Question

µç»¯Ñ§Ô­ÀíÔÚ¹¤ÒµÉú²úÖÐÓÐ×ÅÖØÒªµÄ×÷Óã¬ÇëÀûÓÃËùѧ֪ʶ»Ø´ðÓйØÎÊÌâ¡£
£¨1£©Óõç½âµÄ·½·¨½«Áò»¯ÄÆÈÜÒºÑõ»¯Îª¶àÁò»¯ÎïµÄÑо¿¾ßÓÐÖØÒªµÄʵ¼ÊÒâÒ壬½«Áò»¯Îïת±äΪ¶àÁò»¯ÎïÊǵç½â·¨´¦ÀíÁò»¯Çâ·ÏÆøµÄÒ»¸öÖØÒªÄÚÈÝ¡£Èçͼ£¬Êǵç½â²úÉú¶àÁò»¯ÎïµÄʵÑé×°Öãº

¢ÙÒÑÖªÑô¼«µÄ·´Ó¦Îª£¨x£«1£©S^2£­=S_x£«S^2£­£«2xe^£­£¬ÔòÒõ¼«µÄµç¼«·´Ó¦Ê½ÊÇ________________________________________________________________________¡£
µ±·´Ó¦×ªÒÆx molµç×Óʱ£¬²úÉúµÄÆøÌåÌå»ýΪ____________£¨±ê×¼×´¿öÏ£©¡£
¢Ú½«Na₂S¡¤9H₂OÈÜÓÚË®ÖÐÅäÖÆÁò»¯ÎïÈÜҺʱ£¬Í¨³£ÊÇÔÚµªÆøÆø·ÕÏÂÈܽ⡣ÆäÔ­ÒòÊÇ£¨ÓÃÀë×Ó·´Ó¦·½³Ìʽ±íʾ£©£º___________________________________________________¡£
£¨2£©MnO₂ÊÇÒ»ÖÖÖØÒªµÄÎÞ»ú¹¦ÄܲÄÁÏ£¬ÖƱ¸MnO₂µÄ·½·¨Ö®Ò»ÊÇÒÔʯīΪµç¼«£¬µç½âËữµÄMnSO₄ÈÜÒº£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª______________________¡£ÏÖÒÔǦÐîµç³ØΪµçÔ´µç½âËữµÄMnSO₄ÈÜÒº£¬ÈçͼËùʾ£¬Ç¦Ðîµç³ØµÄ×Ü·´Ó¦·½³ÌʽΪ_______________________________________________£¬µ±Ðîµç³ØÖÐÓÐ4 mol H^£«±»ÏûºÄʱ£¬Ôòµç·ÖÐͨ¹ýµÄµç×ÓµÄÎïÖʵÄÁ¿Îª________£¬MnO₂µÄÀíÂÛ²úÁ¿Îª________g¡£

£¨3£©ÓÃͼµç½â×°ÖÿÉÖƵþßÓо»Ë®×÷ÓõÄFeO₄^2-¡£ÊµÑé¹ý³ÌÖУ¬Á½¼«¾ùÓÐÆøÌå²úÉú£¬Y¼«ÇøÈÜÒºÖð½¥Éú³ÉFeO₄^2-¡£

¢Ùµç½â¹ý³ÌÖУ¬X¼«ÇøÈÜÒºµÄpH________£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£©¡£
¢Úµç½â¹ý³ÌÖУ¬Y¼«·¢ÉúµÄµç¼«·´Ó¦ÎªFe£­6e^£­£«8OH^£­=FeO₄^2-£«4H₂OºÍ________________________________________________________________________£¬
ÈôÔÚX¼«ÊÕ¼¯µ½672 mLÆøÌ壬ÔÚY¼«ÊÕ¼¯µ½168 mLÆøÌ壨¾ùÒÑÕÛËãΪ±ê×¼×´¿öʱÆøÌåÌå»ý£©£¬ÔòYµç¼«£¨Ìúµç¼«£©ÖÊÁ¿¼õÉÙ________g¡£

Answer 1

£¨1£©¢Ù2H₂O£«2e^£­=2OH^£­£«H₂¡ü£¨»ò2H^£«£«2e^£­=H₂¡ü£©¡¡11.2x L
¢Ú2S^2£­£«O₂£«2H₂O=2S¡ý£«4OH^£­
£¨2£©Mn^2£«£­2e^£­£«2H₂O=MnO₂£«4H^£«
Pb£«PbO₂£«2H₂SO₄=2PbSO₄£«2H₂O
2 mol¡¡87
£¨3£©Ôö´ó¡¡4OH^£­£­4e^£­=2H₂O£«O₂¡ü¡¡0.28

£¨1£©µç½âʱ£¬Ë®µçÀëµÄH^£«ÔÚÒõ¼«·¢ÉúµÃµç×Ó»¹Ô­·´Ó¦£¬Éú³ÉH₂¡£¸ù¾Ýµç×ÓÊغã¿ÉÖªÓÐx molµç×ÓתÒÆ£¬²úÉúH₂ 0.5x mol¡£S^2£­¾ßÓнÏÇ¿»¹Ô­ÐÔ£¬Ò×±»¿ÕÆøÖеÄÑõÆøÑõ»¯£¬¹ÊÅäÖÆÈÜҺʱÐèÒªµªÆø×÷±£»¤Æø¡£
£¨2£©¸ù¾ÝÌâÒâMn^2£«Ê§µç×ÓÉú³ÉMnO₂£¬²úÎïÖеÄÑõÔªËØÀ´Ô´ÓÚË®£¬Éú³ÉH^£«£¬ÔÙ½«µç¼«·½³ÌʽÅäƽ¼´¿É¡£
£¨3£©Í¼ÖÐX¼«µÄµç¼«·´Ó¦Îª2H^£«£«2e^£­=H₂¡ü£¨»ò2H₂O£«2e^£­=H₂¡ü£«2OH^£­£©£¬ËùÒÔX¼«ÇøµÄpHÔö´ó£¬ÓÉÉú³ÉÇâÆø672 mL£¬¿ÉÖªµÃµç×ÓÊýΪ0.06 mol£¬Y¼«Éú³ÉÑõÆøΪ168 mL£¬Ê§µç×ÓÊý0.03  mol£¬ÓɵÃʧµç×ÓÊغã¿ÉÖªÌúʧµç×ÓÊýΪ0.03 mol£¬Óɵ缫·´Ó¦¿ÉÖªÌúÈܽâΪ0.005 mol£¬¼´0.28 g¡£