·ÖÎö SnCl2ÔÚÑÎËáÖÐÈܽ⣬ÔÙ¼ÓÈëÎý·Û£¬ÈܽâµÃµ½SnCl2ÈÜÒº¼ÓÈë̼ËáÄƳÁµíÎýÀë×Ó£¬¹ýÂ˵õ½³ÁµíÏ´µÓºó¼ÓÈëÁòËáÈܽâµÃµ½ÁòËáÎýÈÜÒº£¬Õô·¢Å¨ËõÀäÈ´½á¾§£¬¹ýÂËÏ´µÓµÃµ½ÁòËáÎý¾§Ì壬
£¨1£©ÎýÔ×ӵĺ˵çºÉÊýΪ50£¬Óë̼ԪËØÊôÓÚͬһÖ÷×壬´¦ÓÚ¢ôA×壬¸ù¾ÝÔ×ÓÐòÊý¼õ¸÷ÖÜÆÚÈÝÄÉÔªËØÖÖÊýÈ·¶¨ËùÔÚµÄÖÜÆÚ£»
£¨2£©ÓÉÁ÷³Ìͼ¿ÉÖª£¬²Ù×÷¢ñÊÇ´ÓÈÜÒºÖеõ½º¬½á¾§Ë®µÄ¾§Ì壬ֻÄܲÉÈ¡Õô·¢¡¢Å¨Ëõ¡¢ÀäÈ´½á¾§¡¢¹ýÂË¡¢Ï´µÓµÃµ½£»
£¨3£©ÓÉÐÅÏ¢¿ÉÖª£¬SnCl2Ò×Ë®½âÉú³É¼îʽÂÈ»¯ÑÇÎý£¬¼ÓÈëÑÎËᣬÒÖÖÆSn2+Ë®½â£»
£¨4£©ÓÉÐÅÏ¢¿ÉÖª£¬Sn2+Ò×±»Ñõ»¯£¬¼ÓÈëSn·Û³ýµ÷½ÚÈÜÒºpHÍ⣬»¹·ÀÖ¹Sn2+±»Ñõ»¯£»
£¨5£©ËáÐÔÌõ¼þÏ£¬SnSO4»¹¿ÉÒÔÓÃ×÷Ë«Ñõˮȥ³ý¼Á£¬Ë«ÑõË®ÓÐÇ¿Ñõ»¯ÐÔ£¬½«Sn2+Ò×±»Ñõ»¯ÎªSn4+£¬×ÔÉí±»»¹ÔΪˮ£»
£¨6£©¸ù¾Ýµç×ÓתÒÆÊغãÓë·½³Ìʽ¿ÉµÃ¹ØϵʽSn¡«Sn2+¡«2Fe3+¡«2Fe2+¡«$\frac{1}{3}$K2Cr2O7£¬¾Ý´Ë¼ÆË㣮
½â´ð ½â£ºSnCl2ÔÚÑÎËáÖÐÈܽ⣬ÔÙ¼ÓÈëÎý·Û£¬ÈܽâµÃµ½SnCl2ÈÜÒº¼ÓÈë̼ËáÄƳÁµíÎýÀë×Ó£¬¹ýÂ˵õ½³ÁµíÏ´µÓºó¼ÓÈëÁòËáÈܽâµÃµ½ÁòËáÎýÈÜÒº£¬Õô·¢Å¨ËõÀäÈ´½á¾§£¬¹ýÂËÏ´µÓµÃµ½ÁòËáÎý¾§Ì壬
£¨1£©ÎýÔªËØÓë̼ԪËØÊôÓÚͬһÖ÷×壬´¦ÓÚ¢ôA×壬Ô×Ӻ˵çºÉÊýΪ50£¬Ôò£º50-2-8-8-18=14£¬¹ÊSn´¦ÓÚµÚÎåÖÜÆÚ£¬ÔòÔÚÖÜÆÚ±íÖеÄλÖÃΪµÚÎåÖÜÆÚµÚ¢ôA×壬
¹Ê´ð°¸Îª£ºµÚÎåÖÜÆÚµÚ¢ôA×壻
£¨2£©ÓÉÁ÷³Ìͼ¿ÉÖª£¬²Ù×÷¢ñÊÇ´ÓÈÜÒºÖеõ½º¬½á¾§Ë®µÄ¾§Ì壬ֻÄܲÉÈ¡Õô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§¡¢¹ýÂË¡¢Ï´µÓµÃµ½£¬
¹Ê´ð°¸Îª£ºÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§¡¢¹ýÂË¡¢Ï´µÓ£»
£¨3£©ÓÉÐÅÏ¢¿ÉÖª£¬SnCl2Ò×Ë®½âÉú³É¼îʽÂÈ»¯ÑÇÎý£¬´æÔÚƽºâSn Cl2+H2O?Sn£¨OH£©Cl+HCl£¬¼ÓÈëÑÎËᣬʹ¸ÃƽºâÏò×óÒƶ¯£¬ÒÖÖÆSn2+Ë®½â£¬
¹Ê´ð°¸Îª£ºSnCl2Ë®½â£¬·¢ÉúSnCl2+H2O?Sn£¨OH£©Cl+HCl£¬¼ÓÈëÑÎËᣬʹ¸ÃƽºâÏò×óÒƶ¯£¬ÒÖÖÆSn2+Ë®½â£»
£¨4£©ÓÉÐÅÏ¢¿ÉÖª£¬Sn2+Ò×±»Ñõ»¯£¬¼ÓÈëSn·Û³ýµ÷½ÚÈÜÒºpHÍ⣬»¹·ÀÖ¹Sn2+±»Ñõ»¯£¬
¹Ê´ð°¸Îª£º·ÀÖ¹Sn2+±»Ñõ»¯£»
£¨5£©ËáÐÔÌõ¼þÏ£¬SnSO4»¹¿ÉÒÔÓÃ×÷Ë«Ñõˮȥ³ý¼Á£¬Ë«ÑõË®ÓÐÇ¿Ñõ»¯ÐÔ£¬½«Sn2+Ò×±»Ñõ»¯ÎªSn4+£¬×ÔÉí±»»¹ÔΪˮ£¬Àë×Ó·½³ÌʽΪ£ºSn2++H2O2+2H+¨TSn4++2H2O£¬
¹Ê´ð°¸Îª£ºSn2++H2O2+2H+¨TSn4++2H2O£»
£¨6£©ÁîÎý·ÛÖÐÎýµÄÖÊÁ¿·ÖÊýΪx£¬Ôò£º
Sn¡«Sn2+¡«2Fe3+¡«2Fe2+¡«$\frac{1}{3}$K2Cr2O7¼ÆË㣮
119g $\frac{1}{3}$mol
1.226g¡Áx 0.100mol/L¡Á0.032L
¹Ê$\frac{119g}{1.226xg}$=$\frac{\frac{1}{3}mol}{0.100mol/L¡Á0.032L}$
½âµÃx=93.2%£¬
¹Ê´ð°¸Îª£º93.2%£®
µãÆÀ ±¾ÌâSnSO4ÖƱ¸µÄÖ®±ÈΪÔØÌ壬¿¼²éѧÉú¶Ô¹¤ÒÕÁ÷³ÌµÄÀí½â¡¢ÎïÖʵķÖÀëÌá´¿¡¢ÔĶÁÌâÄ¿»ñÈ¡ÐÅÏ¢µÄÄÜÁ¦¡¢³£Óû¯Ñ§ÓÃÓïÊéд¡¢µÎ¶¨Ó¦Óü°ÀûÓùØϵʽ½øÐеļÆËãµÈ£¬ÄѶÈÖеȣ¬¶ÔѧÉúµÄ»ù´¡ÖªÊ¶¼°Âß¼ÍÆÀíÓнϸߵÄÒªÇó£®
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£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
ʵÑé²½Öè | ¼ÓÈëÊÔ¼Á | Àë×Ó·½³Ìʽ |
£¨1£© | Ï¡HCl | CO32-+2H+=CO2¡ü+H2O |
£¨2£© | BaCl2ÈÜÒº | Ba2++SO42-=BaSO4¡ý |
£¨3£© | AgNO3ÈÜÒº | Ag++Cl-=AgCl¡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£º½â´ðÌâ
Àë¡¡×Ó | K+ | Na+ | NH4+ | SO42- | NO3- | Cl- |
Ũ¶È/mol•L-1 | 4¡Á10-6 | 6¡Á10-6 | 2¡Á10-5 | 4¡Á10-5 | 3¡Á10-5 | 2¡Á10-5 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ÎÞÉ«ÈÜÒºÖУºAl3+¡¢Cl-¡¢MnO4-¡¢SO42- | |
B£® | º¬ÓдóÁ¿HCO3-µÄÈÜÒºÖУºNa+¡¢Ca2+¡¢NO3-¡¢Cl- | |
C£® | 0.1mol•L-1AgNO3ÈÜÒº£ºH+¡¢K+¡¢SO42-¡¢Cl- | |
D£® | ʹʯÈï±äºìÉ«µÄÈÜÒº£ºCH3COO-¡¢Cl-¡¢NO3-¡¢K+ |
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