ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÏÂͼÊǶ¡Íé(C4H10)ÁѽâµÄʵÑéÁ÷³Ì¡£Á¬½ÓºÃ×°Öúó£¬Ðè½øÐеÄʵÑé²Ù×÷ÓУº¢Ù¼ìÖÃÕûÌ××°ÖõÄÆøÃÜÐÔ¢ÚÅųö×°ÖÃÖеĿÕÆø¢Û¸øD¡¢G×°ÖüÓÈȵȡ­¡­£»GºóÃæ×°ÖÃÒÔ¼°Ìú¼Ų̈µÈÒÑÊ¡ÂÔ£»CuOÄܽ«ÌþÑõ»¯³ÉCO2ºÍH2O¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)¶¡ÍéÁѽâµÄ¿ÉÄÜ·½³ÌʽΪC4H10CH4+C3H6£¬____________________£»

(2)д³ö¼×ÍéÓëÑõ»¯Í­·´Ó¦µÄ»¯Ñ§·½³Ìʽ_____________________________£»

(3)Èô¶Ô·´Ó¦ºóE×°ÖÃÖеĻìºÏÎï(äåË®×ãÁ¿),ÔÙ°´ÒÔÏÂÁ÷³ÌʵÑ飺

¢Ù·ÖÀë²Ù×÷¢ñºÍ¢òµÄÃû³Æ·Ö±ðÊÇ___________¡¢___________(Ìî×Öĸ) £»

a£®Õô·¢ b£®¹ýÂË c£®·ÖÒº d£®ÕôÁó

¢ÚNa2SO3ÈÜÒºµÄ×÷ÓÃÊÇ(ÓÃÀë×Ó·½³Ìʽ±íʾ)_____________________________________________£»

¢ÛÒÑÖªAµÄ̼ԭ×ÓÊý´óÓÚBµÄ̼ԭ×ÓÊý£¬Çëд³öBµÄÃû³Æ______________________£»

(4)¼Ù¶¨¶¡ÍéÍêÈ«Áѽ⣬Á÷¾­D¡¢G×°ÖÃÖеÄÆøÌåÄÜÍêÈ«·´Ó¦¡£µ±(E+F)×°ÖõÄ×ÜÖÊÁ¿±È·´Ó¦Ç°Ôö¼ÓÁË0.49g£¬G×°ÖõÄÖÊÁ¿±È·´Ó¦¼ä¼õÉÙÁË1.44£¬Ôò¶¡ÍéµÄÁѽâ²úÎïÖм×ÍéºÍÒÒÍéµÄÎïÖʵÄÁ¿Ö®±ÈΪ___________¡£

¡¾´ð°¸¡¿ C4H10C2H6+C2H4 CH4£«4CuO 4Cu£«CO2£«2H2O c d SO32-+Br2+H2O¨TSO42-+2H++2Br- 1£¬2-¶þäåÒÒÍé 1¡Ã2

¡¾½âÎö¡¿(1)¸ù¾ÝÔ­×ÓÊغ㣬×ñÑ­ÌþµÄ×é³É¹æÂÉ£¬Ð´³ö¿ÉÄÜ·¢ÉúµÄ·´Ó¦¡£

(2) Ñõ»¯ÂÁ×÷´ß»¯¼Á,¼ÓÈÈÌõ¼þÏÂ,¼×ÍéºÍÑõ»¯Í­·´Ó¦Éú³É¶þÑõ»¯Ì¼¡¢Ë®ºÍÍ­¡£

£¨3£©¢Ù»ìºÏÎïÖк¬ÓÐä塢ˮ¡¢äå´úÌþ,¼ÓÈëÑÇÁòËáÄÆ,ÑÇÁòËáÄƱ»äåÑõ»¯Éú³ÉÁòËáÄÆ,ͬʱÉú³ÉNaBr,´Ó¶ø³ýÈ¥äå,È»ºó²ÉÓ÷ÖÒº·½·¨·ÖÀë,½«Óлú²ã½øÐзÖÁóµÃµ½ÓлúÎïA¡¢ÓлúÎïB¡£

¢ÚNa2SO3¾ßÓл¹Ô­ÐÔ£¬Äܹ»°ÑäåÀë×ÓÑõ»¯Îªäåµ¥Öʶø³ýÈ¥äå¡£

¢ÛÒÑÖªAµÄ̼ԭ×ÓÊý´óÓÚBµÄ̼ԭ×ÓÊý£¬ËµÃ÷BÖÐ̼ԭ×Ó¸öÊýÊÇ3£¬AÖÐ̼ԭ×Ó¸öÊýÊÇ2¡£

(4) E¡¢FÎüÊÕµÄÊÇÏ©Ìþ,G¼õÉÙµÄÖÊÁ¿ÊÇÑõ»¯Í­ÖеÄÑõÔªËØÖÊÁ¿,¶¡ÍéµÄÁѽâÖÐ,Éú³ÉÒÒÏ©µÄÎïÖʵÄÁ¿ºÍÒÒÍéµÄÎïÖʵÄÁ¿ÏàµÈ,¼×ÍéºÍ±ûÏ©µÄÎïÖʵÄÁ¿ÏàµÈ,ÔÙ½áºÏÔ­×ÓÊغã¼ÆËã¼×ÍéºÍÒÒÍéµÄÎïÖʵÄÁ¿Ö®±È¡£

(1)¶¡ÍéÁѽâµÄ¿ÉÄÜ·½³ÌʽΪC4H10CH4+C3H6£¬¶¡Í黹¿ÉÄÜ´ß»¯ÁÑ»¯ÎªÒÒÏ©ºÍÒÒÍ飺·½³ÌʽΪ. C4H10C2H6+C2H4£»ÕýÈ·´ð°¸£ºC4H10C2H6+C2H4¡£

(2) Ñõ»¯ÂÁ×÷´ß»¯¼Á,¼ÓÈÈÌõ¼þÏÂ,¼×ÍéºÍÑõ»¯Í­·´Ó¦Éú³É¶þÑõ»¯Ì¼¡¢Ë®ºÍÍ­,·´Ó¦·½³ÌʽΪ: CH4£«4CuO 4Cu£«CO2£«2H2O£»ÕýÈ·´ð°¸£ºCH4£«4CuO 4Cu£«CO2£«2H2O¡£

(3)¢Ù¸ù¾Ý·´Ó¦Á÷³Ì¿ÉÖª£¬¼ÓÈëÑÇÁòËáÄÆÈÜÒººó£¬»ìºÏÎï·Ö²ã£¬¿ÉÒÔ½øÐзÖÒº²Ù×÷½øÐзÖÀ룻°ÑÁ½ÖÖ»¥ÈܵÄÓлúÎï·Ö¿ª£¬¿ÉÒÔ²ÉÓÃÕôÁóµÄ·½·¨½øÐУ»ÕýÈ·´ð°¸£ºc£»d¡£

¢ÚNa2SO3¾ßÓл¹Ô­ÐÔ£¬Äܹ»°ÑäåÀë×ÓÑõ»¯Îªäåµ¥Öʶø³ýÈ¥ä壬Àë×Ó·½³Ìʽ£ºSO32-+Br2+H2O¨TSO42-+2H++2Br-£»ÕýÈ·´ð°¸£ºSO32-+Br2+H2O¨TSO42-+2H++2Br- ¡£

¢ÛÒÑÖªAµÄ̼ԭ×ÓÊý´óÓÚBµÄ̼ԭ×ÓÊý£¬ËµÃ÷BÖÐ̼ԭ×Ó¸öÊýÊÇ3£¬AÖÐ̼ԭ×Ó¸öÊýÊÇ2,ÒÒÏ©ºÍäå·¢Éú¼Ó³ÉÉú³É1£¬2-¶þäåÒÒÍ飬ËùÒÔAΪ 1£¬2-¶þäåÒÒÍ飻ÕýÈ·´ð°¸£º1£¬2-¶þäåÒÒÍé¡£

(4)¶¡ÍéµÄÁѽâÖÐÉú³ÉµÄÒÒÏ©ºÍÒÒÍéµÄÎïÖʵÄÁ¿ÏàµÈ£¬Éú³ÉµÄ¼×ÍéºÍ±ûÏ©µÄÎïÖʵÄÁ¿ÏàµÈ,E¡¢FÎüÊÕµÄÊÇÏ©Ìþ,G¼õÉÙµÄÖÊÁ¿ÊÇÑõ»¯Í­ÖеÄÑõÔªËØÖÊÁ¿£»ÉèxΪC2H4µÄÎïÖʵÄÁ¿,yΪC3H6µÄÎïÖʵÄÁ¿£¬ÔòÒÒÍéºÍ¼×ÍéµÄÎïÖʵÄÁ¿·Ö±ðÊÇx¡¢y,28x+42y=0.49,ÒÒÍéºÍ¼×ÍéºÍÑõ»¯Í­·´Ó¦ÐèÒªµÄÑõÔ­×ÓµÄÎïÖʵÄÁ¿Îª:2(2x+y)+(6x+2y)/2=1.44/16,¼ÆËãµÃ³ö:x=0.01mol, y=0.005mol£»Ôò¶¡ÍéµÄÁѽâ²úÎïÖм×ÍéºÍÒÒÍéµÄÎïÖʵÄÁ¿Ö®±ÈΪ0.005£º0.01=1:2£»ÕýÈ·´ð°¸:1:2¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÄÜÔ´Óë²ÄÁÏ¡¢ÐÅÏ¢±»³ÆΪÏÖ´úÉç»á·¢Õ¹µÄÈý´óÖ§Öù£¬»¯Ñ§ÓëÄÜÔ´ÓÐ×ÅÃÜÇÐÁªÏµ¡£

(1)ϱíÖеÄÊý¾Ý±íʾÆÆ»µ1mol»¯Ñ§¼üÐèÏûºÄµÄÄÜÁ¿(¼´¼üÄÜ£¬µ¥Î»ÎªkJ¡¤mol-1)

»¯Ñ§¼ü

H-H

Cl-Cl

H-Cl

¼üÄÜ

436

243

431

Çë¸ù¾ÝÒÔÉÏÐÅÏ¢¿ÉÖª£¬1molÇâÆøÔÚ×ãÁ¿µÄÂÈÆø×ÅȼÉÕÉú³ÉÂÈ»¯ÇâÆøÌå·Å³öÈÈÁ¿___________¡£

(2)ÌìÈ»ÆøÊÇÒ»ÖÖÖØÒªµÄÇé½ÚÄÜÔ´ºÍ»¯¹¤Ô­ÁÏ£¬ÆäÖ÷Òª³É·ÖΪCH4¡£ÒÔCH4¡¢¿ÕÆø¡¢KOHÈÜҺΪԭÁÏ£¬Ê¯Ä«Îªµç¼«¿É¹¹³ÉȼÁϵç³Ø¡£¸Ãµç³ØµÄ¸º¼«·´Ó¦Ê½Îª___________¡£

(3)¹¤ÒµÉÏÓÐÒ»ÖÖ·½·¨ÊÇÓà CO2À´Éú²úȼÁϼ״¼£¬¿ÉÒÔ½«CO2±ä·ÏΪ±¦¡£ÔÚÌå»ýΪ1LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë1molCO2ºÍ3molH2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2+3H2CH3OH(g)+H2O(g),²âµÃCO2(g)ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçÏÂͼËùʾ¡£

¢Ù´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬CH3OHµÄƽ¾ù·´Ó¦ËÙÂÊv(CH3OH)=___________£»H2µÄת»»ÂʦÁ(H2)=___________¡£

¢ÚÈô·´Ó¦CO2+3H2CH3OH(g)+H2O(g)ÔÚËÄÖÖ²»Í¬Çé¿öϵķ´Ó¦ËÙÂÊ·Ö±ðΪ£º

A£®V(CO2)=0.15mol¡¤L-1¡¤min-1B£®V(H2)=0.01mol¡¤L-1¡¤s-1

C£®v(CH3OH)=0.2mol¡¤L-1¡¤min-1D£®v(H2O)=0.45mol¡¤L-1¡¤min-1

¸Ã·´Ó¦½øÐÐÓÉ¿ìµ½ÂýµÄ˳ÐòΪ___________(Ìî×Öĸ)¡£

(4)º£Ë®»¯Ñ§×ÊÔ´µÄÀûÓþßÓзdz£¹ãÀ«µÄÇ°¾°¡£´Óº£Ë®ÖÐÌáÈ¡äåµÄ¹¤ÒµÁ÷³ÌÈçͼ£º

¢ÙÁ÷³Ì¢ÜÖÐÉæ¼°µÄÀë×Ó·´Ó¦ÈçÏ£¬ÇëÔÚÏÂÃæ·½¿òÄÚÌîÈëÊʵ±µÄ»¯Ñ§¼ÆÁ¿Êý£º

¡õBr2+¡õCO32-=¡õBrO3-+¡õBr-+¡õCO2¡ü______

¢ÚÒÔÉÏÎå¸ö¹ý³ÌÖÐÉæ¼°Ñõ»¯»¹Ô­·´Ó¦µÄÓÐ___________¸ö¡£

¢Û²½Öè¢ÛÖÐÒÑ»ñµÃÓÎÀë̬µÄä壬²½Öè¢ÜÓÖ½«Ö®×ª±ä³É»¯ºÏ̬µÄä壬ÆäÄ¿µÄÊÇ___________¡£

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

¾«Ó¢¼Ò½ÌÍø