Question

1£®Ïòº¬Al₂£¨SO₄£©₃ºÍAlCl₃µÄ»ìºÏÈÜÒºÖÐÖðµÎ¼ÓÈë1mol/L Ba£¨OH£©₂ÈÜÒºÖÁ¹ýÁ¿£¬¼ÓÈëBa£¨OH£©₂ÈÜÒºµÄÌå»ýºÍËùµÃ³ÁµíµÄÎïÖʵÄÁ¿µÄ¹ØϵÈçͼ£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

• A£® ͼÖÐAµãÉú³ÉµÄ³ÁµíÊÇBaSO₄ºÍAl£¨OH£©₃
• B£® Ô­»ìºÏÒºÖÐc[Al₂£¨SO₄£©₃]£ºc£¨AlCl₃£©=1£º2
• C£® AB¶Î·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º3Ba²⁺+2Al³⁺+8OH^-+3SO₄^2-¨TBaSO₄¡ý+2AlO₂^-+4H₂O
• D£® ÏòDµãÈÜÒºÖÐͨÈëCO₂ÆøÌ壬Á¢¼´²úÉú°×É«³Áµí

Answer 1

·ÖÎö Ïòº¬Al₂£¨SO₄£©₃ ºÍ AlCl₃µÄ»ìºÏÈÜÒºÓëBa£¨OH£©₂ÈÜÒº·´Ó¦µÄʵÖÊÊÇAl³⁺ÓëOH^-¡¢Ba²⁺ÓëSO₄^2-Ö®¼äµÄÀë×Ó·´Ó¦£¬ÈçÏÂBa²⁺+SO₄^2-¨TBaSO₄¡ý£¬Al³⁺+3OH^-¨TAl£¨OH£©₃¡ý£¬Al£¨OH£©₃+OH^-¨TAlO₂^-+2H₂O£¬¼ÙÉè1molAl₂£¨SO₄£©₃ÖÐSO₄^2-ÍêÈ«±»³ÁµíËùÐèBa£¨OH£©₂Á¿Îª3mol£¬3molBa£¨OH£©₂Ìṩ6molOH^-£¬1molAl₂£¨SO₄£©₃Öк¬ÓÐ2molAl³⁺£¬ÓÉ·´Ó¦Al³⁺+3OH^-¨TAl£¨OH£©₃¡ý¿ÉÖª£¬1molAl₂£¨SO₄£©₃ÖеÄAl³⁺ºÍSO₄^2- ¾ù³ÁµíÍêÈ«£¬¹Ê´ÓÆðµãµ½Aµã£¬¿ÉÒÔÈÏΪÊÇÁòËáÂÁÓëÇâÑõ»¯±µÇ¡ºÃ·¢Éú·´Ó¦Éú³É3molBaSO₄¡¢2molAl£¨OH£©₃³Áµí£¬AµãʱSO₄^2-ÍêÈ«³Áµí£¬A-BΪÂÈ»¯ÂÁÓëÇâÑõ»¯±µµÄ·´Ó¦£¬BµãʱÈÜÒºÖÐAl³⁺ÍêÈ«³Áµí£¬²úÉú³Áµí´ï×î´óÖµ£¬ÈÜÒºÖÐÈÜÖÊΪBaCl₂£¬B-CΪÇâÑõ»¯ÂÁÓëÇâÑõ»¯±µ·´Ó¦£¬CµãʱÇâÑõ»¯ÂÁÍêÈ«Èܽ⣬CµãʱÇâÑõ»¯ÂÁÍêÈ«Èܽ⣬¾Ý´Ë½øÐзÖÎö£®

½â´ð ½â£ºA£®Í¼ÖÐAµãÊÇAl₂£¨SO₄£©₃ºÍBa£¨OH£©₂µÄ·´Ó¦£¬¹ÊÉú³ÉµÄ³ÁµíÊÇBaSO₄ºÍAl£¨OH£©₃£¬¹ÊAÕýÈ·£»
B£®Ç°3LBa£¨OH£©₂ÈÜÒºÓëÈÜÒºÖÐAl₂£¨SO₄£©₃·´Ó¦£¬´Ó3L-6LΪBa£¨OH£©₂ÈÜÒºÓëÈÜÒºÖÐAlCl₃·´Ó¦£¬¶þÕßÏûºÄµÄÇâÑõ»¯±µµÄÎïÖʵÄÁ¿ÏàµÈΪ3L¡Á1mol/L=3mol£¬ÓÉÉú³ÉÁòËá±µ¿ÉÖª3n[Al₂£¨SO₄£©₃]=n[Ba£¨OH£©₂]£¬¹Ên[Al₂£¨SO₄£©₃]=1mol£¬ÓÉÂÈ»¯ÂÁÓëÇâÑõ»¯±µÉú³ÉÇâÑõ»¯ÂÁ¿ÉÖª3n£¨AlCl₃£©=2n[Ba£¨OH£©₂]=6mol£¬¹Ên£¨AlCl₃£©=2mol£¬¹ÊÔ­ÈÜÒºÖÐÔ­»ìºÏÒºÖÐc[Al₂£¨SO₄£©₃]£ºc£¨AlCl₃£©=1£º2£¬¹ÊBÕýÈ·£»
C£®AB¶ÎΪÂÈ»¯ÂÁÓëÇâÑõ»¯±µµÄ·´Ó¦£¬¹ÊÀë×Ó·½³ÌʽΪBa²⁺+SO₄^2-¨TBaSO₄¡ý£¬¹ÊC´íÎó£»
D£®DµãµÄÈÜÒºÖк¬ÓÐAlO₂^-£¬Í¨Èë¶þÑõ»¯Ì¼²úÉúÇâÑõ»¯ÂÁ°×É«³Áµí£¬¹ÊDÕýÈ·£»
¹ÊÑ¡C£®

µãÆÀ ±¾Ì⿼²é»¯Ñ§·´Ó¦µÄÓйØͼÏóÎÊÌâ¡¢ÂÁ»¯ºÏÎïÐÔÖÊ¡¢»ìºÏÎïµÄÓйؼÆËãµÈ£¬ÄѶȽϴ󣬹ؼüÊÇÇå³þ¸÷½×¶Î·¢ÉúµÄ·´Ó¦£¬×¢Òâ´Ó¿ªÊ¼µ½AµãÏ൱ÓÚÊÇÁòËáÂÁÓëÇâÑõ»¯±µµÄ·´Ó¦Éú³ÉÁòËá±µ¡¢ÇâÑõ»¯ÂÁ³Áµí£®