾ù×é³É£¬¼ÆËãxÖµ£®
(2)½«1Ìå»ýµÄ¸ÃβÆøÓð±´ß»¯»¹Ô·¨´¦Àí£¬ÖÁÉÙÏûºÄ¶àÉÙÌå»ýµÄ°±(±ê×¼×´¿öÏÂ)£¿
(1) ÈôÉè¸ÃβÆøÖк¬ÓУ¬b mol NO£¬ÔòÒÀ¾ÝÌâ¸øÐÅÏ¢£º²Î¼ÓÏÂÁз´Ó¦µÄ Ϊ(a£b)mol£®Ôò µÄÎïÖʵÄÁ¿£¬ µÄÎïÖʵÄÁ¿ £®ÒòΪÓÐ £¬ËùÒÔ £®ÕûÀíµÃ (a£3b)(a£«b)=0£¬¼´ºÏÀíµÄ½âΪ£¬ËùÒÔ£®(2) ÒòΪÔÚ°±´ß»¯»¹Ô·¨·´Ó¦ÖУ¬ÖеÄNÓɼۡú0¼Û£¬ÖеÄNÓÉ£3¼Û¡ú0£®¸ù¾ÝµÃʧµç×ÓÏàµÈµÄÔÀíÓУº £¬½âµÃ£¬¼´´¦Àí1Ìå»ýβÆøÐèÏûºÄÌå»ýΪ£® |
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º022
ÏõËṤҵβÆøÖеĵªµÄÑõ»¯Îï(NO¡¢NO2)ÊÇÖ÷ÒªµÄ´óÆøÎÛȾÎïÖ®Ò»£¬Æä³£ÓõÄÖÎÀí·½·¨ÓÐÒÔϼ¸ÖÖ£º
¢ÙNaOHÈÜÒºÎüÊÕ·¨£¬·´Ó¦ÔÀíÈçÏ£º
2NO2+2NaOH====NaNO2+NaNO3+H2O
NO2+NO+2NaOH====2NaNO2+H2O
¢Ú°±´ß»¯»¹Ô·¨£¬·´Ó¦ÔÀíÊÇ£º
NOx+NH3N2+H2O
ÏÖÓÐÒ»¶¨Á¿µÄº¬NO2ºÍNOµÄHNO3¹¤ÒµÎ²Æø(²»º¬ÆäËûÆøÌå)£¬ÈôÓùýÁ¿µÄNaOHÈÜÒºÎüÊÕºó£¬ÈÜÒºÖÐNaNO3ÓëNaNO2µÄÎïÖʵÄÁ¿Ö®±ÈÇ¡ºÃÓëβÆøÖÐNOºÍNO2µÄÎïÖʵÄÁ¿Ö®±ÈÏàµÈ¡£
(1)ÈôÓÃNOx±íʾ¸ÃβÆøÖеªµÄÑõ»¯ÎïµÄƽ¾ù×é³É£¬ÊÔÇóxµÄÖµ¡£
(2)Èô1Ìå»ýµÄ¸ÃβÆøÓð±´ß»¯»¹Ô·¨´¦Àí£¬ÖÁÉÙÏûºÄ¶àÉÙÌå»ýÏàͬ״¿öϵݱÆø?
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º022
ÏõËṤҵβÆøÖеĵªµÄÑõ»¯Îï(NO¡¢NO2)ÊÇÖ÷ÒªµÄ´óÆøÎÛȾÎïÖ®Ò»£¬Æä³£ÓõÄÖÎÀí·½·¨ÓÐÒÔϼ¸ÖÖ£º
¢ÙNaOHÈÜÒºÎüÊÕ·¨£¬·´Ó¦ÔÀíÈçÏ£º
2NO2+2NaOH====NaNO2+NaNO3+H2O
NO2+NO+2NaOH====2NaNO2+H2O
¢Ú°±´ß»¯»¹Ô·¨£¬·´Ó¦ÔÀíÊÇ£º
ÏÖÓÐÒ»¶¨Á¿µÄº¬NO2ºÍNOµÄHNO3¹¤ÒµÎ²Æø(²»º¬ÆäËûÆøÌå)£¬ÈôÓùýÁ¿µÄNaOHÈÜÒºÎüÊÕºó£¬ÈÜÒºÖÐNaNO3ÓëNaNO2µÄÎïÖʵÄÁ¿Ö®±ÈÇ¡ºÃÓëβÆøÖÐNOºÍNO2µÄÎïÖʵÄÁ¿Ö®±ÈÏàµÈ¡£
(1)ÈôÓÃNOx±íʾ¸ÃβÆøÖеªµÄÑõ»¯ÎïµÄƽ¾ù×é³É£¬ÊÔÇóxµÄÖµ¡£
(2)Èô1Ìå»ýµÄ¸ÃβÆøÓð±´ß»¯»¹Ô·¨´¦Àí£¬ÖÁÉÙÏûºÄ¶àÉÙÌå»ýÏàͬ״¿öϵݱÆø?
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºÎïÀí½ÌÑÐÊÒ ÌâÐÍ£º022
¢ÙNaOHÈÜÒºÎüÊÕ·¨£¬·´Ó¦ÔÀíÈçÏ£º
2NO2+2NaOH====NaNO2+NaNO3+H2O
NO2+NO+2NaOH====2NaNO2+H2O
¢Ú°±´ß»¯»¹Ô·¨£¬·´Ó¦ÔÀíÊÇ£º
NOx+NH3N2+H2O
ÏÖÓÐÒ»¶¨Á¿µÄº¬NO2ºÍNOµÄHNO3¹¤ÒµÎ²Æø(²»º¬ÆäËûÆøÌå)£¬ÈôÓùýÁ¿µÄNaOHÈÜÒºÎüÊÕºó£¬ÈÜÒºÖÐNaNO3ÓëNaNO2µÄÎïÖʵÄÁ¿Ö®±ÈÇ¡ºÃÓëβÆøÖÐNOºÍNO2µÄÎïÖʵÄÁ¿Ö®±ÈÏàµÈ¡£
(1)ÈôÓÃNOx±íʾ¸ÃβÆøÖеªµÄÑõ»¯ÎïµÄƽ¾ù×é³É£¬ÊÔÇóxµÄÖµ¡£
(2)Èô1Ìå»ýµÄ¸ÃβÆøÓð±´ß»¯»¹Ô·¨´¦Àí£¬ÖÁÉÙÏûºÄ¶àÉÙÌå»ýÏàͬ״¿öϵݱÆø?
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏõËṤҵβÆøÖеĵªµÄÑõ»¯Îï(NO¡¢NO2)ÊÇÖ÷ÒªµÄ´óÆøÎÛȾÎÆä³£ÓõÄÖÎÀí·½·¨ÓÐÒÔÏÂÁ½ÖÖ(ÒѼò»¯)
¢ÙNaOHÎüÊÕ·¨·´Ó¦ÔÀíÈçÏ£º2 NO2+2NaOH=NaNO3+NaNO2+H2O£»NO+NO2+2NaOH=2NaNO2+ H2O
¢Ú°±´ß»¯»¹Ô·¨ ·´Ó¦ÔÀíÊÇ£ºNO2+NH3¡úN2+H2O
ÏÖÓÐÒ»¶¨Á¿µÄº¬NO2ºÍNOµÄÏõËṤҵβÆø(²»º¬ÆäËûÆøÌå)£»ÈôÓùýÁ¿µÄNaOHÈÜÒºÎüÊÕºóÈÜÒºÖÐ NaNO3Óë NaNO2µÄÎïÖʵÄÁ¿Ö®±ÈÇ¡ºÃÓëβÆøÖÐNOºÍ NO2µÄÎïÖʵÄÁ¿Ö®±ÈÏàµÈ¡£
(1)ÈôÓÃNOx±íʾ¸ÃβÆøÖеªµÄÑõ»¯ÎïµÄƽ¾ù×é³É£¬ÊÔÇóxµÄÖµ¡£
(2)½«1Ìå»ýµÄ¸ÃβÆøÓð±´ß»¯»¹Ô·¨´¦Àí£¬ÖÁÉÙÏûºÄÏàͬ״¿ö϶àÉÙÌå»ýµÄ°±Æø?
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com