Question

1£®¼ÓÈë0.1molµÄMnO₂·ÛÄ©ÓÚ50mL¹ýÑõ»¯ÇâµÄÈÜÒºÖУ¨ÃܶÈΪ 1.1g•mL^-1£©£¬ÔÚ±ê×¼×´¿öÏ·ųöÆøÌåµÄÌå»ýºÍʱ¼äµÄ¹ØϵÈçͼËù Ê¾£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©A¡¢B¡¢C¡¢DËĵ㻯ѧ·´Ó¦ËÙÂÊÓÉ¿ìµ½ÂýµÄ˳ÐòΪD£¾C£¾B£¾A£®
£¨2£©¸Ã·´Ó¦ËÙÂʱ仯µÄÔ­ÒòÊÇËæ×Å·´Ó¦µÄ½øÐУ¬¹ýÑõ»¯ÇâÈÜÒºµÄŨ¶ÈÖð½¥¼õС£¬·´Ó¦ËÙÂÊÒ²Ëæ׿õС£®
£¨3£©¹ýÑõ»¯ÇâµÄ³õʼÎïÖʵÄÁ¿Å¨¶ÈΪ0.107mol•L^-1£®£¨±£ÁôÈýλÓÐЧÊý×Ö£©
£¨4£©·´Ó¦½øÐе½2·ÖÖÓʱ¹ýÑõ»¯ÇâµÄÖÊÁ¿·ÖÊýΪ0.083%£®£¨±£ÁôÈýλÓÐЧÊý×Ö£©

Answer 1

·ÖÎö ·´Ó¦·½³ÌʽΪ£º2H₂O₂$\frac{\underline{\;MnO_2\;}}{\;}$2H₂O+O₂¡ü£¬¸Ã·´Ó¦Îª²»¿ÉÄæ·´Ó¦£¬ÔÚ5minºó£¬ÊÕ¼¯µ½µÄÆøÌåÌå»ý²»ÔÙÔö¼Ó£¬ËµÃ÷¹ýÑõ»¯ÇâÍêÈ«·Ö½â£¬¸ù¾ÝͼÏó¿ÉÖªÉú³ÉµÄÑõÆøµÄÌå»ý£¬¸ù¾Ý·½³Ìʽ¼ÆËã¹ýÑõ»¯ÇâŨ¶È£¬ÒԴ˽â´ð¸ÃÌ⣮

½â´ð ½â£º£¨1£©·´Ó¦ÎïŨ¶È´óС¾ö¶¨·´Ó¦ËÙÂÊ´óС£¬Ëæ×Å·´Ó¦µÄ½øÐУ¬Ë«ÑõË®µÄŨ¶ÈÖð½¥¼õС£¬·´Ó¦ËÙÂÊÒ²Ëæ׿õС£¬A¡¢B¡¢C¡¢DËĵ㻯ѧ·´Ó¦ËÙÂÊÓÉ¿ìµ½ÂýµÄ˳ÐòΪD£¾C£¾B£¾A£¬¹Ê´ð°¸Îª£ºD£¾C£¾B£¾A£»
£¨2£©Ëæ×Å·´Ó¦µÄ½øÐУ¬¹ýÑõ»¯Çâ²»¶ÏÏûºÄ£¬Å¨¶ÈÖð½¥½µµÍ£¬Ôò·´Ó¦ËÙÂÊÖð½¥¼õС£¬¹Ê´ð°¸Îª£ºËæ×Å·´Ó¦µÄ½øÐУ¬¹ýÑõ»¯ÇâÈÜÒºµÄŨ¶ÈÖð½¥¼õС£¬·´Ó¦ËÙÂÊÒ²Ëæ׿õС£»
£¨3£©ÓÉ·´Ó¦·½³ÌʽΪ£º2H₂O₂$\frac{\underline{\;MnO_{2}\;}}{\;}$2H₂O+O₂¡ü£¬¸Ã·´Ó¦Îª²»¿ÉÄæ·´Ó¦£¬ÔÚ5minºó£¬ÊÕ¼¯µ½µÄÆøÌåÌå»ý²»ÔÙÔö¼Ó£¬ËµÃ÷¹ýÑõ»¯ÇâÍêÈ«·Ö½â£¬
ÓÉͼÏó¿ÉÖª£¬Éú³ÉÑõÆøµÄÌå»ýΪ60mL£¬
2H₂O₂$\frac{\underline{\;MnO_{2}\;}}{\;}$2H₂O+O₂¡ü
2mol            22.4L
n£¨H₂O₂£©       0.06L
n£¨H₂O₂£©=$\frac{2mol¡Á0.06L}{22.4L}$=0.00536mol£¬ËùÒÔc£¨H₂O₂£©=$\frac{0.00536mol}{0.05L}$=0.107 mol•L^-1£¬
¹Ê´ð°¸Îª£º0.107 mol•L^-1£»
£¨4£©Éè2minʱ·´Ó¦ÏûºÄµÄ¹ýÑõ»¯ÇâΪxmol£¬
 2H₂O₂$\frac{\underline{\;MnO_2\;}}{\;}$ 2H₂O+O₂¡ü
  2mol                     22.4L
   x                          0.045L
n£¨H₂O₂£©=$\frac{2mol¡Á0.045L}{22.4L}$=0.00402mol£¬
Ôò2minʱÈÜÒºÖÐÊ£ÓàµÄ¹ýÑõ»¯ÇâΪ0.00536mol-0.00402mol=0.00134mol£¬Ôòm£¨H₂O₂£©=nM=0.00134mol¡Á34g/mol=0.04556g£¬Éú³ÉµÄÑõÆøµÄÖÊÁ¿Îªm=0.00402¡Á$\frac{1}{2}$
¡Á32=0.06432g£¬Ôò·´Ó¦½øÐе½2·ÖÖÓʱ¹ýÑõ»¯ÇâµÄÖÊÁ¿·ÖÊýΪ£º$\frac{0.04556}{50¡Á1.1-0.06432}$¡Á100%=0.083%£»
¹Ê´ð°¸Îª£º0.083%£®

µãÆÀ ±¾Ì⿼²é¹ýÑõ»¯Çâ·Ö½âµÄËÙÂÊÇúÏߣ¬ÌâÄ¿ÄѶÈÖеȣ¬Ö÷Òª¿¼²éŨ¶È¶Ô·´Ó¦ËÙÂʵÄÓ°Ïì¡¢Óйط½³ÌʽµÄ¼ÆËãµÈ£¬ÕýÈ··ÖÎöͼÏóµÄÇúÏ߱仯Êǽâ´ð±¾ÌâµÄ¹Ø¼ü£®