ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÔÚÒ»¶¨Î¶ÈÌõ¼þÏ£¬¼×¡¢ÒÒÁ½¸öÈÝ»ýÏàµÈµÄºãÈÝÃܱÕÈÝÆ÷Öоù·¢ÉúÈçÏ·´Ó¦£º3A(g)+B(g)xC(g)+D(s)£¬Ïò¼×ÖÐͨÈë6molAºÍ2molB£¬ÏòÒÒÖÐͨÈë1.5molA¡¢0.5molBºÍ3molCºÍ2molD£¬·´Ó¦Ò»¶Îʱ¼äºó¶¼´ïµ½Æ½ºâ£¬´Ëʱ²âµÃ¼×¡¢ÒÒÁ½ÈÝÆ÷ÖÐCµÄÌå»ý·ÖÊý¶¼Îª0.2£¬ÏÂÁÐÐðÊöÖÐÕýÈ·µÄÊÇ£¨ £©

A.ƽºâʱ¼×ÖÐAµÄÌå»ý·ÖÊýΪ0.4

B.ƽºâʱ¼×¡¢ÒÒÁ½ÈÝÆ÷ÖÐA¡¢BµÄÎïÖʵÄÁ¿Ö®±È²»ÏàµÈ

C.ÈôƽºâʱÁ½ÈÝÆ÷ÖеÄѹǿ²»ÏàµÈ£¬ÔòÁ½ÈÝÆ÷ÖÐѹǿ֮±ÈΪ8£º5

D.Èôƽºâʱ¼×¡¢ÒÒÁ½ÈÝÆ÷ÖÐAµÄÎïÖʵÄÁ¿ÏàµÈ£¬Ôòx=4

¡¾´ð°¸¡¿C

¡¾½âÎö¡¿

A. CµÄÌå»ý·ÖÊý¶¼Îª0.2£¬ÔòA¡¢B¹²Õ¼80%£¬¶øn(A)£ºn(B)=3£º1£¬ËùÒÔƽºâʱ¼×ÖÐAµÄÌå»ý·ÖÊýΪ60%£¬¹ÊA´íÎó£»

B. ¼×¡¢ÒÒÁ½ÈÝÆ÷ÖÐA¡¢BµÄͶÁϱȶ¼µÈÓÚϵÊý±È£¬ËùÒÔƽºâʱ¼×¡¢ÒÒÁ½ÈÝÆ÷ÖÐA¡¢BµÄÎïÖʵÄÁ¿Ö®±ÈÏàµÈ£¬¹ÊB´íÎó£»

C.ÈôƽºâʱÁ½ÈÝÆ÷ÖеÄѹǿ²»ÏàµÈ£¬Ôò¼×¡¢ÒÒΪµÈ±ÈµÈЧµÄµÈЧƽºâ£¬x=4£¬¼×µÄͶÁÏÊÇ6mol+2mol=8mol£¬ÒÒµÄͶÁÏÊÇ1.5mol+0.5mol+3mol=5mol£¬ÔòÁ½ÈÝÆ÷ÖÐѹǿ֮±ÈΪ8£º5£¬¹ÊCÕýÈ·£»

D. Èôƽºâʱ¼×¡¢ÒÒÁ½ÈÝÆ÷ÖÐAµÄÎïÖʵÄÁ¿ÏàµÈ£¬ÔòΪµÈÁ¿µÈЧµÄµÈЧƽºâ£¬¼´Âú×㣬Ôòx=2£¬¹ÊD´íÎó£»

Ñ¡C¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿°´ÒªÇóÍê³ÉÏÂÁÐÌî¿Õ

I.£¨1£©¸ø¶¨Ìõ¼þϵÄÏÂÁÐËÄÖÖÎïÖÊ£º

a£®10gÄÊÆø

b£®º¬ÓÐ40molµç×ÓµÄNH3

c£®±ê×¼×´¿öÏÂ8.96LCO2

d£®±ê×¼×´¿öÏÂ112gҺ̬ˮ

ÔòÉÏÊöÎïÖÊÖÐËùº¬·Ö×ÓÊýÄ¿ÓɶൽÉÙµÄ˳ÐòÊÇ________________£¨ÌîÐòºÅ£©¡£

£¨2£©±ê×¼×´¿öÏ£¬0.51gijÆøÌåµÄÌå»ýΪ672mL£¬Ôò¸ÃÆøÌåĦ¶ûÖÊÁ¿Îª______¡£

£¨3£©½«100mL H2SO4ºÍHClµÄ»ìºÏÈÜÒº·Ö³ÉÁ½µÈ·Ý£¬Ò»·ÝÖмÓÈ뺬0.2molNaOHÈÜҺʱǡºÃÖкÍÍêÈ«£¬ÏòÁíÒ»·ÝÖмÓÈ뺬0.05molBaCl2ÈÜҺʱǡºÃ³ÁµíÍêÈ«£¬ÔòÔ­ÈÜÒºÖÐc(Cl£­)=____ mol/L¡£

II£®ÏÖÓÐÒÔÏÂÎïÖÊ£º¢ÙÂÁ£»¢Ú¶þÑõ»¯¹è£»¢ÛÒºÂÈ£»¢ÜNaOHÈÜÒº£»¢ÝҺ̬HCl£»¢ÞNaHCO3¾§Ì壻¢ßÕáÌÇ£»¢àÈÛÈÚNa2O£»¢áNa2O2¹ÌÌ壻¢âCO2¡£»Ø´ðÏÂÁÐÎÊÌ⣨ÓÃÏàÓ¦ÎïÖʵÄÐòºÅÌîд£©£º

£¨1£©ÆäÖпÉÒÔµ¼µçµÄÓÐ__________¡£

£¨2£©ÊôÓÚµç½âÖʵÄÓÐ_______£¬·Çµç½âÖÊÓÐ__________¡£

£¨3£©Ð´³öÏò¢áÓë¢â·´Ó¦µÄ»¯Ñ§·½³Ìʽ___________¡£

£¨4£©Ð´³ö¢ÙÓë¢ÜµÄÀë×Ó·½³Ìʽ_____________¡£

£¨5£©Ð´³ö¢ÝµÄË®ÈÜÒºÓë¢ÞµÄË®ÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ___________________ ¡£

£¨6£©Ð´³ö¢ÚÓë¢Ü·´Ó¦µÄÀë×Ó·½³Ìʽ_______________________________ ¡£

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

¾«Ó¢¼Ò½ÌÍø