ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÔÚÒ»¶¨Ìõ¼þÏ£¬½«3molAºÍ1molBÁ½ÖÖÆøÌå»ìºÏÓڹ̶¨ÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÈçÏ·´Ó¦£º3A(g)+B(g) xC(g)+2D(g)¡£2minÄ©¸Ã·´Ó¦´ïµ½Æ½ºâ£¬Éú³É0.8mol D£¬²¢²âµÃCµÄŨ¶ÈΪ0.2mol/L¡£ÏÂÁÐÅжϴíÎóµÄÊÇ

A. x=1

B. BµÄת»¯ÂÊΪ80%

C. 2minÄÚAµÄƽ¾ù·´Ó¦ËÙÂÊΪ0.3 mol¡¤L-1¡¤min-1

D. Èô»ìºÏÆøÌåµÄÃܶȲ»±ä£¬Ò²²»ÄÜ˵Ã÷¸Ã·´Ó¦´ïµ½Æ½ºâ״̬

¡¾´ð°¸¡¿B

¡¾½âÎö¡¿

A. ƽºâʱÉú³ÉµÄCµÄÎïÖʵÄÁ¿Îª0.2molL-1¡Á2L=0.4mol£¬ÎïÖʵÄÁ¿Ö®±ÈµÈÓÚ»¯Ñ§¼ÆÁ¿ÊýÖ®±È£¬¹Ê0.4mol£º0.8mol=x£º2£¬½âµÃx=1£¬AÏîÕýÈ·£»
B. 2minÄ©¸Ã·´Ó¦´ïµ½Æ½ºâ£¬Éú³É0.8mol D£¬ÓÉ·½³Ìʽ3A£¨g£©+B£¨g£©xC£¨g£©+2D£¨g£©¿ÉÖª£¬²Î¼Ó·´Ó¦µÄBµÄÎïÖʵÄÁ¿Îª£º0.8mol¡Á=0.4mol£¬¹ÊBµÄת»¯ÂÊΪ¡Á100%=40%£¬BÏî´íÎó£»
C. 2minÄÚÉú³É0.8mol D£¬¹Ê2 minÄÚDµÄ·´Ó¦ËÙÂÊv£¨D£©= =0.2 mol£¨Lmin£©-1£¬ËÙÂÊÖ®±ÈµÈÓÚ»¯Ñ§¼ÆÁ¿ÊýÖ®±È£¬¹Êv£¨A£©=v£¨D£©=¡Á0.2 mol£¨Lmin£©-1=0.3 mol£¨Lmin£©-1£¬CÏîÕýÈ·£»
D. ÈÝÆ÷µÄÈÝ»ý²»±ä£¬»ìºÏÆøÌåµÄÖÊÁ¿²»±ä£¬ÃܶÈΪ¶¨Öµ£¬Ê¼ÖÕ²»±ä£¬¹Ê»ìºÏÆøÌåµÄÃܶȲ»±ä£¬²»ÄÜ˵Ã÷·´Ó¦´ïµ½Æ½ºâ״̬£¬DÏîÕýÈ·£»
´ð°¸Ñ¡B¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿SO2ºÍCO¾ùΪȼúÑÌÆøÖеÄÖ÷ÒªÎÛȾÎ¶Ô¶þÕßµÄÖÎÀí±¸ÊÜÖõÄ¿¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÏÂÁÐÊÂʵÖУ¬²»ÄÜÓÃÓڱȽÏÁòÔªËغÍ̼ԪËطǽðÊôÐÔÇ¿ÈõµÄÊÇ___________£¨ÌîÑ¡Ïî×Öĸ£©¡£

A£®SO2ÄÜʹËáÐÔKMnO4ÈÜÒºÍÊÉ«¶øCO2²»ÄÜ

B£®ÏàͬÌõ¼þÏ£¬ÁòËáµÄËáÐÔÇ¿ÓÚ̼Ëá

C£®CS2ÖÐÁòÔªËØÏÔ-2¼Û

D£®ÏàͬÌõ¼þÏ£¬SO3µÄ·Ðµã¸ßÓÚCO2

£¨2£©ÓÐÈËÉè¼Æͨ¹ýÁòÑ­»·Íê³É¶þÕßµÄ×ۺϴ¦Àí£¬Ô­ÀíΪ

i£®2CO(g) +SO2 (g)S(l)+2CO2(g) ¡÷H1=-37.0 kJ¡¤mol-1

ii£®S(l) + 2H2O(g)2H2(g) + SO2(g) ¡÷H2 =-45.4 kJ¡¤mol-1

1 mol COºÍË®ÕôÆøÍêÈ«·´Ó¦Éú³ÉH2ºÍCO2µÄÈÈ»¯Ñ§·½³ÌʽΪ__________________________¡£

£¨3£©T¡æ£¬Ïò5LºãÈÝÃܱÕÈÝÆ÷ÖгäÈë2 mol COºÍ1 mol SO2£¬·¢Éú·´Ó¦i¡£10min´ïµ½Æ½ºâʱ£¬²âµÃS(l)µÄÎïÖʵÄÁ¿Îª0.8mol¡£

¢Ù010 minÄÚ£¬ÓÃCO±íʾ¸Ã·´Ó¦ËÙÂÊv(CO)=____________________¡£

¢Ú·´Ó¦µÄƽºâ³£ÊýΪ______________________¡£

£¨4£©ÆðʼÏòÃܱÕÈÝÆ÷ÖгäÈëÒ»¶¨Á¿µÄS(1)ºÍH2O(g)£¬·¢Éú·´Ó¦ii¡£H2O(g)µÄƽºâת»¯ÂÊÓëζȣ¨T£©ºÍѹǿ£¨p£©µÄ¹ØϵÈçͼËùʾ¡£

¢ÙM¡¢PÁ½µãµÄƽºâת»¯ÂÊ£ºa(M)___________a(P)£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°=¡±£©£¬ÀíÓÉΪ_____________________¡£

¢ÚN¡¢PÁ½µãµÄƽºâ³£Êý£ºK(N)___________K(P)£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°=¡±£©¡£

¡¾ÌâÄ¿¡¿¡°·Ö×Óɸ¡±ÊÇÒ»ÖÖ¾ßÓжà¿×½á¹¹µÄÂÁ¹èËáÑÎ(NaAlSiO4¡¤nH2O)£¬ÆäÖÐÓÐÐí¶àÁý×´¿×ѨºÍͨµÀ£¬ÄÜÈÃÖ±¾¶±È¿×ѨСµÄ·Ö×Óͨ¹ý¶ø½«´óµÄ·Ö×ÓÁôÔÚÍâÃ棬¹Ê´ËµÃÃû¡£ÀûÓÃÂÁ»Ò(Ö÷Òª³É·ÖΪAl¡¢Al2O3¡¢AlN¡¢FeOµÈ)ÖƱ¸¡°·Ö×Óɸ¡±µÄÒ»ÖÖ¹¤ÒÕÁ÷³ÌÈçÏ£º

£¨1£©¡°·Ö×Óɸ¡±µÄ»¯Ñ§Ê½ÓÃÑõ»¯ÎïÐÎʽ¿É±íʾΪ_______________¡£

£¨2£©ÂÁ»ÒË®½â²úÉúµÄÆøÌåΪ________(Ìѧʽ)£»¡°Ë®½â¡±ÔÚ¼ÓÈÈÌõ¼þ϶ø²»ÔÚÊÒÎÂϽøÐеÄÔ­ÒòÊÇ________________________¡£

£¨3£©¡°ËáÈÜ¡±Ê±£¬·¢ÉúÑõ»¯»¹Ô­·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________________________¡£

£¨4£©¸Ã¹¤ÒÕÖÐÂËÔüµÄÑÕɫΪ________________________¡£

£¨5£©Ä³Ñ§Ï°Ð¡×éÉè¼ÆʵÑéÄ£Äâ´ÓŨËõº£Ë®(º¬Ca2+¡¢Mg2+¡¢SO42-)ÖÐÌáÈ¡ÊÔ¼Á¼¶NaCl£º

¢ÙʵÑéÖÐÈôÏòŨËõº£Ë®ÖмÓÈëµÄÊÇNa2CO3ŨÈÜÒº£¬ÔòÓÐÄÑÈܵÄMg2(OH)2CO3Éú³É£¬Í¬Ê±ÓÐÆøÌåÒݳö¡£¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________¡£

¢Ú¸ÃѧϰС×é·¢ÏÖÉÏÊöʵÑ鼴ʹBaCl2ÓÃÁ¿²»×㣬µÚ¢ó²½³ÁµíÖÐÒÀÈ»º¬ÓÐÉÙÁ¿BaCO3¡£´Óƽºâ½Ç¶È·ÖÎöÆäÔ­Òò£º_____________________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø