Question

10£®ÉßÎÆʯÊÇÒ»ÖÖ¸»Ã¾¹èËáÑοóÎïµÄ×ܳƣ¬ÒòÆäÑÕÉ«ÇàÂÌÏà¼ä¿áËÆÉßƤ¶øµÃÃû£®ÉßÎÆʯ¿ÉÒÔ¿´×÷ÓÉMgO¡¢FeOºÍFe₂O₃¡¢Al₂O₃¡¢SiO₂×é³É£®¹¤ÒµÉÏÓÉÉßÎÆʯÖÆÈ¡¼îʽ̼Ëáþ²úÆ·µÄÁ÷³ÌÈçÏ£º

£¨1£©ÉßÎÆʯ¼ÓÑÎËáÈܽâºó£¬ÈÜÒºÀï³ýÁËMg²⁺¡¢Al³⁺Í⣬»¹º¬ÓеĽðÊôÀë×ÓÊÇFe³⁺¡¢Fe²⁺£®
£¨2£©Ïò¢ñÖмÓÈëH₂O₂µÄ×÷ÓÃÊǽ«Fe²⁺Ñõ»¯³ÉFe³⁺»ò½«Fe£¨OH£©₂Ñõ»¯³ÉFe£¨OH£©₃£¬ÓйØÀë×Ó·½³Ìʽ2Fe²⁺+H₂O₂+2H⁺=2Fe³⁺+2H₂O£¬¼ÓÈëCa£¨OH£©₂ʱ£¬ÐèÒª¿ØÖÆÈÜÒºpHÔÚ7¡«8Ö®¼ä£¨ÓйØÇâÑõ»¯Îï³ÁµíµÄpH¼ûÏÂ±í£©£®

ÇâÑõ»¯Îï	Fe£¨OH£©3	Al£¨OH£©3	Mg£¨OH£©2
¿ªÊ¼³ÁµípH	1.5	3.3	9.4

ÈôpH£¾8¿ÉÄܵ¼Öµĺó¹ûAl£¨OH£©₃Èܽ⡢Mg²⁺³Áµí£®
£¨3£©³ÁµíÎïA¿É×÷ΪÌáÈ¡ºìÉ«ÑÕÁϵÄÔ­ÁÏ£¬ÈôÏò³ÁµíÎïAÖмÓÈëNaOHÈÜÒº£¬È»ºó¹ýÂË¡¢Ï´µÓ¡¢£¨¸ÉÔ×ÆÉÕ£¨ÌîдʵÑé²Ù×÷Ãû³Æ£©£¬¼´¿É»ñµÃºìÉ«ÑÕÁÏ£¬ÊµÏÖ·ÏÎïµÄ×ÛºÏÀûÓã®
£¨4£©Èô²úÆ·µÄ»¯Ñ§Ê½ÓÃaMgCO₃•bMg£¨OH£©₂•cH₂O±íʾ£¬ÏÖ³ÆÈ¡18.2g²úÆ·²¢Ê¹Ö®ÍêÈ«·Ö½â£¬ÊÕ¼¯µ½3.36L CO₂£¨±ê×¼×´¿öÏ£©ºÍ8.0g MgO£¬Í¨¹ý¼ÆËãÈ·¶¨²úÆ·µÄ»¯Ñ§Ê½ÖУºa=3£¬b=1£¬c=3£®

Answer 1

·ÖÎö £¨1£©ÉßÎÆʯ¿ÉÒÔ¿´×÷ÓÉMgO¡¢FeOºÍFe₂O₃¡¢Al₂O₃¡¢SiO₂×é³É£¬ÉßÎÆʯ¼ÓÑÎËáÈܽâºó£¬MgO¡¢FeO¡¢Fe₂O₃¡¢Al₂O₃ºÍHCl·´Ó¦Èܽ⣬¶øSiO₂ºÍHCl²»·´Ó¦£¬²»ÄÜÈܽ⣻
£¨2£©Ïò¢ñÖмÓÈëH₂O₂£¬½«Fe²⁺Ñõ»¯³ÉFe³⁺»ò½«Fe£¨OH£©₂Ñõ»¯³ÉFe£¨OH£©₃£¬ÕâʱÈÜÒºÖеÄFe³⁺¡¢Al³⁺ÊÇÔÓÖÊ£¬µ«³ýÈ¥Fe³⁺¡¢Al³⁺µÄ¹ý³ÌÖÐÒ»¶¨²»Äܽ«Mg²⁺Ëðʧ£¬ÎªÁË´ïµ½Õâ¸öÄ¿µÄ£¬¸ù¾ÝÌâÖÐÌáʾӦÓõÄÔ­ÀíÊÇFe³⁺+3H₂O?Fe£¨OH£©₃+3H⁺ºÍAl³⁺+3H₂O?Al£¨OH£©₃+3H⁺£¬Ö»Òª¼ÓÈëÒ»ÖÖÄÜÏûºÄH⁺µÄÎïÖʼ´¿ÉʹÉÏÊöË®½âƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬³ýÈ¥Fe³⁺¡¢Al³⁺£¬¸Ã·½°¸Ê¹ÓüÓCa£¨OH£©₂µÄ·½·¨£¬µ«´Ë²½¼ÓÈëCa£¨OH£©₂²»ÄÜÌ«¶à£¬Ð轫ÈÜÒºµÄpH¿ØÖÆÔÚ7¡«8£»ÈôpHСÓÚ7£¬Al³⁺²»ÄÜÍêÈ«³Áµí£»ÈôpH¹ý¸ßµ¼ÖÂÉú³ÉµÄAl£¨OH£©₃ÓÖ·¢Éú·´Ó¦Al£¨OH£©₃+OH^-=AlO₂^-+2H₂O¶øÈܽ⣬ͬʱMg²⁺Ҳת»¯Îª³Áµí¶øËðʧ£»
£¨3£©ÖÆ´¿¾»Fe₂O₃£¬Ó¦ÏȽ«ÆäÖк¬ÓеÄÉÙÁ¿Al£¨OH£©₃³ýÈ¥£¬³ýÈ¥Al£¨OH£©₃µÄ·½·¨ÊÇÀûÓÃËüÄÜÈÜÓÚÇ¿¼îµÄÐÔÖÊ£»
£¨4£©È·¶¨²úÆ·aMgCO₃•bMg£¨OH£©₂•cH₂OÖÐa¡¢b¡¢cµÄÖµ£¬ÒÀ¾ÝÐ趨µÄÊý¾ÝÑùÆ·ÖÊÁ¿£¬½áºÏÔªËØÊغã¼ÆËãµÃµ½£®

½â´ð ½â£º£¨1£©ÉßÎÆʯ¼ÓÑÎËáÈܽ⣬MgO¡¢FeO¡¢Fe₂O₃¡¢Al₂O₃ºÍHCl·´Ó¦Èܽ⣬·´Ó¦·½³Ìʽ·Ö±ðΪ£ºMgO+2HCl=MgCl₂+H₂O£¬FeO+2HCl=FeCl₂+H₂O£¬Fe₂O₃+6HCl=2FeCl₃+3H₂O£¬Al₂O₃+6HCl=2AlCl₃+3H₂O£®¶øSiO₂²»ÓëÑÎËá·´Ó¦£¬ÒÔ³ÁµíµÄÐÎʽ³ýÈ¥£¬¹ÊÈÜÒºÀï³ýÁËMg²⁺¡¢Al³⁺Í⣬»¹º¬ÓеĽðÊôÀë×ÓÊÇFe³⁺¡¢Fe²⁺£»
¹Ê´ð°¸Îª£ºFe³⁺¡¢Fe²⁺£»
£¨2£©ÕâʱÈÜÒºÖеÄFe³⁺¡¢Fe²⁺¡¢Al³⁺ÊÇÔÓÖÊ£¬µ«³ýÈ¥Fe³⁺¡¢Fe²⁺¡¢Al³⁺µÄ¹ý³ÌÖÐÒ»¶¨²»Äܽ«Mg²⁺Ëðʧ£¬ÎªÁË´ïµ½Õâ¸öÄ¿µÄ£¬ÏȼÓÈëÑõ»¯¼Á£¬½«Fe²⁺Ñõ»¯³ÉFe³⁺»ò½«Fe£¨OH£©₂Ñõ»¯³ÉFe£¨OH£©₃£¬H₂O₂µÄÑõ»¯ÐÔ£¾Fe³⁺µÄÑõ»¯ÐÔ£¬Ïò¢ñÖмÓÈëH₂O₂£¬·¢Éú2Fe²⁺+H₂O₂+2H⁺=2Fe³⁺+2H₂O£»¸ù¾ÝÌâÖÐÌáʾӦÓõÄÔ­ÀíÊÇFe³⁺+3H₂O?Fe£¨OH£©₃+3H⁺ºÍAl³⁺+3H₂O?Al£¨OH£©₃+3H⁺£¬Ö»Òª¼ÓÈëÒ»ÖÖÄÜÏûºÄH⁺µÄÎïÖʼ´¿ÉʹÉÏÊöË®½âƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬³ýÈ¥Fe³⁺¡¢Al³⁺£¬¸Ã·½°¸Ê¹ÓüÓCa£¨OH£©₂µÄ·½·¨£¬µ«´Ë²½¼ÓÈëCa£¨OH£©₂²»ÄÜÌ«¶à£¬Ð轫ÈÜÒºµÄpH¿ØÖÆÔÚ7¡«8£»ÈôpHСÓÚ7£¬Al³⁺²»ÄÜÍêÈ«³Áµí£»ÈôpH¹ý¸ßµ¼ÖÂÉú³ÉµÄAl£¨OH£©₃ÓÖ·¢Éú·´Ó¦Al£¨OH£©₃+OH^-=AlO₂^-+2H₂O¶øÈܽ⣬ͬʱMg²⁺Ҳת»¯Îª³Áµí¶øËðʧ£»
¹Ê´ð°¸Îª£º½«Fe²⁺Ñõ»¯³ÉFe³⁺»ò½«Fe£¨OH£©₂Ñõ»¯³ÉFe£¨OH£©₃£»2Fe²⁺+H₂O₂+2H⁺=2Fe³⁺+2H₂O£»Al£¨OH£©₃Èܽ⡢Mg²⁺³Áµí£»
£¨3£©ÖÆ´¿¾»Fe₂O₃£¬Ó¦ÏȽ«ÆäÖк¬ÓеÄÉÙÁ¿Al£¨OH£©₃³ýÈ¥£¬³ýÈ¥Al£¨OH£©₃µÄ·½·¨ÊÇÀûÓÃËüÄÜÈÜÓÚÇ¿¼îµÄÐÔÖÊ£®
Al£¨OH£©₃ÄÜÈÜÓÚÇ¿¼î£»¹Ê¿ÉAÖмÓÈëÇâÑõ»¯ÄÆ£¬ÇâÑõ»¯ÂÁÓëÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉÆ«ÂÁËáÄÆ£¬Æ«ÂÁËáÄÆÓë¶þÑõ»¯Ì¼·´Ó¦Éú³ÉÇâÑõ»¯ÂÁ³ÁµíºÍ̼ËáÇâÄÆ£¬¹ýÂ˵õ½ÇâÑõ»¯Ìú³Áµí£¬£¨¸ÉÔ×ÆÉÕÇâÑõ»¯Ìú³ÁµíµÃµ½Ñõ»¯Ìú£»Àë×Ó·½³ÌʽΪ£ºAlO₂^-+CO₂+2H₂O=Al£¨OH£©₃¡ý+HCO₃^-£»2Fe£¨OH£©3$\frac{\underline{\;\;¡÷\;\;}}{\;}$Fe₂O₃+3H₂O
¹Ê´ð°¸Îª£ºNaOH£»£¨¸ÉÔ×ÆÉÕ£»
£¨4£©m£¨ÑùÆ·£©=18.2g£¬m£¨CO₂£©=6.6g£¬m£¨MgO£©=8.0g£¬¼îʽ̼Ëáþ·Ö½â£ºaMgCO₃?bMg£¨OH£©₂?cH₂O $\frac{\underline{\;¸ßÎÂ\;}}{\;}$£¨a+b£©MgO+aCO₂¡ü+£¨b+c£©H₂O¡ü£¬
¸ù¾ÝÖÊÁ¿ÊغãµÃ£ºm£¨H₂O£©=18.2g-6.6g-8.0g=3.6g£¬
Ôòm£¨MgO£©¨T0.2mol£¬
n£¨CO₂£©¨T0.15mol£¬
n£¨H₂O£©¨T0.2mol£¬
µÃ£ºa£ºb£ºc=0.15£º0.05£º0.15=3£º1£º3£¬
¹Ê´ð°¸Îª£º3£»1£»3£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÔªËØ»¯ºÏÎïÖнðÊôÂÁ¡¢ÌúµÄÐÔÖÊ£¬½âÌâʱ²»½öÐèÊìÖªÎïÖʼäµÄ·´Ó¦Ô­Àí£¬»¹ÐèÁé»îÔËÓÃÓÚʵÑéµÄÉè¼Æ£¬ÌâÄ¿ÄѶȽϴó£®