ÌâÄ¿ÄÚÈÝ
£¨12·Ö£©ÀûÓÃÌìÈ»ÆøºÏ³É°±µÄ¹¤ÒÕÁ÷³ÌʾÒâÈçÏ£º
![]()
![]()
ÒÀ¾ÝÉÏÊöÁ÷³Ì£¬Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©ÌìÈ»ÆøÍÑÁòʱµÄ»¯Ñ§·½³ÌʽÊÇ_______________________¡£
£¨2£©K2CO3£¨aq£©ºÍCO2·´Ó¦ÔÚ¼ÓѹϽøÐУ¬¼ÓѹµÄÀíÂÛÒÀ¾ÝÊÇ____________£¨¶àÑ¡¿Û·Ö£©¡£
£¨a£©ÏàËÆÏàÈÜÔÀí £¨b£©ÀÕÉ³ÌØÁÐÔÀí £¨c£©Ëá¼îÖкÍÔÀí
£¨3£©ÓÉKHCO3·Ö½âµÃµ½µÄCO2¿ÉÒÔÓÃÓÚ_______________________£¨Ð´³öCO2µÄÒ»ÖÖÖØÒªÓÃ;£©¡£
£¨4£©Õû¸öÁ÷³ÌÓÐÈý´¦Ñ»·£¬Ò»ÊÇFe£¨OH£©3Ñ»·£¬¶þÊÇK2CO3£¨aq£©Ñ»·£¬ÇëÔÚÏÂͼÖбê³öÉÏÊöÁ÷³ÌͼµÚÈý´¦Ñ»·£¨Ñ»··½Ïò¡¢Ñ»·ÎïÖÊ£©¡£
![]()
![]()
£¨5£©ÔÚÒ»¶¨Î¶ȺÍѹǿµÄÃܱպϳɷ´Ó¦Æ÷ÖУ¬ H2ºÍN2»ìºÏÆøÌ寽¾ùÏà¶Ô·Ö×ÓÖÊÁ¿Îª8.5£¬µ±¸Ã·´Ó¦´ïµ½Æ½ºâʱ£¬²â³öƽºâ»ìºÏÆøµÄƽ¾ùʽÁ¿Îª10£¬Çë¼ÆËã´ËʱH2µÄת»¯ÂÊ£¨Ð´³ö¼ÆËã¹ý³Ì£©£º
£¨1£©3H2S£«2Fe£¨OH£©3
Fe2S3+6H2O£¨2·Ö£©
£¨2£©b£¨2·Ö£©
£¨3£©Éú²ú´¿¼î£Û»ò×÷ÖÆÀä¼ÁµÈ£¬ÆäËüºÏÀí´ð°¸Ò²¸ø·Ö£Ý£¨2·Ö£©
£¨4£©
£¨2·Ö£©
£¨5£©Éè³äÈëÆøÌå×ÜÁ¿Îª1mol£¬µªÆøÎªx£¬ÔòÇâÆøÎª£¨1-x£©¡£
ÔòÓУº 28x+2(1-x)=8.5½âµÃ£ºN2£ºx=0.25mol H2£º1mol-0.25mol=0.75mol
ÓÖÉèÆ½ºâʱN2ת»¯y£¬Ôò£º
N2 + 3H2
2NH3
Æðʼ 0.25mol 0.75mol 0
±ä»¯ y 3y 2y
ƽºâ (0.25-y)mol (0.75-3y)mol 2ymol
ÔòÓУº
½âµÃ£ºy=0.075mol
ÔòÇâÆøµÄת»¯ÂÊΪ£º
£¨ÆäËüºÏÀí½â·¨²ÎÕÕ¸ø·Ö£©£¨4·Ö£©