£¨10·Ö£©ÀûÓÃÌìÈ»ÆøºÏ³É°±µÄ¹¤ÒÕÁ÷³ÌʾÒâͼÈçÏ£º

ÒÀ¾ÝÉÏÊöÁ÷³Ì£¬Íê³ÉÏÂÁÐÌî¿Õ£º

£¨1£©ÌìÈ»ÆøÍÑÁòʱµÄ»¯Ñ§·½³ÌʽÊÇ                                         ¡£

£¨2£©Í¼ÖÐCH4µÄµÚÒ»´Îת»¯¹ý³ÌÖеĻ¯Ñ§·½³ÌʽÊÇ                           ¡£

£¨3£©Õû¸öÁ÷³ÌÓÐÈý´¦Ñ­»·£¬Ò»ÊÇK2CO3(aq)Ñ­»·£¬¶þÊÇN2ºÍH2Ñ­»·£¬ÈýÊÇ        £¨Ìѧʽ£©Ñ­»·¡£

£¨4£©K2CO3£¨aq£©ºÍCO2·´Ó¦ÔÚ¼ÓѹϽøÐУ¬¼ÓѹµÄÀíÂÛÒÀ¾ÝÊÇ      £¨¶àÑ¡¿Û·Ö£©¡£

a.ìØÔöÔ­Àí           b.ÀÕÏÄÌØÁÐÔ­Àí            c.Ëá¼îÖкÍÔ­Àí

 

£¨1£©3H2S£«2Fe(OH)3=== Fe2S3£«6H2O   £¨3·Ö£©

£¨2£©CH4+H2OCO+3H2   £¨3·Ö£©

£¨3£©Fe(OH)3£¨2·Ö£©

£¨4£©b£¨2·Ö£©

½âÎö:ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2009?÷ÖÝÄ£Ä⣩ÀûÓÃÌìÈ»ÆøºÏ³É°±µÄ¹¤ÒÕÁ÷³ÌʾÒâͼÈçͼ1Ëùʾ£º

ÒÀ¾ÝÉÏÊöÁ÷³Ì£¬Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©ÌìÈ»ÆøÍÑÁòʱµÄ»¯Ñ§·½³ÌʽÊÇ
3H2S+2Fe£¨OH£©3¨TFe2S3+6H2O
3H2S+2Fe£¨OH£©3¨TFe2S3+6H2O
£®
£¨2£©K2CO3£¨aq£©ºÍCO2·´Ó¦ÔÚ¼ÓѹϽøÐУ¬¼ÓѹµÄÀíÂÛÒÀ¾ÝÊÇ
b
b
£¨¶àÑ¡¿Û·Ö£©£®
£¨a£©ÏàËÆÏàÈÜÔ­Àí  £¨b£©ÀÕɳÌØÁÐÔ­Àí  £¨c£©Ëá¼îÖкÍÔ­Àí
£¨3£©ÓÉKHCO3·Ö½âµÃµ½µÄCO2¿ÉÒÔÓÃÓÚ
Éú²ú´¿¼î£¨»ò×÷ÖÆÀä¼ÁµÈ£©
Éú²ú´¿¼î£¨»ò×÷ÖÆÀä¼ÁµÈ£©
£¨Ð´³öCO2µÄÒ»ÖÖÖØÒªÓÃ;£©£®
£¨4£©Õû¸öÁ÷³ÌÓÐÈý´¦Ñ­»·£¬Ò»ÊÇFe£¨OH£©3Ñ­»·£¬¶þÊÇK2CO3£¨aq£©Ñ­»·£¬ÇëÔÚͼ2Öбê³öÉÏÊöÁ÷³ÌͼµÚÈý´¦Ñ­»·£¨Ñ­»··½Ïò¡¢Ñ­»·ÎïÖÊ£©£®
£¨5£©ÔÚÒ»¶¨Î¶ȺÍѹǿµÄÃܱպϳɷ´Ó¦Æ÷ÖУ¬H2ºÍN2»ìºÏÆøÌåƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿Îª8.5£¬µ±¸Ã·´Ó¦´ïµ½Æ½ºâʱ£¬²â³öƽºâ»ìºÏÆøµÄƽ¾ùʽÁ¿Îª10£¬Çë¼ÆËã´ËʱH2µÄת»¯ÂÊ£¨Ð´³ö¼ÆËã¹ý³Ì£©£º
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÀûÓÃÌìÈ»ÆøºÏ³É°±µÄ¹¤ÒÕÁ÷³ÌʾÒâÈçÏ£º
ÒÀ¾ÝÉÏÊöÁ÷³Ì£¬Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©ÌìÈ»ÆøÍÑÁòʱµÄ»¯Ñ§·½³ÌʽÊÇ
3H2S+2Fe£¨OH£©3¨TFe2S3+6H2O
3H2S+2Fe£¨OH£©3¨TFe2S3+6H2O
£®
£¨2£©n mol CH4¾­Ò»´Îת»¯ºó²úÉúCO 0.9n mol¡¢²úÉúH2
2.7n
2.7n
mol£¨Óú¬nµÄ´úÊýʽ±íʾ£©
£¨3£©K2CO3£¨aq£©ºÍ CO2·´Ó¦ÔÚ¼ÓѹϽøÐУ¬¼ÓѹµÄÀíÂÛÒÀ¾ÝÊÇ
b
b
£¨¶àÑ¡¿Û·Ö£©
£¨a£©ÏàËÆÏàÈÜÔ­Àí    £¨b£©ÀÕɳÌØÁÐÔ­Àí    £¨c£©Ëá¼îÖкÍÔ­Àí
£¨4£©ÓÉKHCO3·Ö½âµÃµ½µÄCO2¿ÉÒÔÓÃÓÚ
Éú²ú´¿¼î
Éú²ú´¿¼î
£¨Ð´³öCO2µÄÒ»ÖÖÖØÒªÓÃ;£©£®
£¨5£©Õû¸öÁ÷³ÌÓÐÈý´¦Ñ­»·£¬Ò»ÊÇFe£¨OH£©3Ñ­»·£¬¶þÊÇK2CO3£¨aq£©Ñ­»·£¬ÇëÔÚÉÏÊöÁ÷³ÌͼÖбê³öµÚÈý´¦Ñ­»·£¨Ñ­»··½Ïò¡¢Ñ­»·ÎïÖÊ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÀûÓÃÌìÈ»ÆøºÏ³É°±µÄ¹¤ÒÕÁ÷³ÌʾÒâͼÈçÏ£º

ÒÀ¾ÝÉÏÊöÁ÷³Ì£¬Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©Í¼ÖÐCH4µÄµÚÒ»´Îת»¯¹ý³ÌÖеĻ¯Ñ§·½³ÌʽÊÇ
CH4+H2O?CO+3H2
CH4+H2O?CO+3H2
£®
£¨2£©ÍÑÁò¹ý³ÌÖУ¬ÈôÓÐn mol Fe2O3?H2Oת»¯£¬ÔòÉú³ÉSµÄÎïÖʵÄÁ¿Îª
n
n
mol£¨Óú¬nµÄ´úÊýʽ±íʾ£©£®
£¨3£©Õû¸öÁ÷³ÌÓÐÈý¸öÑ­»·£ºÒ»ÊÇK2CO3£¨aq£©Ñ­»·£¬¶þÊÇN2ºÍH2Ñ­»·£¬µÚÈý¸öÑ­»·Öб»Ñ­»·ÎïÖÊÊÇ
Fe2O3?H2O
Fe2O3?H2O
£®
£¨4£©¸ÄÓùýÁ¿NaOHÈÜÒºÎüÊÕÌìÈ»ÆøÖеÄÁò»¯Ç⣬ÒÔʯī×÷µç¼«µç½âÎüÊÕºóËùµÃÈÜÒº¿É»ØÊÕÁò£¬Æäµç½â×Ü·´Ó¦·½³Ìʽ£¨ºöÂÔÑõÆøµÄÑõ»¯»¹Ô­£©Îª
Na2S+2H2O
 Í¨µç 
.
 
2NaOH+S+H2¡ü
Na2S+2H2O
 Í¨µç 
.
 
2NaOH+S+H2¡ü
£¬¸Ã·½·¨µÄÓŵãÊÇ
¸Ã·½·¨NaOH¿ÉÒÔÑ­»·ÀûÓã¬Í¬Ê±»ñµÃ¸±²úÆ·ÇâÆø
¸Ã·½·¨NaOH¿ÉÒÔÑ­»·ÀûÓã¬Í¬Ê±»ñµÃ¸±²úÆ·ÇâÆø
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012ÄêËս̰æ¸ßÖл¯Ñ§Ñ¡ÐÞ2 2.1°±µÄºÏ³ÉÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

ÀûÓÃÌìÈ»ÆøºÏ³É°±µÄ¹¤ÒÕÁ÷³ÌʾÒâͼÈçÏ£º

ÒÀ¾ÝÉÏÊöÁ÷³Ì£¬Íê³ÉÏÂÁÐÌî¿Õ£º

(1)n mol CH4¾­Ò»´Îת»¯ºó²úÉúCO 0.9n mol£¬²úÉúH2________mol(Óú¬nµÄ´úÊýʽ±íʾ)¡£

(2)K2CO3ÈÜÒººÍCO2·´Ó¦ÔÚ¼ÓѹÌõ¼þϽøÐУ¬¼ÓѹµÄÀíÂÛÒÀ¾ÝÊÇ________(Ìî×Öĸ´úºÅ)¡£

a£®ÏàËÆÏàÈÜÔ­Àí¡¡b£®Æ½ºâÒƶ¯Ô­Àí¡¡c£®Ëá¼îÖкÍÔ­Àí

(3)ÓÉKHCO3·Ö½âµÃµ½µÄCO2¿ÉÒÔÓÃÓÚ________(д³öCO2µÄÒ»ÖÖÖØÒªÓÃ;)¡£

(4)Õû¸öÁ÷³ÌÓÐÈý´¦Ñ­»·£¬Ò»ÊÇFe(OH)3Ñ­»·£¬¶þÊÇK2CO3ÈÜҺѭ»·£¬ÈýÊÇ________Ñ­»·¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸