ÌâÄ¿ÄÚÈÝ


X2ÊÇÈËÌåºôÎüÖв»¿ÉȱÉÙµÄÆøÌ壬YÊǵؿÇÖк¬Á¿×î¶àµÄ½ðÊôÔªËØ£¬ZÔªËØµÄÑæÉ«Îª»ÆÉ«£¬W2ÊÇÒ»ÖÖ»ÆÂÌÉ«ÆøÌå¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

   £¨1£©Ð´³öZµÄÔªËØ·ûºÅ        £¬»­³öXµÄÔ­×ӽṹʾÒâͼ            £»

£¨2£©¹¤ÒµÉÏÀûÓÃW2ºÍÏûʯ»ÒÖÆÈ¡Æ¯°×·Û£¬ÔòƯ°×·ÛµÄÓÐЧ³É·ÖÊÇ           £»

   £¨3£©ZÔªËØ×é³ÉµÄÁ½ÖÖÑÎA¡¢BÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒÔÏ໥ת»¯£¬A¡¢B¶¼¿ÉÓë×ãÁ¿ÑÎËá·´Ó¦ÇÒ²úÎïÍêȫһÑù£¬ÆäÖÐAÊÇ·¢½Í·ÛµÄÖ÷Òª³É·ÖÖ®Ò»£¬ÔòBͨ¹ý»¯ºÏ·´Ó¦×ª»¯ÎªAµÄ»¯Ñ§·½³Ìʽ                            £»

   £¨4£©¹¤ÒµÉÏÊÇͨ¹ýµç½âY2X3Ò±Á¶Yµ¥ÖÊ£¬ÈôÉú³ÉÁË0.1 mol Yµ¥ÖÊ£¬Ôò×ªÒÆ     molµç×Ó£»

£¨5£©½ñÓÐMgSO4ºÍY2(SO4)3µÄ»ìºÏÈÜÒº£¬ÒÑÖªc£¨SO42£­£© = 1.5mol¡¤L£­1¡£ÔòÈÜÒºÖÐc£¨Y3+£©=         mol¡¤L£­1¡£ÆäÖУ¬ÍùÒ»¶¨Ìå»ý¸Ã»ìºÏÈÜÒºÖмÓÈëZµ¥ÖʵÄÎïÖʵÄÁ¿ÓëÉú³É³ÁµíµÄÎïÖʵÄÁ¿µÄ¹ØÏµÈçͼ£º


£¨1£©Na    

    £¨2£©Ca(ClO)2

    £¨3£©Na2CO3+CO2+H2O==2NaHCO3   (·½³ÌʽûÅ䯽¿Û1·Ö)

    £¨4£©0.3

    £¨5£©0.6


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¼×´¼¿É×÷ΪȼÁÏµç³ØµÄÔ­ÁÏ¡£ÒÔCH4ºÍH2OΪԭÁÏ£¬Í¨¹ýÏÂÁз´Ó¦·´Ó¦À´ÖƱ¸¼×´¼¡£

I£ºCH4 ( g ) + H2O ( g ) £½CO ( g ) + 3H2 (g )    ¡÷H £½+206.0 kJ¡¤mol£­1

II£ºCO ( g ) + 2H2 ( g ) £½ CH3OH ( g )        ¡÷H £½¡ª129.0 kJ¡¤mol£­1

£¨1£©CH4(g)ÓëH2O(g)·´Ó¦Éú³ÉCH3OH (g)ºÍH2(g)µÄÈÈ»¯Ñ§·½³ÌʽΪ         ¡£

£¨2£©½«1.0 mol CH4ºÍ2.0 mol H2O ( g )ͨÈëÈÝ»ýΪ100 LµÄ·´Ó¦ÊÒ£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦I£¬²âµÃÔÚÒ»¶¨µÄѹǿÏÂCH4µÄת»¯ÂÊÓëζȵÄ

¹ØÏµÈçÓÒͼ¡£

¢Ù¼ÙÉè100 ¡æÊ±´ïµ½Æ½ºâËùÐèµÄʱ¼äΪ5 min£¬

ÔòÓÃH2±íʾ¸Ã·´Ó¦µÄƽ¾ù·´Ó¦ËÙÂÊΪ      ¡£

¢Ú100¡æÊ±·´Ó¦IµÄƽºâ³£ÊýΪ             ¡£

£¨3£©ÔÚѹǿΪ0.1 MPa¡¢Î¶ÈΪ300¡æÌõ¼þÏ£¬½«a mol COÓë

3a mol H2µÄ»ìºÏÆøÌåÔÚ´ß»¯¼Á×÷ÓÃÏ·¢Éú·´Ó¦IIÉú³É¼×´¼£¬Æ½ºâºó½«ÈÝÆ÷µÄÈÝ»ýѹËõµ½Ô­À´µÄl/2£¬ÆäËûÌõ¼þ²»±ä£¬¶ÔƽºâÌåϵ²úÉúµÄÓ°ÏìÊÇ      ¡ø     (Ìî×ÖĸÐòºÅ)¡£

A£®c ( H2 )¼õÉÙ   B£®Õý·´Ó¦ËÙÂÊ¼Ó¿ì£¬Äæ·´Ó¦ËÙÂʼõÂý   C£®CH3OH µÄÎïÖʵÄÁ¿Ôö¼Ó        D£®ÖØÐÂÆ½ºâc ( H2 )/ c (CH3OH )¼õС    E£®Æ½ºâ³£ÊýKÔö´ó

£¨4£©Ð´³ö¼×´¼¡ª¿ÕÆø¡ªKOHÈÜÒºµÄȼÁÏµç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½£º                   ¡£

£¨5£©¼×´¼¶ÔË®ÖÊ»áÔì³ÉÒ»¶¨µÄÎÛȾ£¬ÓÐÒ»Öֵ绯ѧ·¨¿ÉÏû³ý

ÕâÖÖÎÛȾ£¬ÆäÔ­ÀíÊÇ£ºÍ¨µçºó£¬½«Co2+Ñõ»¯³ÉCo3+£¬

È»ºóÒÔCo3+×öÑõ»¯¼Á°ÑË®Öеļ״¼Ñõ»¯³ÉCO2¶ø¾»»¯¡£

ʵÑéÊÒÓÃÓÒͼװÖÃÄ£ÄâÉÏÊö¹ý³Ì£º

¢Ù д³öÑô¼«µç¼«·´Ó¦Ê½                          ¡£

¢Ú д³ö³ýÈ¥¼×´¼µÄÀë×Ó·½³Ìʽ                   ¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø