Question

18£®Á×ÊǵؿÇÖк¬Á¿½ÏΪ·á¸»µÄ·Ç½ðÊôÔªËØ£¬Ö÷ÒªÒÔÄÑÈÜÓÚË®µÄÁ×ËáÑÎÈçCa₃£¨PO₄£©₂µÈÐÎʽ´æÔÚ£®ËüµÄµ¥Öʺͻ¯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÖÐÓÐ×ÅÖØÒªµÄÓ¦Óã®
£¨1£©°×Á×£¨P₄£©¿ÉÓÉCa₃£¨PO₄£©₂¡¢½¹Ì¿ºÍSiO₂ÔÚµç¯ÖиßΣ¨¡«1550¡æ£©ÏÂͨ¹ýÏÂÃæÈý¸ö·´Ó¦¹²È۵õ½£®
¢Ù4Ca₃£¨PO₄£©₂£¨s£©+10C£¨s£©¨T12CaO£¨s£©+2P₄£¨s£©+10CO₂£¨g£©¡÷H₁=+Q₁kJ•mol^-1
¢ÚCaO£¨s£©+SiO₂£¨s£©¨TCaSiO₃£¨s£©¡÷H₂=-Q₂ kJ•mol^-1
¢ÛCO₂£¨g£©+C£¨s£©¨T2CO£¨g£©¡÷H₃=+Q₃kJ•mol^-1
ÒÑÖª£ºCaSiO₃µÄÈ۵㣨1546¡æ£©±ÈSiO₂µÍ£®
д³öÓÉÁ×Ëá¸Æ¿óÖÆÈ¡°×Á××ܵķ´Ó¦·½³Ìʽ2Ca₃£¨PO₄£©₂+6SiO₂+10C$\frac{\underline{\;¸ßÎÂ\;}}{\;}$6CaSiO₃+P₄+10CO£®
£¨2£©°×Á×ÔÚÈȵÄŨÇâÑõ»¯¼ØÈÜÒºÖÐÆ绯µÃµ½Ò»ÖÖ´ÎÁ×ËáÑΣ¨KH₂PO₂£©ºÍÒ»ÖÖÆøÌåPH₃£¨Ð´»¯Ñ§Ê½£©£®
£¨3£©Á×µÄÖØÒª»¯ºÏÎïNaH₂PO₄¿Éͨ¹ýH₃PO₄ÓëNaOHÈÜÒº·´Ó¦»ñµÃ£®¹¤ÒµÉÏΪÁËʹ·´Ó¦µÄÖ÷Òª²úÎïÊÇNaH₂PO₄£¬Í¨³£½«pH¿ØÖÆÔÚ2.1¡«7.2Ö®¼ä £¨ÒÑÖªÁ×ËáµÄ¸÷¼¶µçÀë³£ÊýΪ£ºK₁=7.1¡Á10^-3K₂=6.3¡Á10^-8K₃=4.2¡Á10^-13 lg7.1¡Ö0.9  lg6.3¡Ö0.8  lg4.2¡Ö0.6£©£®Na₂HPO₄ ÈÜÒºÏÔ¼îÐÔ£¬ÈôÏòÆäÈÜÒºÖмÓÈë×ãÁ¿µÄCaCl₂ ÈÜÒº£¬ÈÜÒºÔòÏÔËáÐÔ£¬ÆäÔ­ÒòÊÇ3Ca²⁺+2HPO₄^-=Ca₃£¨PO₄ £©₂¡ý+2H⁺ £¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£®
£¨4£©°×Á×Öж¾ºó¿ÉÓÃCuSO₄ÈÜÒº½â¶¾£¬½â¶¾Ô­Àí¿ÉÓÃÏÂÁл¯Ñ§·½³Ìʽ±íʾ£º11P ₄+60CuSO₄+96H₂O¨T20Cu₃P+24H₃PO₄+60H₂SO₄60molCuSO₄ÄÜÑõ»¯°×Á×µÄÎïÖʵÄÁ¿ÊÇ3mol£®

Answer 1

·ÖÎö £¨1£©ÒÀ¾ÝÈÈ»¯Ñ§·½³ÌʽºÍ¸Ç˹¶¨ÂɼÆËãËùÐ軯ѧ·½³Ìʽ£»
£¨2£©Æ绯·´Ó¦Á×ÔªËØÉý¸ßΪ´ÎÁ×ËáÑΣ¬Á×ÔªËØ»¯ºÏ¼ÛΪ+1¼Û£¬½µµÍµÄÖ»ÄÜΪ¸º¼Û£¬ÔòΪÇ⻯Î
£¨3£©H₃PO₄?H⁺+H₂PO₄^-K₁=7.1¡Á10^-3£»H₂PO₄^-?H⁺+HPO₄^2-K₂=6.3¡Á10^-8£»HPO₄^2-?H⁺+HPO₄^3-K₃=4.2¡Á10^-13£»ÉÏÒ»¼¶µçÀë²úÉúµÄH⁺¶ÔÏÂÒ»¼¶µçÀëÓÐÒÖÖÆ×÷Óã»Îª»ñµÃ¾¡¿ÉÄÜ´¿µÄNaH₂PO₄£¬ÐèH₂PO₄^-Ũ¶È×î´ó£¬¸ù¾ÝK₁¡¢K₂¡¢ÒÔ¼°pH=-lgc£¨H⁺£©¼ÆË㣻
ÈôÏòNa₂HPO₄ÈÜÒºÖмÓÈë×ãÁ¿µÄCaCl₂ÈÜÒº£¬HPO₄^2-Àë×ÓºÍCa²⁺Àë×Ó·´Ó¦Éú³É³Áµí£¬´Ù½øHPO₄^2-µÄµçÀ룬ÈÜÒºÏÔʾËáÐÔ£»
£¨4£©CuÔªËصĻ¯ºÏ¼ÛÓÉ+2¼Û½µµÍµ½+1¼Û£¬CuSO₄ÊÇÑõ»¯¼Á£¬P₄²¿·ÖÁ×ÔªËØÓÉ0¼Û½µµÍµ½-3¼Û£¬²¿·ÖÁ×ÔªËØÓÉ0¼ÛÉý¸ßµ½+5¼Û£¬Á×ÔªËصĻ¯ºÏ¼Û¼ÈÉý¸ßÓÖ½µµÍ£¬ËùÒÔP₄¼ÈÊÇÑõ»¯¼ÁÓÖÊÇ»¹Ô­¼Á£¬ÈôÓÐ11molP₄²Î¼Ó·´Ó¦£¬ÆäÖÐ5molµÄP₄×öÑõ»¯¼Á£¬60molÁòËáÍ­×öÑõ»¯¼Á£¬Ö»ÓÐ6molµÄP₄×ö»¹Ô­¼Á£¬½áºÏµç×ÓÊغãÀ´½â´ð£®

½â´ð ½â£º£¨1£©¢Ù4Ca₃£¨PO₄£©₂£¨s£©+10C£¨s£©=12CaO£¨s£©+2P₄£¨s£©+10CO₂£¨g£©¡÷H₁=+Q₁kJ•mol^-1
¢ÚCaO£¨s£©+SiO₂£¨s£©=CaSiO₃£¨s£©¡÷H₂=-Q₂ kJ•mol^-1
¢ÛCO₂£¨g£©+C£¨s£©=2CO£¨g£©¡÷H₃=+Q₃kJ•mol^-1
ÒÀ¾ÝÈÈ»¯Ñ§·½³ÌʽºÍ¸Ç˹¶¨ÂɼÆËãµÃµ½¢Ù¡Á$\frac{1}{2}$+¢Û+¢Ú¡Á6µÃµ½»¯Ñ§·½³ÌʽΪ£º2Ca₃£¨PO₄£©₂+6SiO₂+10C$\frac{\underline{\;¸ßÎÂ\;}}{\;}$6CaSiO₃+P₄+10CO£¬
¹Ê´ð°¸Îª£º2Ca₃£¨PO₄£©₂+6SiO₂+10C$\frac{\underline{\;¸ßÎÂ\;}}{\;}$6CaSiO₃+P₄+10CO£»
£¨2£©°×Á×ÔÚÈȵÄŨÇâÑõ»¯¼ØÈÜÒºÖÐ᪻¯µÃµ½Ò»ÖÖ´ÎÁ×ËáÑΣ¨KH₂PO₂£©ºÍÒ»ÖÖÆøÌ壬Æ绯·´Ó¦Á×ÔªËØÉý¸ßΪ´ÎÁ×ËáÑΣ¬Á×ÔªËØ»¯ºÏ¼ÛΪ+1¼Û£¬½µµÍµÄÖ»ÄÜΪ¸º¼Û£¬ÔòΪÇ⻯ÎïPH₃£»
¹Ê´ð°¸Îª£ºPH₃£»
£¨3£©Îª»ñµÃ¾¡¿ÉÄÜ´¿µÄNaH₂PO₄£¬ÐèÅäÖÆÁ×Ëá¡¢Á×Ëá¶þÇâÄƵĻº³åÈÜÒº£¬µ±ÈÜҺȫ²¿ÎªÁ×ËáÈÜҺʱ£¬Á×ËáÒÔµÚÒ»²½µçÀëΪÖ÷£¬ËùÒÔH₃PO₄?H⁺+H₂PO₄^-K₁=7.1¡Á10^-3£»PH=-lgc£¨H⁺£©=3-lg7.1¡Ö2.1£¬
µ±ÈÜҺȫ²¿ÎªNaH₂PO₄ÈÜҺʱ£¬H₂PO₄^-?H⁺+HPO₄^2-K₂=6.3¡Á10^-8£¬ËùÒÔpH=-lgc£¨H⁺£©=8-lg6.3¡Ö7.2£¬
ËùÒÔpHÓ¦¿ØÖƽéÓÚ2.1¡«7.2Ö®¼ä£¬
HPO₄^2-Àë×Ó¼ÈÄÜ·¢ÉúµçÀëÓÖÄÜ·¢ÉúË®½â£¬µçÀ뷴ӦʽΪHPO₄^2-?PO₄^3-+H⁺£¬Ë®½â·´Ó¦Ê½ÎªHPO₄^2-+H₂O?H₂PO₄^-+OH^-£¬ÈÜÒº³Ê¼îÐÔ£¬ËµÃ÷Ë®½â³Ì¶È´óÓÚµçÀë³Ì¶È£»
ÓÉÓÚHPO₄^2-Àë×ÓºÍCa²⁺Àë×Ó·´Ó¦Éú³É³Áµí£º3Ca²⁺+2HPO₄^2-¨TCa₃£¨PO₄£©₂¡ý+2H⁺£¬´Ù½øHPO₄^2-µÄµçÀ룬ÈÜÒºÔòÏÔËáÐÔ£»
¹Ê´ð°¸Îª£º2.1¡«7.2£¬3Ca²⁺+2HPO₄^-=Ca₃£¨PO₄ £©₂¡ý+2H⁺£»
£¨4£©CuÔªËصĻ¯ºÏ¼ÛÓÉ+2¼Û½µµÍµ½+1¼Û£¬CuSO₄ÊÇÑõ»¯¼Á£¬P₄²¿·ÖÁ×ÔªËØÓÉ0¼Û½µµÍµ½-3¼Û£¬²¿·ÖÁ×ÔªËØÓÉ0¼ÛÉý¸ßµ½+5¼Û£¬Á×ÔªËصĻ¯ºÏ¼Û¼ÈÉý¸ßÓÖ½µµÍ£¬ËùÒÔP₄¼ÈÊÇÑõ»¯¼ÁÓÖÊÇ»¹Ô­¼Á£¬ÈôÓÐ11molP₄²Î¼Ó·´Ó¦£¬ÆäÖÐ5molµÄP₄×öÑõ»¯¼Á£¬60molÁòËáÍ­×öÑõ»¯¼Á£¬Ö»ÓÐ6molµÄP₄×ö»¹Ô­¼Á£¬
ÔòÓɵç×ÓÊغã¿ÉÖª£¬ÓÐ60molµÄCuSO₄²Î¼Ó·´Ó¦£¬60molCuSO₄µÃµ½60molµç×Ó£¬1molP₄²Î¼Ó·´Ó¦Ê§È¥20molµç×Ó£¬ËùÒÔ60molCuSO₄ÄÜÑõ»¯°×Á×µÄÎïÖʵÄÁ¿ÊÇ$\frac{60}{20}$=3mol£¬
¹Ê´ð°¸Îª£º3mol£®

µãÆÀ ±¾Ì⿼²éËá¼î»ìºÏµÄ¶¨ÐÔÅжϺͼÆËã¡¢¸Ç˹¶¨ÂɵÄÓ¦Óá¢Ñõ»¯»¹Ô­·´Ó¦µç×ÓתÒƼÆËã¡¢Àë×Ó·½³ÌʽµÄÊéдµÈ£¬ÌâÄ¿Éæ¼°µÄ֪ʶµã½Ï¶à£¬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦ºÍ¼ÆËãÄÜÁ¦µÄ¿¼²é£¬ÌâÄ¿ÄѶÈÖеȣ®