·ÖÎö £¨1£©Ì¼ËظַÅÈëŨÁòËáÖУ¬ÔÚ¼ÓÈÈʱ·¢Éú·´Ó¦£º2Fe+6H2SO4£¨Å¨£©$\frac{\underline{\;\;¡÷\;\;}}{\;}$Fe2£¨SO4£©3+3SO2¡ü+6H2O£¬C+2H2SO4£¨Å¨£©$\frac{\underline{\;\;¡÷\;\;}}{\;}$CO2¡ü+2SO2¡ü+2H2O£¬Ëæ×Å·´Ó¦µÄ½øÐУ¬µ±ËáÈÜÒº±äÏ¡ºó»á·¢Éú·´Ó¦£ºFe+H2SO4=FeSO4+H2¡ü£®·´Ó¦ºóµÃµ½µÄÈÜÒºXÖк¬ÓÐÁòËáÌú¡¢¼°¹ýÁ¿µÄÁòËᣬÆøÌåYÖк¬ÓÐSO2¡¢CO2ºÍË®ÕôÆø£®ÔÚ¼ìÑéʱҪעÒâÅųý SO32-¡¢CO32- µÈÀë×ÓµÄÓ°Ï죬ÏȼÓÑÎËáÅųý¸ÉÈÅÀë×Ó£¬ÔÙ¼ÓÂÈ»¯±µ¼ìÑéÁòËá¸ùÀë×Ó£»
£¨2£©Ì¼ÄܺÍŨÁòËá·´Ó¦Éú³É¶þÑõ»¯Áò¡¢¶þÑõ»¯Ì¼ºÍË®£»
£¨3£©ÎÞË®ÁòËáÍÓöµ½Ë®ÕôÆø±äÀ¶É«·ÖÎö£»
£¨4£©È·ÈÏÆøÌåYÖк¬ÓÐCO2£¬Ó¦ÍêÈ«Åųý¶þÑõ»¯ÁòµÄ¸ÉÈÅ£¬¸ù¾Ý¶þÑõ»¯ÁòµÄƯ°×ÐÔ½â´ð£»
£¨5£©A£®ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴµÓºóδ¸ÉÔï¼´ÓÃÀ´ÅäÖÆÈÜÒº£¬ÓÉÓÚ²»Ó°ÏìÈÜÖʵÄÎïÖʵÄÁ¿ºÍÈÜÒºµÄÌå»ý£»
B£®¶¨ÈÝʱ£¬¹Û²ìÒºÃ温Êӿ̶ÈÏߣ¬ÔòÈÜÒºµÄÌå»ýƫС£»
C£®Ò¡ÔȺó£¬ÒºÃæµÍÓڿ̶ÈÏߣ¬Ã»ÓÐÔÙ¼ÓÕôÁóË®£¬²Ù×÷ºÏÀí£»
D£®ÓÃÕôÁóˮϴµÓÉÕ±ºÍ²£Á§°ô£¬²¢½«Ï´µÓҺתÒÆÈëÈÝÁ¿Æ¿ÖУ¬²Ù×÷ÕýÈ·£»
£¨6£©¸ÃʵÑéÖÐÒѾ֪µÀµÄÊý¾ÝÓÐÆøÌåÁ÷ËÙa L/min¡¢ËáÐÔKMnO4ÈÜÒºµÄÌå»ýb L£¬ÆäŨ¶ÈΪc mol/L£®Èô´ÓÆøÌåͨÈëµ½×ÏÉ«Ç¡ºÃÍÊÈ¥£¬ÓÃʱ5·ÖÖÓ£¬ÔòÆøÌåµÄÌå»ýÊÇV=5aL£¬¸ù¾ÝSO2ÓëKMnO4·´Ó¦µÄ·½³Ìʽ£º2KMnO4+5SO2+2H2O=K2SO4+2MnSO4+2H2SO4¿ÉÖªn£¨SO2£©=$\frac{5}{2}$n£¨KMnO4£©=2.5bcmol£¬ÒԴ˼ÆËã´Ë´ÎÈ¡Ñù´¦µÄ¿ÕÆøÖжþÑõ»¯Áòº¬Á¿£®
½â´ð ½â£º£¨1£©¢Ù¼ìÑé SO42- ʱҪ±ÜÃâ SO32- µÄÓ°Ï죬¾Í²»Òª¼ÓÈëÏ¡ÏõËᣬÒòΪϡÏõËáÄÜ°ÑSO32- Ñõ»¯ÎªSO42-£¬´Ó¶ø¸ÉÈÅSO42- µÄ¼ìÑ飻ҪÏȼÓÈë¹ýÁ¿Ï¡ÑÎËᣬÒÔ³ýÈ¥ SO32-¡¢CO32- µÈÀë×ÓµÄÓ°Ï죬ͬʱ£¬µ±¼ÓÈëÏ¡ÑÎËáʱûÓгÁµí£¬¿ÉÅųýCl-µÄÓ°Ï죮µ±È»£¬ÈôÈÜÒºÖв»º¬SO32-Àë×Ó£¬Ôò¼ÓÈëÏ¡ÏõËáÒ²¿É£®¼ìÑéSO42-¿ÉÓÃBaCl2ÈÜÒº£¬¿´ÓÐûÓа×É«³Áµí²úÉú£®×ÛÉÏ·ÖÎö¿ÉÖª£¬¼ìÑéijÈÜÒºÖÐÊÇ·ñº¬ÓÐSO42-µÄ²Ù×÷·½·¨ÊÇ£ºÊ×ÏÈÔÚÊÔÒºÖмÓÈëÑÎËáËữ£¬ÔÙ¼ÓÈëBaCl2ÈÜÒº£¬ÈôÓÐBaSO4°×É«³Áµí²úÉú£¬ÔòÖ¤Ã÷ÓÐSO42-£¬¼ìÑéÈÜÒºXÖк¬ÓÐSO42-µÄÊÔ¼ÁÊÇHCl¡¢BaCl2£¬
¹Ê´ð°¸Îª£ºHCl¡¢BaCl2£»
£¨2£©Ì¼ÄܺÍŨÁòËá·´Ó¦Éú³É¶þÑõ»¯Áò¡¢¶þÑõ»¯Ì¼ºÍË®£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºC+2H2SO4£¨Å¨£©$\frac{\underline{\;\;¡÷\;\;}}{\;}$CO2¡ü+2SO2¡ü+2H2O£¬
¹Ê´ð°¸Îª£ºC+2H2SO4£¨Å¨£©$\frac{\underline{\;\;¡÷\;\;}}{\;}$CO2¡ü+2SO2¡ü+2H2O£»
£¨3£©×°ÖÃFµÄ×÷ÓÃÊǼìÑéÉú³ÉÎïÖеÄË®ÕôÆø£¬ÎÞË®ÁòËáÍÓöµ½Ë®ÕôÆø±äÀ¶É«£¬
¹Ê´ð°¸Îª£º¼ìÑéÊÇ·ñÓÐË®ÕôÆø²úÉú£»
£¨4£©È·ÈÏÆøÌåYÖк¬ÓÐCO2£¬Ó¦ÍêÈ«Åųý¶þÑõ»¯ÁòµÄ¸ÉÈÅ£¬µ±BÖÐÆ·ºìÈÜÒº²»ÍÊÉ«£¬CÖÐʯ»ÒË®±ä»ë×Ç£¬¿É˵Ã÷º¬ÓÐCO2£¬
¹Ê´ð°¸Îª£ºBÖÐÆ·ºì²»ÍÊÉ«£¬CÖгÎÇåʯ»ÒË®±ä»ë×Ç£»
£¨5£©A£®ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴµÓºóδ¸ÉÔï¼´ÓÃÀ´ÅäÖÆÈÜÒº£¬ÓÉÓÚ²»Ó°ÏìÈÜÖʵÄÎïÖʵÄÁ¿ºÍÈÜÒºµÄÌå»ý£¬ËùÒÔ¶ÔÅäÖƵÄÈÜÒºµÄŨ¶È²»²úÉúÈκÎÓ°Ï죬¹Ê²»Ñ¡£»
B£®¶¨ÈÝʱ£¬¹Û²ìÒºÃ温Êӿ̶ÈÏߣ¬ÔòÈÜÒºµÄÌå»ýƫС£¬µ¼ÖÂÅäÖƵÄÈÜÒºµÄŨ¶ÈÆ«´ó£¬¹ÊÑ¡£»
C£®Ò¡ÔȺó£¬ÒºÃæµÍÓڿ̶ÈÏߣ¬Ã»ÓÐÔÙ¼ÓÕôÁóË®£¬Ôò²»»á¶ÔÈÜÒºµÄŨ¶È²úÉúÎó²î£¬¹Ê²»Ñ¡£»
D£®ÓÃÕôÁóˮϴµÓÉÕ±ºÍ²£Á§°ô£¬²¢½«Ï´µÓҺתÒÆÈëÈÝÁ¿Æ¿ÖУ¬²Ù×÷ÕýÈ·ÎÞÎ󣬹ʲ»Ñ¡£»
¹Ê´ð°¸Îª£ºB£»
£¨6£©¸ÃʵÑéÖÐÒѾ֪µÀµÄÊý¾ÝÓÐÆøÌåÁ÷ËÙa L/min¡¢ËáÐÔKMnO4ÈÜÒºµÄÌå»ýb L£¬ÆäŨ¶ÈΪc mol/L£®Èô´ÓÆøÌåͨÈëµ½×ÏÉ«Ç¡ºÃÍÊÈ¥£¬ÓÃʱ5·ÖÖÓ£¬ÔòÆøÌåµÄÌå»ýÊÇV=5aL£¬¸ù¾ÝSO2ÓëKMnO4·´Ó¦µÄ·½³Ìʽ£º2KMnO4+5SO2+2H2O=K2SO4+2MnSO4+2H2SO4¿ÉÖªn£¨SO2£©=$\frac{5}{2}$n£¨KMnO4£©=2.5bcmol£¬Ôò´Ë´ÎÈ¡Ñù´¦µÄ¿ÕÆøÖжþÑõ»¯Áòº¬Á¿Îª$\frac{2.5bcmol¡Á64g/mol}{5aL}$=$\frac{32bc}{a}$g/L£¬
¹Ê´ð°¸Îª£º$\frac{32bc}{a}$£®
µãÆÀ ±¾Ì⿼²éÎïÖʵÄÐÔÖÊʵÑé¼°ÖƱ¸ÊµÑ飬Ϊ¸ßƵ¿¼µã£¬°ÑÎÕ»¯Ñ§ÊµÑé»ù±¾²Ù×÷¡¢ÌúÖʲÄÁÏÓëÈÈŨÁòËáµÄ·´Ó¦µÄʵÑé·½°¸µÄÉè¼Æ¡¢Àë×ӵļìÑé·½·¨¡¢ÎïÖʳɷֵÄÈ·¶¨¼°»ìºÏÎïÖÐ×é³É³É·ÖµÄº¬Á¿µÄ¼ÆËãµÄ֪ʶΪ½â´ðµÄ¹Ø¼ü£¬²àÖØ·ÖÎöÓëʵÑéÄÜÁ¦µÄ×ۺϿ¼²é£¬ÌâÄ¿ÄѶÈÖеȣ®
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£ºÑ¡ÔñÌâ
A£® | t¡æʱ£¬AgClÔÚË®ÖеÄÈܽâ¶È±ÈÔÚÏ¡ÑÎËáÖÐС | |
B£® | t¡æʱ£¬AgClµÄÈܽâ¶È´óÓÚAg2CrO4 | |
C£® | ÔÚ±¥ºÍAg2CrO4ÈÜÒºÖмÓÈëÉÙÁ¿K2CrO4£¬¿ÉʹÈÜÒºÓÉYµãÒÆÖÁXµã | |
D£® | ÏòͬŨ¶ÈNaClºÍK2CrO4»ìºÏÒºÖУ¬µÎ¼Ó0.1mol•L-1AgNO3ÈÜÒº£¬ÏÈÉú³É°×É«³Áµí |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | CO2+2OH-¡úCO32-+H2O | B£® | Al2O3+2OH-+3H2O¡ú2[Al£¨OH£©4]- | ||
C£® | 2Al+2OH-+6H2O¡ú2[Al£¨OH£©4]-+3H2¡ü | D£® | Al3++4OH-¡ú[Al£¨OH£©4]- |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | B£® | C£® | D£® |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ¢Ù¢Ú¢Û¢Ü¢Ý | B£® | Ö»ÓÐ¢Ý | C£® | ¢Û¢Ü¢Ý | D£® | Ö»ÓÐ¢Ù¢Ú¢Û¢Ý |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ¢Ú¢Ý | B£® | ¢Ù¢Ý | C£® | ¢Ù¢Ú¢Û¢à | D£® | ¢Ü¢Ý¢Þ¢ß |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
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