ÌâÄ¿ÄÚÈÝ
°±ÆøÔÚ¹¤Å©ÒµÉú²úÖÐÓÐÖØÒªÓ¦Óá£
£¨1£©¢ÙµªÆøÓÃÓÚ¹¤ÒµºÏ³É°±£¬Ð´³öµªÆøµÄµç×Óʽ £»
¢ÚNH3µÄÎȶ¨ÐÔ±ÈPH3 £¨Ìîд¡°Ç¿¡±»ò¡°Èõ¡±£©¡£
£¨2£©ÈçÏÂͼËùʾ£¬ÏòNaOH¹ÌÌåÉϵμ¸µÎŨ°±Ë®£¬Ñ¸ËÙ¸ÇÉϸǣ¬¹Û²ìÏÖÏó¡£
¢ÙŨÑÎËáÒºµÎ¸½½ü»á³öÏÖ°×ÑÌ£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ ¡£
¢ÚŨÁòËáÒºµÎÉÏ·½Ã»ÓÐÃ÷ÏÔÏÖÏó£¬Ò»¶Îʱ¼äºóŨÁòËáµÄÒºµÎÖÐÓа×É«¹ÌÌ壬¸Ã¹ÌÌå¿ÉÄÜÊÇ £¨Ð´»¯Ñ§Ê½£¬Ò»ÖÖ¼´¿É£©¡£
¢ÛFeSO4ÒºµÎÖÐÏȳöÏÖ»ÒÂÌÉ«³Áµí£¬¹ýÒ»¶Îʱ¼äºó±ä³ÉºìºÖÉ«£¬·¢ÉúµÄ·´Ó¦°üÀ¨
Fe2++2NH3¡¤H2O=Fe(OH)2¡ý+ 2NH4+ ºÍ ¡£
£¨3£©¿ÕÆø´µÍÑ·¨ÊÇÄ¿Ç°Ïû³ýNH3¶ÔË®ÌåÎÛȾµÄÖØÒª·½·¨¡£ÔÚÒ»¶¨Ìõ¼þÏ£¬ÏòË®ÌåÖмÓÈëÊÊÁ¿NaOH¿ÉʹNH3µÄÍѳýÂÊÔö´ó£¬ÓÃƽºâÒƶ¯ÔÀí½âÊÍÆäÔÒò ¡£
£¨4£©ÔÚ΢ÉúÎï×÷ÓÃÏ£¬µ°°×ÖÊÔÚË®Öзֽâ²úÉúµÄ°±Äܹ»±»ÑõÆøÑõ»¯Éú³ÉÑÇÏõËᣨHNO2£©£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ £¬Èô·´Ó¦ÖÐÓÐ0.3 molµç×Ó·¢ÉúתÒÆʱ£¬Éú³ÉÑÇÏõËáµÄÖÊÁ¿Îª g£¨Ð¡Êýµãºó±£ÁôÁ½Î»ÓÐЧÊý×Ö£©¡£
£¨1£©¢ÙµªÆøÓÃÓÚ¹¤ÒµºÏ³É°±£¬Ð´³öµªÆøµÄµç×Óʽ £»
¢ÚNH3µÄÎȶ¨ÐÔ±ÈPH3 £¨Ìîд¡°Ç¿¡±»ò¡°Èõ¡±£©¡£
£¨2£©ÈçÏÂͼËùʾ£¬ÏòNaOH¹ÌÌåÉϵμ¸µÎŨ°±Ë®£¬Ñ¸ËÙ¸ÇÉϸǣ¬¹Û²ìÏÖÏó¡£
¢ÙŨÑÎËáÒºµÎ¸½½ü»á³öÏÖ°×ÑÌ£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ ¡£
¢ÚŨÁòËáÒºµÎÉÏ·½Ã»ÓÐÃ÷ÏÔÏÖÏó£¬Ò»¶Îʱ¼äºóŨÁòËáµÄÒºµÎÖÐÓа×É«¹ÌÌ壬¸Ã¹ÌÌå¿ÉÄÜÊÇ £¨Ð´»¯Ñ§Ê½£¬Ò»ÖÖ¼´¿É£©¡£
¢ÛFeSO4ÒºµÎÖÐÏȳöÏÖ»ÒÂÌÉ«³Áµí£¬¹ýÒ»¶Îʱ¼äºó±ä³ÉºìºÖÉ«£¬·¢ÉúµÄ·´Ó¦°üÀ¨
Fe2++2NH3¡¤H2O=Fe(OH)2¡ý+ 2NH4+ ºÍ ¡£
£¨3£©¿ÕÆø´µÍÑ·¨ÊÇÄ¿Ç°Ïû³ýNH3¶ÔË®ÌåÎÛȾµÄÖØÒª·½·¨¡£ÔÚÒ»¶¨Ìõ¼þÏ£¬ÏòË®ÌåÖмÓÈëÊÊÁ¿NaOH¿ÉʹNH3µÄÍѳýÂÊÔö´ó£¬ÓÃƽºâÒƶ¯ÔÀí½âÊÍÆäÔÒò ¡£
£¨4£©ÔÚ΢ÉúÎï×÷ÓÃÏ£¬µ°°×ÖÊÔÚË®Öзֽâ²úÉúµÄ°±Äܹ»±»ÑõÆøÑõ»¯Éú³ÉÑÇÏõËᣨHNO2£©£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ £¬Èô·´Ó¦ÖÐÓÐ0.3 molµç×Ó·¢ÉúתÒÆʱ£¬Éú³ÉÑÇÏõËáµÄÖÊÁ¿Îª g£¨Ð¡Êýµãºó±£ÁôÁ½Î»ÓÐЧÊý×Ö£©¡£
£¨1£©¢Ù£¨1·Ö£© ¢ÚÇ¿£¨1·Ö£©
£¨2£©¢ÙNH3+HCl=NH4Cl£¨1·Ö£© ¢ÚNH4HSO4»ò(NH4)2SO4£¨1·Ö£©
¢Û4Fe(OH)2+O2+2H2O=4Fe(OH)3£¨2·Ö£©
£¨3£©°±ÔÚË®ÖдæÔÚƽºâ£ºNH3+H2ONH3¡¤H2ONH4+ +OH-£¨1·Ö£¬Ð´³ÉNH3+H2ONH4+ +OH-²»¿Û·Ö£©£¬¼ÓÈëNaOHºóOH-Ũ¶ÈÔö´óƽºâÄæÏòÒƶ¯£¨1·Ö£©£¬¹ÊÓÐÀûÓÚ°±µÄÍѳý£¨2·Ö£©
£¨4£©2NH3+3O2=2HNO2+2H2O£¨2·Ö£¬²»Ð´Î¢ÉúÎï²»¿Û·Ö£© 2.35 g£¨2·Ö£©
£¨2£©¢ÙNH3+HCl=NH4Cl£¨1·Ö£© ¢ÚNH4HSO4»ò(NH4)2SO4£¨1·Ö£©
¢Û4Fe(OH)2+O2+2H2O=4Fe(OH)3£¨2·Ö£©
£¨3£©°±ÔÚË®ÖдæÔÚƽºâ£ºNH3+H2ONH3¡¤H2ONH4+ +OH-£¨1·Ö£¬Ð´³ÉNH3+H2ONH4+ +OH-²»¿Û·Ö£©£¬¼ÓÈëNaOHºóOH-Ũ¶ÈÔö´óƽºâÄæÏòÒƶ¯£¨1·Ö£©£¬¹ÊÓÐÀûÓÚ°±µÄÍѳý£¨2·Ö£©
£¨4£©2NH3+3O2=2HNO2+2H2O£¨2·Ö£¬²»Ð´Î¢ÉúÎï²»¿Û·Ö£© 2.35 g£¨2·Ö£©
ÊÔÌâ·ÖÎö£º£¨1£©¢Ùµª·Ö×ÓÄÚÁ½NÔ×Ó¼äÐγÉ3¶Ô¹²Óõç×Ó£¬µç×ÓÊýΪ£º
¢ÚNÔªËطǽðÊôÐÔÇ¿ÓÚPÔªËØ£¬ËùÒÔNH3µÄÎȶ¨ÐÔ±ÈPH3Ç¿¡£
£¨2£©¢ÙŨÑÎËá»Ó·¢³öµÄHClÓëŨ°±Ë®»Ó·¢³öµÄNH3·´Ó¦£¬Éú³ÉµÄ°×ÑÌΪNH4Cl£¬»¯Ñ§·½³ÌʽΪ£ºNH3+HCl=NH4Cl¡£
¢ÚŨ°±Ë®»Ó·¢³öµÄNH3£¬½øÈëŨÁòËáÓëÁòËá·´Ó¦Éú³É£ºNH4HSO4»ò(NH4)2SO4¡£
¢Û×îÖÕ±äΪºìºÖÉ«ÊÇÉú³ÉÁËFe(OH)3£¬»¯Ñ§·½³ÌʽΪ£º4Fe(OH)2+O2+2H2O=4Fe(OH)3¡£
£¨3£©NH3?H2OΪÈõ¼î£¬´æÔÚµçÀëƽºâ£ºNH3+H2O NH3?H2ONH4++OH?£¬¼ÓÈëNaOHºó£¬Ôö¼ÓÁËOH?Ũ¶È£¬Ê¹Æ½ºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬ÓÐÀûÓÚ°±µÄÍѳý¡£
£¨4£©NH3ºÍO2·´Ó¦Éú³ÉÑÇÏõËáºÍË®£¬Åäƽ¿ÉµÃ»¯Ñ§·½³Ìʽ£º2NH3+3O22HNO2+2H2O£»¸ù¾Ý»¯ºÏ¼ÛµÄ±ä»¯¿ÉÖª£º6¡Án£¨HNO2£©=0.3mol£¬n£¨HNO2£©=0.05mol£¬Ôòm£¨HNO2£©=2.35g¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿