Question

3£®Ä³»¯Ñ§Ñо¿Ð¡×é¶ÔCuµÄ³£¼û»¯ºÏÎïµÄÐÔÖʽøÐÐʵÑé̽¾¿£¬Ñо¿µÄÎÊÌâºÍ¹ý³ÌÈçÏ£º
¢ÙÔÚÖÜÆÚ±íÖÐCuºÍAlλÖñȽϽӽü£¬Al£¨OH£©₃¾ßÓÐÁ½ÐÔ£¬Cu£¨OH£©₂ÊÇ·ñÒ²ÓÐÁ½ÐÔ£¿
¢Úͨ³£Çé¿öÏ£¬+2¼ÛFe²»ÈçÊ®3¼ÛFeÎȶ¨£¬Ê®l¼ÛCuÒ²Ò»¶¨²»Èç+2¼ÛCuÎȶ¨Âð£¿
¢ÛCuOÓÐÑõ»¯ÐÔ£¬Äܱ»H₂¡¢CO»¹Ô­£¬CuOÒ²Äܱ»NH₃»¹Ô­Âð£¿
£¨¢ñ£©ÎªÌ½¾¿ÎÊÌâ¢Ù£¬³ýCu£¨OH£©₂Íâ±ØÐëÑ¡ÓõÄÊÔ¼ÁΪab£¨ÌîÐòºÅ£©£º
a¡¢Ï¡H₂SO₄    b¡¢Å¨µÄNaOHÈÜÒº    c¡¢Å¨°±Ë®   d£®Å¨µÄNa₂CO₃ÈÜÒº
£¨¢ò£©ÎªÌ½¾¿ÎÊÌâ¢Ú£¬½øÐÐÈçÏÂʵÑ飺
£¨1£©½«CuO·ÛÄ©¼ÓÈÈÖÁ1000¡æÒÔÉÏÍêÈ«·Ö½â³ÉºìÉ«µÄCu₂O·ÛÄ©£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ4CuO$\frac{\underline{\;¸ßÎÂ\;}}{\;}$2Cu₂O+O₂¡ü£®¸Ã·´Ó¦ËµÃ÷£ºÔÚ¸ßÎÂÌõ¼þÏ£¬+1¼ÛµÄCu±È+2¼ÛCu¸üÎȶ¨£¨Ìî¡°Îȶ¨¡±»ò¡°²»Îȶ¨¡±£©£®
£¨2£©ÏòCu₂OÖмÓÊÊÁ¿Ï¡ÁòËᣬµÃµ½À¶É«ÈÜÒººÍÒ»ÖÖºìÉ«¹ÌÌ壬ÓÉ´Ë¿ÉÖª£¬ÔÚËáÐÔÈÜÒºÖУ¬+1¼ÛCu±È+2¼ÛCu¸ü²»Îȶ¨£¨Ìî¡°Îȶ¨¡±»ò¡°²»Îȶ¨¡±£©£®
£¨¢ó£©Îª½â¾öÎÊÌâ¢Û£¬Í¬Ê±²â¶¨CuµÄ½üËÆÏà¶ÔÔ­×ÓÖÊÁ¿£¬ËûÃÇÉè¼ÆÁËÈçÏÂʵÑ飨¼Ð³Ö×°ÖÃδ»­£©£º
£¨1£©×°ÖÃCÖÐ˵Ã÷CuOÄܹ»±»NH₃»¹Ô­µÄÏÖÏóΪ¹ÌÌåÓɺÚɫת±äºìÉ«£¬NH₃ºÍCuO·´Ó¦Éú³ÉÍ­£¬Í¬Ê±Éú³ÉÒ»ÖÖÎÞÎÛȾµÄÆøÌ壬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ3CuO+2NH₃$\frac{\underline{\;¸ßÎÂ\;}}{\;}$N₂+3Cu+3H₂O£®
£¨2£©²â¶¨·´Ó¦ÎïCuOºÍÉú³ÉÎïH₂OµÄÖÊÁ¿£¨m£¨CuO£©¡¢m£¨H₂O£©£©£¬Áгö¼ÆËãCuµÄÏà¶ÔÔ­×ÓÖÊÁ¿µÄ±í´ïʽ$\frac{18m£¨CuO£©}{m£¨{H}_{2}O£©}$-16£»×°ÖÃD²»¿ÉÒÔ£¨Ìî¡°¿ÉÒÔ¡±»ò¡°²»¿ÉÒÔ¡±£©»»³É×°ÖÃE£»×°ÖÃEµÄ×÷Ó㺳ä·ÖÎüÊÕ·´Ó¦¶àÓàµÄNH₃²¢·ÀÖ¹¿ÕÆøÖÐË®·Ö½øÈ룮
£¨3£©ÏÂÁÐÇé¿ö½«Ê¹²â¶¨½á¹ûÆ«´óµÄÊÇ£¨a£©¡¢£¨c£©£®
£¨a£©CuOδȫ²¿»¹Ô­ÎªCu  £¨b£©CuOÊܳ±   £¨c£©CuOÖлìÓÐCu£®

Answer 1

·ÖÎö £¨I£©¸ù¾ÝÖ¤Ã÷Al£¨OH£©₃³ÊÁ½ÐÔµÄÊÔ¼ÁÑ¡Ôñ£»
£¨II£©£¨1£©¸ù¾ÝÑõ»¯»¹Ô­·´Ó¦ÅжÏÁíÒ»ÖÖÉú³ÉÎ¸ù¾Ý·´Ó¦Îï¡¢Éú³ÉÎï¡¢·´Ó¦Ìõ¼þд³öÏàÓ¦µÄ·½³Ìʽ£»¸ù¾Ýת»¯¹ØϵÅжÏÎïÖʵÄÎȶ¨ÐÔ£»
£¨2£©¸ù¾Ýת»¯¹ØϵÅжÏÎïÖʵÄÎȶ¨ÐÔ£»
£¨III£©£¨1£©Ñõ»¯Í­ÎªºÚÉ«¹ÌÌ壬±»»¹Ô­ÎªÍ­µ¥ÖʱäΪºìÉ«£»¸ù¾ÝÔªËØÊغã½áºÏÌâÒâÅжÏÉú³ÉÎﲢд³ö·´Ó¦·½³Ìʽ£»
£¨2£©¸ù¾ÝÌâ¸øµÄÒªÇóÀ´²âÁ¿Ñõ»¯Í­ºÍË®µÄÖÊÁ¿£»¸ù¾Ý°±ÆøµÄÐÔÖÊÅжÏ×°ÖÃEµÄ×÷Óã»
£¨3£©ÒªÊ¹²â¶¨½á¹ûÆ«´ó£¬Ôòm£¨H₂O£©ÒªÆ«Ð¡£¬¾Ý´Ë¶Ô¸÷Ñ¡Ïî½øÐÐÅжϣ®

½â´ð ½â£º£¨I£©Ö¤Ã÷Al£¨OH£©₃³ÊÁ½ÐÔµÄÊÔ¼ÁÊÇÇ¿ËáºÍÇ¿¼î£¬ÒªÏëÖ¤Ã÷Cu£¨OH£©₂³ÊÁ½ÐÔ£¬Ò²±ØÐëÑ¡Ç¿ËáºÍÇ¿¼îÈÜÒº£®ÁòËáÊÇÇ¿ËᣬËùÒÔ¿ÉÒÔÑ¡È¡£»ÇâÑõ»¯ÄÆÊÇÇ¿¼î£¬ËùÒÔ¿ÉÑ¡È¡£»°±Ë®ÊÇÈõ¼î£¬ËùÒÔ²»¿ÉÑ¡£®¹ÊÑ¡a¡¢b£»
£¨II£©£¨1£©¸ù¾ÝÑõ»¯»¹Ô­·´Ó¦£¬Ñõ»¯Í­×÷Ñõ»¯¼ÁºÍ»¹Ô­¼Á£¬µÃµç×ÓÉú³ÉÑõ»¯ÑÇÍ­£¬Ê§µç×ÓÉú³ÉÑõÆø£¬ËùÒÔ·´Ó¦ÎïÊÇÑõ»¯Í­£¬Éú³ÉÎïÊÇÑõ»¯ÑÇÍ­ºÍÑõÆø£¬·´Ó¦Ìõ¼þÊǸßΣ¬ËùÒÔ·½³ÌʽΪ£º4CuO$\frac{\underline{\;¸ßÎÂ\;}}{\;}$2Cu₂O+O₂¡ü£» ÎïÖÊÓɲ»Îȶ¨ÎïÖÊת»¯ÎªÎȶ¨ÎïÖʵÄÇãÏò£¬ËùÒÔÔÚ¸ßÎÂÌõ¼þÏ£¬Ê®l¼ÛµÄCu±ÈÊ®2¼ÛCu¸üÎȶ¨£¬
¹Ê´ð°¸Îª£º4CuO$\frac{\underline{\;¸ßÎÂ\;}}{\;}$2Cu₂O+O₂¡ü£»Îȶ¨£»
£¨2£©ÎïÖÊÓɲ»Îȶ¨ÎïÖÊת»¯ÎªÎȶ¨ÎïÖʵÄÇãÏò£¬ËùÒÔÔÚËáÐÔÈÜÒºÖУ¬+1¼ÛCu±È+2¼ÛCu¸ü²»Îȶ¨£®¹Ê´ð°¸Îª£º²»Îȶ¨£®
£¨III£©£¨1£©×°ÖÃCÖкÚÉ«¹ÌÌå±äΪºìÉ«£¬ÔòÖ¤Ã÷CuO±»°±Æø»¹Ô­£¬¸ù¾ÝÔªËØÊغã½áºÏÌâÒâÖª£¬Éú³ÉÖеÄÆøÌåÊǵªÆø£¬Í¬Ê±ÓÐË®Éú³É£¬ËùÒÔ·´Ó¦·½³ÌʽΪ3CuO+2NH₃$\frac{\underline{\;¸ßÎÂ\;}}{\;}$N₂ ¡ü+3Cu+3H₂O£¬¹Ê´ð°¸Îª£º¹ÌÌåÓɺÚɫת±äºìÉ«£»3CuO+2NH₃$\frac{\underline{\;¸ßÎÂ\;}}{\;}$N₂+3Cu+3H₂O£»£¨2£©ÉèÍ­µÄÏà¶ÔÔ­×ÓÖÊÁ¿Îªx
      2NH₃+3CuO$\frac{\underline{\;¸ßÎÂ\;}}{\;}$N₂+3Cu+3H₂O
           3£¨x+16£©54
           m£¨CuO£©       m£¨H₂O£©  
$\frac{3£¨x+16£©}{m£¨CuO£©}=\frac{54}{m£¨{H}_{2}O£©}$£¬½âµÃx=$\frac{18m£¨CuO£©}{m£¨{H}_{2}O£©}$-16£¬Å¨ÁòËá¿ÉÒÔÎüÊÕ¶àÓàµÄ°±Æø£¬·ÀÖ¹ÎÛȾ´óÆø£¬»¹¿ÉÒÔ·ÀÖ¹¿ÕÆøÖÐË®ÕôÆø½øÈë×°ÖÃD£¬Ó°Ïì²â¶¨½á¹û£¬
¹Ê´ð°¸Îª£º$\frac{18m£¨CuO£©}{m£¨{H}_{2}O£©}$-16£»²»¿ÉÒÔ£»³ä·ÖÎüÊÕ·´Ó¦¶àÓàµÄNH₃²¢·ÀÖ¹¿ÕÆøÖÐË®·Ö½øÈ룻
£¨3£©ÒªÊ¹²â¶¨½á¹ûÆ«´ó£¬Ôòm£¨H₂O£©ÒªÆ«Ð¡£¬ÆäÖУ¨a£©µ¼ÖÂm£¨H₂O£©Æ«Ð¡¡¢£¨b£©µ¼ÖÂm£¨H₂O£©Æ«´ó¡¢£¨c£©Ï൱ÓÚm£¨H₂O£©Æ«Ð¡£¬¹ÊÑ¡£¨a£©£¨c£©£¬
¹Ê´ð°¸Îª£º£¨a£©£¨c£©£®

µãÆÀ ±¾Ì⿼²éÁËCuµÄ³£¼û»¯ºÏÎïµÄÐÔÖʵÄʵÑé̽¾¿£¬²ÉÓüÙÉèµÄ·½·¨ÅжÏÍ­µÄ³£¼û»¯ºÏÎïµÄÐÔÖÊ£¬È»ºó×÷ʵÑéÑéÖ¤£»Òª×¢ÒâµÄÊÇ£ºÓÐÓж¾ÆøÌå²úÉúʱ£¬Ò»¶¨ÓÐβÆø´¦Àí×°Ö㬲»ÄÜÖ±½ÓÅÅ¿Õ£®