Question

13£®H₂O₂ÊÇÖØÒªµÄ»¯Ñ§ÊÔ¼Á£¬ÔÚʵÑéÊÒºÍÉú²úʵ¼ÊÖÐÓ¦Óù㷺£®
£¨1£©H₂O₂²»Îȶ¨£¬ÆäÉú²ú¹ý³ÌÒªÑϸñ±ÜÃâ»ìÈëFe²⁺ºÍFe³⁺£¬·ñÔò»á·¢ÉúÈçÏÂÁ½¸ö·´Ó¦£¬¼Ó¿ìÆä·Ö½â£¬ÇÒFe²⁺ºÍFe³⁺¸÷×ÔµÄÁ¿±£³Ö²»±ä£®
¢Ù2Fe²⁺+H₂O₂+2H⁺=2Fe³⁺+2H₂O
¢Ú2Fe³⁺+H₂O₂=O₂¡ü+2Fe²⁺+2H⁺£¨ÌîÀë×Ó·½³Ìʽ£©£®
£¨2£©H₂O₂µÄË®ÈÜÒºÏÔÈõËáÐÔ£¬Ð´³öÒ»¸öÓÃÑÎËáÖÆÈ¡H₂O₂µÄ»¯Ñ§·½³ÌʽNa₂O₂+2HCl¨T2NaCl+H₂O₂£®
£¨3£©H₂O₂ÔÚʹÓÃÖÐͨ³£²»»á´øÀ´ÎÛȾ£®
¢Ù¹¤ÒµÉÏÀûÓÃCl₂+H₂O₂=2HCl+O₂³ýÂÈ£¬ÔÚ´Ë·´Ó¦Öб»Ñõ»¯µÄÎïÖÊÊÇH₂O₂£®
¢ÚÏû³ýij·ÏÒºÖеÄÇ軯Îº¬CN^-£©£¬Íê³ÉÏÂÁз´Ó¦£ºCN^-+H₂O₂+H₂O=HCO₃^-+NH₃¡ü
£¨4£©ÊµÑéÊÒ³£ÓÃËáÐÔKMnO₄ÈÜÒº²â¶¨H₂O₂µÄŨ¶È£®È¡10.00mL H₂O₂ÈÜÒºÑùÆ·£¨ÃܶȽüËÆΪ1g/mL£©ÓÚ׶ÐÎÆ¿ÖУ¬¼ÓË®µ½50.00mL£¬µ±µÎÈë0.10mol/L KMnO₄ ÈÜÒº40.00mLºó£¬Ç¡ºÃÍêÈ«·´Ó¦£®ÔòÔ­ÑùÆ·ÖÐH₂O₂µÄÖÊÁ¿·ÖÊýÊÇ3.4%£®

Answer 1

·ÖÎö £¨1£©Fe²⁺ºÍFe³⁺¸÷×ÔµÄÁ¿±£³Ö²»±ä£¬¿ÉÖªµÚ¶þ¸öÀë×Ó·´Ó¦ÖÐÌúÀë×Óת»¯ÎªÑÇÌúÀë×Ó£»
£¨2£©ÓÃÑÎËáÖÆÈ¡H₂O₂£¬ÀûÓÃÑÎËáÓë¹ýÑõ»¯ÄÆ·¢ÉúÇ¿ËáÖÆÈ¡ÈõËáµÄ·´Ó¦£»
£¨3£©¢ÙCl₂+H₂O₂=2HCl+O₂ÖУ¬ClÔªËصĻ¯ºÏ¼Û½µµÍ£¬OÔªËصĻ¯ºÏ¼ÛÉý¸ß£»
¢Ú½áºÏÔ­×Ó¡¢µçºÉÊغã·ÖÎö£»
£¨4£©·¢Éú2MnO₄^-+5H₂O₂+6H⁺=2Mn²⁺+8H₂O+5O₂¡ü£¬½áºÏ·´Ó¦¼ÆË㣮

½â´ð ½â£º£¨1£©¹ýÑõ»¯ÇâÓÐÑõ»¯ÐÔ£¬ÄÜÑõ»¯Fe²⁺£¬·¢Éú2Fe²⁺+H₂O₂+2H⁺¨T2Fe³⁺+2H₂O£¬ÓÖÓл¹Ô­ÐÔ£¬Äܱ»Fe³⁺Ñõ»¯·¢Éú2Fe³⁺+H₂O₂=2Fe²⁺+O₂¡ü+2H⁺£¬
¹Ê´ð°¸Îª£º2Fe³⁺+H₂O₂=O₂¡ü+2Fe²⁺+2H⁺£»
£¨2£©¸ù¾ÝÑÎËáÄÜÓë¹ýÑõ»¯ÄÆ·´Ó¦Éú³ÉÂÈ»¯ÄƺÍH₂O₂£¬·½³ÌʽΪNa₂O₂ +2HCl=2NaCl+H₂O₂£¬
¹Ê´ð°¸Îª£ºNa₂O₂+2HCl¨T2NaCl+H₂O₂£»
£¨3£©¢ÙCl₂+H₂O₂¨T2HCl+O₂£¬¸Ã·´Ó¦ÖÐÔªËصĻ¯ºÏ¼Û±ä»¯Îª£ºH₂O₂¡úO₂£¬OÔªËØÓÉ-1¼Û¡ú0¼Û£¬Ê§µç×Ó£¬ËùÒÔH₂O₂ÊÇ»¹Ô­¼Á£¬»¹Ô­¼Á±»Ñõ»¯·¢ÉúÑõ»¯·´Ó¦£¬
¹Ê´ð°¸Îª£ºH₂O₂£»
¢ÚÓÉÔ­×ÓÊغãºÍµçºÉÊغãµÃ£ºCN^-+H₂O₂=HCO₃^-+NH₃¡ü£¬¹Ê´ð°¸Îª£ºHCO₃^-£»
£¨4£©ÏûºÄ¸ßÃÌËá¸ùµÄÎïÖʵÄÁ¿Îª0.1mol/L¡Á0.04L=0.004mol£¬ÉèË«ÑõË®µÄÎïÖʵÄÁ¿Îªn£¬Ôò
2MnO₄^-+5H₂O₂+6H⁺=2Mn²⁺+8H₂O+5O₂¡ü
2        5
0.004mol n
½âµÃn=0.01mol£¬
ËùÒÔË«ÑõË®µÄÖÊÁ¿Îª0.01mol¡Á34g/mol=0.34g£¬
10.00mL¡¢ÃܶȽüËÆΪ1g/mLµÄH₂O₂ÈÜÒºÑùÆ·µÄÖÊÁ¿Îª10g£¬Ë«ÑõË®µÄÖÊÁ¿·ÖÊý=$\frac{0.34g}{10g}$¡Á100%=3.4%£¬
¹Ê´ð°¸Îª£º3.4%£®

µãÆÀ ±¾Ì⿼²éÑõ»¯»¹Ô­·´Ó¦µÄ¼ÆË㣬Ϊ¸ßƵ¿¼µã£¬°ÑÎÕ·´Ó¦ÖÐÔªËصĻ¯ºÏ¼Û±ä»¯¡¢ÎïÖʵÄÁ¿¹ØϵΪ½â´ðµÄ¹Ø¼ü£¬²àÖØ·ÖÎöÓë¼ÆËãÄÜÁ¦µÄ¿¼²é£¬×¢ÒâÊغ㷨ӦÓã¬ÌâÄ¿ÄѶȲ»´ó£®