Question

11£®»¯¹¤Ô­ÁϺ췯ÄÆ£¨ÖظõËáÄÆ£ºNa₂Cr₂O₇•2H₂O£©Ö÷ÒªÊÇÒÔ¸õÌú¿ó£¨Ö÷Òª³É·ÝΪ FeO•Cr₂O₃£¬»¹º¬ÓÐA1₂O₃¡¢SiO₂ µÈÔÓÖÊ£©ÎªÖ÷ÒªÔ­ÁÏÉú²ú£¬ÆäÖ÷Òª¹¤ÒÕÁ÷³ÌÈçÏ£¨Èçͼ1£©£º

²½Öè¢ÙÖÐÖ÷Òª·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º4FeO•Cr₂O₃+8Na₂CO₃+7O₂=8Na₂CrO₄+2Fe₂O₃+8CO₂
£¨1£©ÔÓÖÊ A1₂O₃¡¢SiO₂ÔÚ¢ÙÖÐת»¯ºó¾­¼ÓË®¹ýÂ˽øÈëÂËÒº£¬Ð´³öA1₂O₃ÔÚ¢ÙÖÐת»¯µÄ»¯Ñ§·´Ó¦·½³ÌʽAl₂O₃+Na₂CO₃$\frac{\underline{\;¸ßÎÂ\;}}{\;}$2NaAlO₂+CO₂¡ü£®
£¨2£©Óû¯Ñ§Æ½ºâÒƶ¯Ô­Àí˵Ã÷¢ÛÖÐÖó·ÐµÄ×÷ÓÃÊÇË®½âÎüÈÈ£¬Öó·Ð´Ù½øË®½âƽºâAlO₂^-+2H₂O?Al£¨OH£©₃+OH^-¡¢SiO₃^2-+2H₂O?H₂SiO₃+2OH^-ÏòÓÒÒƶ¯£¬Éú³ÉÇâÑõ»¯ÂÁ¡¢¹èËá³Áµí£»H⁺½«Ë®½âÉú³ÉµÄÇâÑõ»¯ÂÁ³ÁµíÈܽ⣬ÔÓÖÊAl³⁺ÓëNa₂CrO₄ÄÑÒÔ·ÖÀ루ÓÃÀë×Ó·½³Ìʽ½áºÏÎÄ×Ö˵Ã÷£©
£¨3£©¢ÝÖÐËữÊÇʹ CrO₄^2һת»¯Îª Cr₂O₇^2Ò»£¬Èô1LËữºóËùµÃÈÜÒºÖк¬¸õÔªËصÄÖÊÁ¿Îª 28.6g£¬CrO₄^2-ÓÐ$\frac{10}{11}$ת»¯Îª Cr₂O₇^2-£®
¢ÙËữºóËùµÃÈÜÒºÖРc£¨Cr₂O₇^2- £©=0.25mol•L^-1
¢Ú¹¤ÒµÉÏÒ²¿ÉÉè¼Æͼ2ʾװÖã¨Á½¼«¾ùΪ¶èÐԵ缫£©µç½âNa₂CrO₄ÈÜÒºÖÆÈ¡ Na₂Cr₂O₇£¬Í¼2ÖÐÓÒ²àµç¼«µÄµç¼«·´Ó¦Ê½Îª2H₂O-4e^-=O₂+4H⁺»ò4OH^--4e^-=2H₂O+O₂¡ü£®Èôµç½âÇ°Á½²àÈÜÒºµÄÖÊÁ¿ÏàµÈ£¬Ôòµ±µç½â¹ý³ÌÖÐתÒÆÁË1mol µç×ÓʱÁ½²àÈÜÒºµÄÖÊÁ¿²îΪ53g£®

Answer 1

·ÖÎö ¸õÌú¿ó£¨Ö÷Òª³É·ÖΪFeO•Cr₂O₃£¬»¹º¬ÓÐAl₂O₃¡¢SiO₂µÈÔÓÖÊ£©¼ÓÈë̼ËáÄÆìÑÉÕ£¬¿ÉÉú³ÉNa₂CrO₄¡¢Fe₂O₃¡¢NaAlO₂¡¢Na₂SiO₃µÈ£¬¼ÓÈëË®½þÈ¡£¬µÃµ½ÂËÔü1ΪFe₂O₃£¬¹ýÂ˺ó½«ÂËҺϡÊÍ¡¢Öó·Ð¡¢µ÷½ÚpH7¡«8µÃµ½ÂËÔü2ΪAl£¨OH£©₃¡¢H₂SiO₃£¬¹ýÂ˺ó½«ÂËÒºËữ£¬·¢ÉúÁË·´Ó¦2Na₂CrO₄+H₂SO₄?Na₂SO₄+Na₂Cr₂O₇+H₂O£¬Éú³ÉÁËNa₂Cr₂O₇ºÍNa₂SO₄£¬Õô·¢½á¾§µÃµ½ÁòËáÄÆ£¬ÀäÈ´½á¾§µÃµ½Na₂Cr₂O₇•2H₂O£¬
£¨1£©ÔÓÖÊAl₂O₃ÔÚ¢ÙÖÐת»¯Éú³ÉNaAlO₂£»
£¨2£©ÂËÒºÖк¬ÓÐNaAlO₂¡¢Na₂SiO₃µÈ£¬¶¼ÎªÇ¿¼îÈõËáÑΣ¬¼ÓÈÈÖó·Ð£¬´Ù½øË®½â£¬Èôµ÷½ÚpH¹ýµÍ£¬H⁺½«Ë®½âÉú³ÉµÄÇâÑõ»¯ÂÁ³ÁµíÈܽ⣬ÔÓÖÊAl³⁺ÓëNa₂CrO₄ÄÑÒÔ·ÖÀ룻
£¨3£©¢ÙËữʱ·¢ÉúµÄ·´Ó¦Îª£º2CrO₄^2-+2H⁺?Cr₂O₇^2-+H₂O£¬¸ù¾ÝÊغãÁйØϵʽ£º2Cr¡«2CrO₄^2-¡«Cr₂O₇^2-£¬ÁÐʽ¼ÆËãµÃn£¨Cr₂O₇^2-£©ºÍn£¨CrO₄^2-£©Ê£Ó࣬½ø¶øÇó³öËữºóËùµÃÈÜÒºÖÐc£¨Cr₂O₇^2-£©£»
¢Ú¸ù¾Ý2CrO₄^2-+2H⁺?Cr₂O₇^2-+H₂OÉè¼ÆͼʾװÖ㨾ùΪ¶èÐԵ缫£©£¬ÓÒ²àµç¼«µç½âNa₂CrO₄ÈÜÒºÖÆÈ¡Na₂Cr₂O₇£¬ËµÃ÷Ôڸõ缫ÊÇÑô¼«£¬Á¬½ÓµçÔ´µÄÕý¼«£¬µç¼«·´Ó¦Ê½Îª2H₂O-4e^-=O₂+4H£¬Òõ¼«µç¼«·´Ó¦2H⁺+2e^-=H₂¡ü£¬¾Ý´Ë¼ÆË㣮

½â´ð ½â£º¸õÌú¿ó£¨Ö÷Òª³É·ÖΪFeO•Cr₂O₃£¬»¹º¬ÓÐAl₂O₃¡¢SiO₂µÈÔÓÖÊ£©¼ÓÈë̼ËáÄÆìÑÉÕ£¬¿ÉÉú³ÉNa₂CrO₄¡¢Fe₂O₃¡¢NaAlO₂¡¢Na₂SiO₃µÈ£¬¼ÓÈëË®½þÈ¡£¬µÃµ½ÂËÔü1ΪFe₂O₃£¬¹ýÂ˺ó½«ÂËҺϡÊÍ¡¢Öó·Ð¡¢µ÷½ÚpH7¡«8µÃµ½ÂËÔü2ΪAl£¨OH£©₃¡¢H₂SiO₃£¬¹ýÂ˺ó½«ÂËÒºËữ£¬·¢ÉúÁË·´Ó¦2Na₂CrO₄+H₂SO₄?Na₂SO₄+Na₂Cr₂O₇+H₂O£¬Éú³ÉÁËNa₂Cr₂O₇ºÍNa₂SO₄£¬Õô·¢½á¾§µÃµ½ÁòËáÄÆ£¬ÀäÈ´½á¾§µÃµ½Na₂Cr₂O₇•2H₂O£¬
£¨1£©ÔÓÖÊAl₂O₃ÔÚ¢ÙÖÐת»¯µÄ»¯Ñ§·´Ó¦·½³ÌʽΪAl₂O₃+Na₂CO₃$\frac{\underline{\;¸ßÎÂ\;}}{\;}$2NaAlO₂+CO₂¡ü£¬
¹Ê´ð°¸Îª£ºAl₂O₃+Na₂CO₃$\frac{\underline{\;¸ßÎÂ\;}}{\;}$2NaAlO₂+CO₂¡ü£»
£¨2£©ÂËÒºÖк¬ÓÐNaAlO₂¡¢Na₂SiO₃µÈ£¬¶¼ÎªÇ¿¼îÈõËáÑΣ¬¼ÓÈÈÖó·Ð£¬´Ù½øË®½âƽºâAlO₂^-+2H₂O?Al£¨OH£©₃+OH^-¡¢SiO₃^2-+2H₂O?H₂SiO₃+2OH^-ÏòÓÒÒƶ¯£¬Éú³ÉÇâÑõ»¯ÂÁ¡¢¹èËá³Áµí£¬Èôµ÷½ÚpH¹ýµÍ£¬H⁺½«Ë®½âÉú³ÉµÄÇâÑõ»¯ÂÁ³ÁµíÈܽ⣬ÔÓÖÊAl³⁺ÓëNa₂CrO₄ÄÑÒÔ·ÖÀ룬
¹Ê´ð°¸Îª£ºË®½âÎüÈÈ£¬Öó·Ð´Ù½øË®½âƽºâAlO₂^-+2H₂O?Al£¨OH£©₃+OH^-¡¢SiO₃^2-+2H₂O?H₂SiO₃+2OH^-ÏòÓÒÒƶ¯£¬Éú³ÉÇâÑõ»¯ÂÁ¡¢¹èËá³Áµí£» H⁺½«Ë®½âÉú³ÉµÄÇâÑõ»¯ÂÁ³ÁµíÈܽ⣬ÔÓÖÊAl³⁺ÓëNa₂CrO₄ÄÑÒÔ·ÖÀ룻
£¨3£©¢ÙËữʱ·¢ÉúµÄ·´Ó¦Îª£º2CrO₄^2-+2H⁺?Cr₂O₇^2-+H₂O£¬Èô1LËữºóËùµÃÈÜÒºÖк¬¸õÔªËصÄÖÊÁ¿Îª28.6g£¬CrO₄^2-ÓÐ$\frac{10}{11}$ת»¯ÎªCr₂O₇^2-£¬ËµÃ÷¸õÔªËØÓÐ$\frac{10}{11}$ת»¯ÎªCr₂O₇^2-
¸ù¾ÝÊغãÁйØϵʽ£º2Cr¡«2CrO₄^2-¡«Cr₂O₇^2-
                  2            1
                  $\frac{28.6g¡Á\frac{10}{11}}{52g/mol}$   n£¨Cr₂O₇^2-£©
µÃn£¨Cr₂O₇^2-£©=0.25mol£¬
¢ÙËữºóËùµÃÈÜÒºÖÐc£¨Cr₂O₇^2-£©=$\frac{n}{V}$=0.25mol•L^-1£¬
¹Ê´ð°¸Îª£º0.25mol•L^-1£»
¢Ú¸ù¾Ý2CrO₄^2-+2H⁺?Cr₂O₇^2-+H₂OÉè¼ÆͼʾװÖ㨾ùΪ¶èÐԵ缫£©µç½âNa₂CrO₄ÈÜÒºÖÆÈ¡Na₂Cr₂O₇£¬ÐèҪͨ¹ýµç½âÉú³ÉH⁺Ìá¸ßÈÜÒºµÄËáÐÔ£¬ËµÃ÷Ôڸõ缫ÊÇÑô¼«£¬Á¬½ÓµçÔ´µÄÕý¼«£¬µç¼«·´Ó¦Ê½Îª2H₂O-4e^-=O₂+4H⁺£¬Í¨µçÍê³Éºó£¬ÈôתÒÆ0.1molµç×Ó£¬Ñô¼«µç¼«·´Ó¦£º2H₂O-4e^-=O₂+4H⁺»ò4OH^--4e^-=2H₂O+O₂¡ü£¬Òõ¼«µç¼«·´Ó¦2H⁺+2e^-=H₂¡ü£¬2CrO₄^2-+2H⁺=Cr₂O₇^2-+H₂O£¬ÉèÔ­À´ÈÜÒºÖÊÁ¿ÏàͬΪm£¬ÒÀ¾ÝÈÜÒºÖеçºÉÊغ㣬ÔòÒõ¼«ÇøºÍÑô¼«ÇøµÄÖÊÁ¿²î=m-0.5mol¡Á2g/mol+1mol¡Á23g/mol-£¨m-0.25mol¡Á32g/mol-1mol¡Á23g/mol£©
=53g£¬
¹Ê´ð°¸Îª£º2H₂O-4e^-=O₂+4H⁺»ò4OH^--4e^-=2H₂O+O₂¡ü£»53£®

µãÆÀ ±¾Ì⿼²éÁËÎïÖÊÖƱ¸Á÷³ÌºÍ·½°¸µÄ·ÖÎöÅжϣ¬ÎïÖÊÐÔÖʵÄÓ¦Óã¬Îª¸ßƵ¿¼µã£¬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦¡¢ÊµÑéÄÜÁ¦µÄ¿¼²é£¬×¢Òâ°ÑÎÕÌâ¸ÉÐÅÏ¢µÄ·ÖÎöÀí½â£¬²Ù×÷²½ÖèµÄ×¢ÒâÎÊÌâºÍ»ù±¾²Ù×÷·½·¨ÊǽâÌâ¹Ø¼ü£¬ÌâÄ¿ÄѶÈÖеȣ®