Question

9£®Ä³Ñо¿Ð¡×éΪÁË̽¾¿Ò»ÖÖ¿¹ËáÒ©X£¨½öº¬ÎåÖÖ¶ÌÖÜÆÚÔªËØ£©µÄ×é³ÉºÍÐÔÖÊ£¬Éè¼Æ²¢Íê³ÉÁËÈçÏÂʵÑ飺

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©°×É«³ÁµíBÖнðÊôÔªËصÄÀë×ӽṹʾÒâͼΪ£¬AµÄµç×ÓʽΪ£®
£¨2£©²½Öè3ÖÐÉú³É³ÁµíCµÄÀë×Ó·½³ÌʽΪAlO₂^-+CO₂+2H₂O=Al£¨OH£©₃¡ý+HCO₃^-£®
£¨3£©XµÄ»¯Ñ§Ê½ÎªAlMg₃£¨CO₃£©£¨OH£©₇£¬¸Ã¿¹ËáÒ©²»ÊʺÏθÀ£ÑñÖØÖ¢»¼ÕßʹÓã¬Ô­ÒòÊǺÍθËá·´Ó¦Éú³É¶þÑõ»¯Ì¼ÆøÌåÔì³Éθ´©¿×£®
£¨4£©Éè¼ÆʵÑé·½°¸ÑéÖ¤ÈÜÒºCÖеĺ¬Ì¼Àë×ÓÈ¡ÉÙÁ¿ÈÝÒ×ÉÙÐíÓÚÊÔ¹ÜÖУ¬¼ÓÈëÂÈ»¯¸ÆÎÞÏÖÏó£¬ÔÚ¼ÓÈëÑÎËáÉú³ÉÎÞÉ«ÎÞζµÄÆøÌ壬ͨÈë³ÎÇåʯ»ÒË®±ä»ë×Ç£¬Ö¤Ã÷º¬ÓÐ̼ËáÇâ¸ùÀë×Ó£®

Answer 1

·ÖÎö ¿¹ËáÒ©X¼ÓÈë¹ýÁ¿ÑÎËáÉú³ÉÎÞÉ«ÆøÌåAºÍÎÞÉ«ÈÜÒºA£¬ÈÜÒºAÖмÓÈë¹ýÁ¿ÇâÑõ»¯ÄÆÈÜÒº¹ýÂ˵õ½°×É«³ÁµíBºÍÈÜÒºB£¬ÈÜÒºBÖмÓÈë×ãÁ¿AµÃµ½°×É«³ÁµíC£¬³ÁµíC¼ÓÈë¹ýÁ¿ÇâÑõ»¯ÄÆÈÜÒºÍêÈ«Èܽ⣬˵Ã÷³Á³ÁµÄCΪAl£¨OH£©₃£¬ÔòÈÜÒºBº¬ÓÐÆ«ÂÁËáÄÆ£¬AΪCO2£¬ÎÞÉ«ÈÜÒºCΪ̼ËáÇâÄÆÈÜÒº£¬¿¹ËáÒ©X½öº¬ÎåÖÖ¶ÌÖÜÆÚÔªËØ£¬ÅжϳÁµíBÖ»ÄÜΪMg£¨OH£©₂£¬ÎÞÉ«ÈÜÒºAΪAlCl₃ÈÜÒººÍMgCl₂ÈÜÒº£¬¿¹ËáÒ©X¼ÓÈë¹ýÁ¿ÑÎËáÉú³ÉÎÞÉ«ÆøÌå¶þÑõ»¯Ì¼£¬ËµÃ÷º¬ÓÐCO₃^2-£¬Al³⁺£¬Mg²⁺£¬ÆäÖÐn£¨Al³⁺£©=n£¨Al£¨OH£©₃£©=$\frac{0.78g}{78g/mol}$=0.01mol£¬n£¨Mg²⁺£©=n£¨Mg£¨OH£©₂£©=$\frac{1.74g}{58g/mol}$=0..03mol£¬n£¨CO₃^2-£©=n£¨CO₂£©=$\frac{0.224L}{22.4L/mol}$=0.01mol£¬ÔòµÃµ½Ê£Óಿ·ÖÖÊÁ¿=2.78g-0.01mol¡Á27g/mol-0.03mol¡Á24g/mol-0.01mol¡Á60g/mol=1.19g£¬ÒÀ¾ÝµçºÉÊغ㣬ÎïÖÊÖк¬ÓÐ7¸ö¸ºµçºÉ£¬ÔòÒ»¸öµÄÖÊÁ¿=$\frac{1.19g}{7}$=0.17g£¬ÅжÏΪOH^-£¬µÃµ½»¯Ñ§Ê½ÎªAlMg₃£¨CO₃£©£¨OH£©₇£¬¾Ý´Ë·ÖÎö»Ø´ðÎÊÌ⣮

½â´ð ½â£º¿¹ËáÒ©X¼ÓÈë¹ýÁ¿ÑÎËáÉú³ÉÎÞÉ«ÆøÌåAºÍÎÞÉ«ÈÜÒºA£¬ÈÜÒºAÖмÓÈë¹ýÁ¿ÇâÑõ»¯ÄÆÈÜÒº¹ýÂ˵õ½°×É«³ÁµíBºÍÈÜÒºB£¬ÈÜÒºBÖмÓÈë×ãÁ¿AµÃµ½°×É«³ÁµíC£¬³ÁµíC¼ÓÈë¹ýÁ¿ÇâÑõ»¯ÄÆÈÜÒºÍêÈ«Èܽ⣬˵Ã÷³Á³ÁµÄCΪAl£¨OH£©₃£¬ÔòÈÜÒºBº¬ÓÐÆ«ÂÁËáÄÆ£¬AΪCO2£¬ÎÞÉ«ÈÜÒºCΪ̼ËáÇâÄÆÈÜÒº£¬¿¹ËáÒ©X½öº¬ÎåÖÖ¶ÌÖÜÆÚÔªËØ£¬ÅжϳÁµíBÖ»ÄÜΪMg£¨OH£©₂£¬ÎÞÉ«ÈÜÒºAΪAlCl₃ÈÜÒººÍMgCl₂ÈÜÒº£¬¿¹ËáÒ©X¼ÓÈë¹ýÁ¿ÑÎËáÉú³ÉÎÞÉ«ÆøÌå¶þÑõ»¯Ì¼£¬ËµÃ÷º¬ÓÐCO₃^2-£¬Al³⁺£¬Mg²⁺£¬ÆäÖÐn£¨Al³⁺£©=n£¨Al£¨OH£©₃£©=$\frac{0.78g}{78g/mol}$=0.01mol£¬n£¨Mg²⁺£©=n£¨Mg£¨OH£©₂£©=$\frac{1.74g}{58g/mol}$=0..03mol£¬n£¨CO₃^2-£©=n£¨CO₂£©=$\frac{0.224L}{22.4L/mol}$=0.01mol£¬ÔòµÃµ½Ê£Óಿ·ÖÖÊÁ¿=2.78g-0.01mol¡Á27g/mol-0.03mol¡Á24g/mol-0.01mol¡Á60g/mol=1.19g£¬ÒÀ¾ÝµçºÉÊغ㣬ÎïÖÊÖк¬ÓÐ7¸ö¸ºµçºÉ£¬ÔòÒ»¸öµÄÖÊÁ¿=$\frac{1.19g}{7}$=0.17g£¬n£¨OH-£©=$\frac{1.19g}{17g/mol}$=0.07mol£¬ÅжÏΪOH^-£¬µÃµ½»¯Ñ§Ê½ÎªAlMg₃£¨CO₃£©£¨OH£©₇£¬
£¨1£©ÉÏÊö·ÖÎö¿ÉÖª°×É«³ÁµíBÖнðÊôÔªËØΪþԪËØ£¬Ã¾Àë×ÓºËÄÚÓÐ12¸öÖÊ×Ó£¬ºËÍâÓÐÁ½¸öµç×Ӳ㣬µÚÒ»²ã2¸öµç×Ó£¬µÚ¶þ²ã8¸öµç×Ó£¬Ã¾Àë×ӽṹʾÒâͼΪ£¬AΪ¶þÑõ»¯Ì¼ÊôÓÚ¹²¼Û»¯ºÏÎ¶þÑõ»¯Ì¼ÎªÖ±ÏßÐͽṹ£¬·Ö×ÓÖдæÔÚÁ½¸ö̼ÑõË«¼ü£¬¶þÑõ»¯Ì¼µÄµç×ÓʽΪ£¬
¹Ê´ð°¸Îª£º£»£»
£¨2£©²½Öè3ÖÐÉú³É³ÁµíCΪAl£¨OH£©₃£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪAlO₂^-+CO₂+2H₂O=Al£¨OH£©₃¡ý+HCO₃^-£¬¹Ê´ð°¸Îª£ºAlO₂^-+CO₂+2H₂O=Al£¨OH£©₃¡ý+HCO₃^-£»
£¨3£©ÉÏÊö·ÖÎö¼ÆËã¿ÉÖªXΪAlMg₃£¨CO₃£©£¨OH£©₇£¬XºÍθËá·´Ó¦Éú³É¶þÑõ»¯Ì¼ÈÝÒ×Ôì³Éθ´©¿×£¬ÒýÆðΣÏÕ£¬
¹Ê´ð°¸Îª£ºAlMg₃£¨CO₃£©£¨OH£©₇£»ºÍθËá·´Ó¦Éú³É¶þÑõ»¯Ì¼ÈÝÒ×Ôì³Éθ´©¿×£»
£¨4£©Éè¼ÆʵÑé·½°¸ÑéÖ¤ÈÜÒºCÖеĺ¬Ì¼Àë×ÓΪHCO₃^-£¬È¡ÉÙÁ¿ÈÝÒ×ÉÙÐíÓÚÊÔ¹ÜÖУ¬¼ÓÈëÂÈ»¯¸ÆÎÞÏÖÏó£¬ÔÚ¼ÓÈëÑÎËáÉú³ÉÎÞÉ«ÎÞζµÄÆøÌ壬ͨÈë³ÎÇåʯ»ÒË®±ä»ë×Ç£¬Ö¤Ã÷º¬ÓÐ̼ËáÇâ¸ùÀë×Ó£¬
¹Ê´ð°¸Îª£ºÈ¡ÉÙÁ¿ÈÝÒ×ÉÙÐíÓÚÊÔ¹ÜÖУ¬¼ÓÈëÂÈ»¯¸ÆÎÞÏÖÏó£¬ÔÚ¼ÓÈëÑÎËáÉú³ÉÎÞÉ«ÎÞζµÄÆøÌ壬ͨÈë³ÎÇåʯ»ÒË®±ä»ë×Ç£¬Ö¤Ã÷º¬ÓÐ̼ËáÇâ¸ùÀë×Ó£®

µãÆÀ ±¾Ì⿼²éÁËÎïÖÊת»¯µÄ¶¨Á¿¹Øϵ¼ÆËãÅжϣ¬ÎïÖÊÐÔÖÊ£¬Àë×ÓÐÔÖʺͷ´Ó¦ÏÖÏó¼°ÆäÖÊÁ¿±ä»¯·ÖÎöÊǽâÌâ¹Ø¼ü£¬ÌâÄ¿ÄѶȽϴó£®