Question

18£®Q¡¢W¡¢X¡¢Y¡¢ZÊÇÖÜÆÚ±íÇ°36ºÅÔªËØÖеÄÎåÖÖ³£¼ûÔªËØ£¬ÆäÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£®WÔ­×Ó×îÍâ²ãµç×ÓÊýÓëºËÍâµç×Ó×ÜÊýÖ®±ÈΪ3£º4£¬Q¡¢YµÄÑõ»¯ÎïÊǵ¼ÖÂËáÓêµÄÖ÷ÒªÎïÖÊ£¬XÊǵؿÇÖк¬Á¿×î¸ßµÄ½ðÊôÔªËØ£¬ZÄÜÐγɺìÉ«£¨»òשºìÉ«£©µÄZ₂OºÍºÚÉ«µÄZOÁ½ÖÖÑõ»¯Î
ÒÔÏÂÎÊÌâ¾ùÓû¯Ñ§ÓÃÓï»Ø´ð£º
£¨1£©Qµ¥Öʵĵç×ÓʽΪ£®W¡¢X¡¢YµÄÀë×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪS^2-£¾O^2-£¾Al³⁺£¨ÓÃÀë×Ó·ûºÅ»Ø´ð£©
£¨2£©XºÍY×é³ÉµÄ»¯ºÏÎï·ÅÈëË®Öз´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇAl₂S₃+6H₂O=2Al£¨OH£©₃¡ý+3H₂S¡ü
£¨3£©Na₂YÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪc£¨Na⁺£©£¾c£¨S^2-£©£¾c£¨OH^-£©£¾c£¨HS^-£©£¾c£¨H⁺£©
£¨4£©ZOÔÚ¸ßÎÂϱ»QµÄ¼òµ¥Æø̬Ç⻯ÎﻹԭΪZµ¥ÖÊ£¬ÇÒÉú³ÉµÄÆäËü²úÎï²»»á¶Ô»·¾³²úÉúÎÛȾ£¬Ð´³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ3CuO+2NH₃$\frac{\underline{\;¸ßÎÂ\;}}{\;}$3Cu+N₂+3H₂O
£¨5£©ÊÒÎÂÏ£¬ZCl₂ÈÜÒºÖлìÓÐFeCl₃ÔÓÖÊʱ£¬¿É¼ÓÈëCuO£¨ÌîÊÔ¼Á£©µ÷½ÚpH=4£¬ÔÙ¹ýÂË£®£¨ÒÑÖª£ºFe£¨OH£©₃µÄK_sp=10^-35£¬»¯Ñ§ÉÏÈÏΪ²ÐÁôÔÚÈÜÒºÖеÄÀë×ÓŨ¶ÈСÓÚ1¡Á10^-5mol/Lʱ£¬³Áµí¾Í´ïÍêÈ«£®£©
£¨6£©¹¤ÒµÉÏ¿ÉÓøßÎÂÌõ¼þÏÂZ₂Y+O₂=2Z+YO₂Ò±Á¶½ðÊôZ£¬Éú³É1molZʱתÒÆ3molµç×Ó£®

Answer 1

·ÖÎö Q¡¢W¡¢X¡¢Y¡¢ZÊÇÖÜÆÚ±íÇ°36ºÅÔªËØÖеÄÎåÖÖ³£¼ûÔªËØ£¬ÆäÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬WÔ­×Ó×îÍâ²ãµç×ÓÊýÓëºËÍâµç×Ó×ÜÊýÖ®±ÈΪ3£º4£¬¹ÊWΪÑõÔªËØ£¬Q¡¢YµÄÑõ»¯ÎïÊǵ¼ÖÂËáÓêµÄÖ÷ÒªÎïÖÊ£¬¹ÊQ¡¢YΪS»òNÔªËØ£¬ÒÀ¾ÝÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬¹ÊQΪµªÔªËØ£¬YΪÁòÔªËØ£¬XÊǵؿÇÖк¬Á¿×î¸ßµÄ½ðÊôÔªËØ£¬¹ÊXÊÇÂÁÔªËØ£¬ZÄÜÐγɺìÉ«£¨»òשºìÉ«£©µÄZ₂OºÍºÚÉ«µÄZOÁ½ÖÖÑõ»¯Î¹ÊZÊÇÍ­ÔªËØ£¬¾Ý´Ë½â´ð¸÷СÌâ¼´¿É£®

½â´ð ½â£ºÒÀ¾Ý·ÖÎö¿ÉÖª£ºQΪN£¬WΪO£¬XΪAl£¬YΪS£¬ZΪCu£¬
£¨1£©QΪµªÔªËØ£¬µªÆø·Ö×ÓÖдæÔÚµªµªÈý¼ü£¬µªÔ­×Ó×îÍâ²ãΪ8µç×Ó£¬N₂µÄµç×ÓʽΪ£º£¬ÔªËØÀë×Óµç×Ó²ãÊýÔ½¶à£¬°ë¾¶Ô½´ó£¬ºËÍâµç×ÓÅŲ¼ÏàͬµÄÀë×Ó£¬Ô­×ÓÐòÊýԽС£¬Àë×Ӱ뾶Խ´ó£¬¹ÊO¡¢Al¡¢SµÄÀë×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ£ºS^2-£¾O^2-£¾Al³⁺£¬¹Ê´ð°¸Îª£º£»S^2-£¾O^2-£¾Al³⁺£»
£¨2£©AlºÍS×é³ÉµÄ»¯ºÏÎï·ÅÈëË®Öз¢ÉúË«Ë®½â·´Ó¦£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£ºAl₂S₃+6H₂O=2Al£¨OH£©₃¡ý+3H₂S¡ü£¬¹Ê´ð°¸Îª£ºAl₂S₃+6H₂O=2Al£¨OH£©₃¡ý+3H₂S¡ü£»
£¨3£©Na₂SÈÜÒºÖдæÔÚS^2-΢ÈõµÄË®½â£¬ÈÜÒº³Ê¼îÐÔ£¬¼´£ºS^2-+H₂O?HS^-+OH^-£¬¹ÊÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ£ºc£¨Na⁺£©£¾c£¨S^2-£©£¾c£¨OH^-£©£¾c£¨HS^-£©£¾c£¨H⁺£©£¬¹Ê´ð°¸Îª£ºc£¨Na⁺£©£¾c£¨S^2-£©£¾c£¨OH^-£©£¾c£¨HS^-£©£¾c£¨H⁺£©£»
£¨4£©CuOÔÚ¸ßÎÂÏÂÓë°±ÆøÉú³ÉµªÆø£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º3CuO+2NH₃$\frac{\underline{\;¸ßÎÂ\;}}{\;}$3Cu+N₂+3H₂O£¬¹Ê´ð°¸Îª£º3CuO+2NH₃$\frac{\underline{\;¸ßÎÂ\;}}{\;}$3Cu+N₂+3H₂O£»
£¨5£©Fe£¨OH£©₃µÄKsp=10^-35£¬»¯Ñ§ÉÏÈÏΪ²ÐÁôÔÚÈÜÒºÖеÄÀë×ÓŨ¶ÈСÓÚ1¡Á10^-5mol/Lʱ£¬³Áµí¾Í´ïÍêÈ«£¬¹Ê´ËʱÈÜÒºÖÐc£¨OH^-£©=$\root{3}{\frac{1{0}^{-35}}{1{0}^{-5}}}$=10^-10£¬¹Ê´Ëʱc£¨H⁺£©=$\frac{1{0}^{-14}}{1{0}^{-10}}$=10^-4£¬¹Ê´ËʱpHֵΪ4£¬µ±CuCl₂ÈÜÒºÖлìÓÐFeCl₃ÔÓÖÊʱ£¬¿É¼ÓÈëÑõ»¯Í­µ÷½ÚpHΪ4£¬È»ºóÔÙ¹ýÂË£¬¹Ê´ð°¸Îª£ºCuO£»4£»
£¨6£©¹¤ÒµÉÏ¿ÉÓøßÎÂÌõ¼þÏÂCu₂S+O₂¨T2Cu+SO₂À´Ò±Á¶½ðÊôCu£¬·´Ó¦ÖÐCuµÄ»¯ºÏ¼ÛÓÉ+1¼Û±äΪ0¼Û£¬2molCuÔªËØתÒÆ2molµç×Ó£¬OÔªËØÓÉ0¼Û±äΪ-2¼Û£¬1molÑõÆøתÒÆ4molµç×Ó£¬¹ÊÉú³É2molCuʱתÒƵç×ÓÊýΪ6mol£¬Éú³É1molCuתÒÆ3molµç×Ó£¬¹Ê´ð°¸Îª£º3£®

µãÆÀ ±¾ÌâÊǶÔÎïÖʽṹµÄ¿¼²é£¬ÍƶÏÔªËØÊǽâÌâ¹Ø¼ü£¬Éæ¼°µç×Óʽ¡¢Àë×Ӱ뾶´óС±È½Ï¡¢Àë×ÓŨ¶È´óС±È½Ï¡¢Ñõ»¯»¹Ô­·´Ó¦µÈ£¬ÐèҪѧÉú¾ß±¸ÔúʵµÄ»ù´¡ÖªÊ¶£¬ÄѶÈÖеȣ®