ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿¾Ý±¨µÀ£¬Ä³ÖÖÒÔ¼×´¼ÎªÔ­ÁÏ£¬ÒÔKOHΪµç½âÖʵÄÓÃÓÚÊÖ»úµÄ¿É³äµçµÄ¸ßЧȼÁϵç³Ø£¬³äÒ»´Îµç¿ÉÁ¬ÐøʹÓýϳ¤Ê±¼ä¡£ÏÂͼÊÇÒ»¸öµç»¯Ñ§¹ý³ÌµÄʾÒâͼ¡£ÒÑÖª¼×³ØµÄ×Ü·´Ó¦Ê½Îª£º2CH3OH+3O2+4KOH2K2CO3+6H2O¡£

ÇëÌî¿Õ£º

(1)³äµçʱ£¬¢ÙȼÁϵç³ØµÄ¸º¼«ÓëµçÔ´________¼«ÏàÁ¬¡£

¢ÚÑô¼«µÄµç¼«·´Ó¦Ê½Îª£º________________________¡£

(2)·Åµçʱ£º¸º¼«µÄµç¼«·´Ó¦Ê½Îª£º__________________¡£

(3)Ôڴ˹ý³ÌÖÐÈôÍêÈ«·´Ó¦£¬ÒÒ³ØÖÐA¼«µÄÖÊÁ¿Ôö¼Ó648 g£¬Ôò¼×³ØÖÐÀíÂÛÉÏÏûºÄO2____________L(±ê×¼×´¿ö)¡£

(4)ÈôÔÚ³£Î³£Ñ¹Ï£¬1 g CH3OHȼÁÏÉú³ÉCO2ºÍҺ̬H2Oʱ·ÅÈÈ22.68 kJ£¬±íʾ¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£º____________________¡£

¡¾´ð°¸¡¿¸º 4OH£­-4e-=2H2O+O2¡ü CH3OH£«8OH£­-6e£­=CO32-£«6H2O 33.6 CH3OH(l)+O2(g)=CO2(g)+2H2O(l) ¦¤H=£­725.76 kJ¡¤mol£­1

¡¾½âÎö¡¿

(1)³äµçʱ£¬Ô­µç³Ø¸º¼«ÓëµçÔ´¸º¼«ÏàÁ¬£¬×÷Òõ¼«£¬·¢Éú»¹Ô­·´Ó¦£»Ñô¼«Ê§µç×Ó·¢ÉúÑõ»¯·´Ó¦£»

(2)·Åµçʱ£¬¸º¼«Éϼ״¼Ê§µç×Ó·¢ÉúÑõ»¯·´Ó¦£»

(3)ÒÒ³ØÖÐA¼«ÎªÒõ¼«£¬ÒøÀë×ӵõç×Ó·¢Éú»¹Ô­·´Ó¦£¬¸ù¾ÝתÒƵç×ÓÏàµÈ¼ÆËãÑõÆøµÄÌå»ý£»

(4)¸ù¾ÝÎïÖÊ·´Ó¦·Å³öµÄÈÈÁ¿Óë·´Ó¦µÄÎïÖʶàÉÙ³ÊÕý±È¼ÆËã·´Ó¦ÈÈ£¬È»ºóÊéдÈÈ»¯Ñ§·½³Ìʽ¡£

(1)¢Ù³äµçʱ£¬È¼Áϵç³Ø¸º¼«ÓëµçÔ´¸º¼«ÏàÁ¬£»

¢ÚÑô¼«ÉÏÇâÑõ¸ùÀë×Óʧµç×Ó·¢ÉúÑõ»¯·´Ó¦£¬µç¼«·´Ó¦Ê½Îª£º4OH--4e-=2H2O+O2¡ü£»

(2)·Åµçʱ£¬¼×´¼Ê§µç×ÓºÍÇâÑõ¸ùÀë×Ó·´Ó¦Éú³É̼Ëá¸ùÀë×ÓºÍË®£¬ËùÒԵ缫·´Ó¦Ê½Îª£ºCH3OH£«8OH--6e-=CO32-£«6H2O£»

(3)ÒÒ³ØÖÐAµç¼«ÉÏAg+µÃµç×Ó·¢Éú»¹Ô­·´Ó¦£¬µ±ÒÒ³ØÖÐAµç¼«µÄÖÊÁ¿Ôö¼Ó648g£¬n(Ag)==6mol£¬Ôò¸ù¾Ýͬһ±ÕºÏ»Ø·Öеç×ÓתÒÆÊýÄ¿ÏàµÈ¿ÉÖª£ºÔڼ׳ØÖÐÀíÂÛÉÏÏûºÄO2Ìå»ýV(O2)=33.6L¡£

(4)ÈôÔÚ³£Î³£Ñ¹Ï£¬1g CH3OHȼÁÏÉú³ÉCO2ºÍҺ̬H2Oʱ·ÅÈÈ22.68 kJ£¬ÓÉÓÚ1mol¼×´¼µÄÖÊÁ¿ÊÇ32g£¬ËùÒÔ1mol CH3OHȼÁÏÉú³ÉCO2ºÍҺ̬H2Oʱ·ÅÈÈ22.68 kJ¡Á32=725.76 kJ£¬ËùÒÔ±íʾ¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCH3OH(l)+O2(g)=CO2(g)+2H2O(l) ¦¤H=£­725.76 kJ¡¤mol£­1¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø