ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿Ìú¼°Æ仯ºÏÎïÔÚÉú²úÉú»îÖÐÓ¦Óù㷺£¬ÈçÌúºì£¨Fe2O3£©¿É×÷ΪÑÕÁÏ£¬µç×Ó¹¤Òµ³£ÓÃÒ»¶¨Å¨¶ÈµÄFeCl3ÈÜÒº¸¯Ê´·óÓÐͲµÄ¾øÔµ°å£¬ÖƳÉÓ¡Ë¢Ïß·°å¡£ aFe2£¨SO4£© 3¡¤b£¨NH4£© 2SO4¡¤cH2O£¬£¨ÁòËáÌú泥©³£ÓÃÓÚÉú»îÒûÓÃË®¡¢¹¤ÒµÑ»·Ë®µÄ¾»»¯´¦Àí¡£
£¨1£©ÏÖÓÐÒ»º¬ÓÐFe2O3ºÍFe3O4µÄ»ìºÏÎïÑùÆ·£¬²âµÃn£¨Fe£©:n£¨O£©=1:1.375,Ôò¸ÃÑùÆ·ÖÐFe2O3µÄÎïÖʵÄÁ¿·ÖÊýΪ___________¡££¨½á¹û±£Áô2λÓÐЧÊý×Ö£©
£¨2£©CuOºÍFe2O3µÄ»ìºÏÎï9.6 gÔÚ¸ßÎÂÏÂÓë×ãÁ¿µÄCO³ä·Ö·´Ó¦£¬·´Ó¦ºóÈ«²¿ÆøÌåÓÃ100mL 1.2mol/L Ba£¨OH£©2 ÈÜÒºÎüÊÕ£¬Éú³É15.76 g°×É«³Áµí¡£ÔòÎüÊÕÆøÌåºóÈÜÒºÖеÄÈÜÖʵĻ¯Ñ§Ê½Îª__________£¬»ìºÏÎïÖÐCuOºÍFe2O3µÄÎïÖʵÄÁ¿Ö®±ÈΪ___________¡£
£¨3£©³ÆȡijÁòËáÌúï§ÑùÆ·7.00 g£¬½«ÆäÈÜÓÚË®ÅäÖƳÉ100 mLÈÜÒº£¬·Ö³ÉÁ½µÈ·Ý£¬ÏòÆäÖÐÒ»·ÝÖмÓÈë×ãÁ¿NaOHÈÜÒº£¬¹ýÂËÏ´µÓµÃµ½1.07 g³Áµí£»ÏòÁíÒ»·ÝÈÜÒºÖмÓÈ뺬0.025 molBa £¨NO3£©2µÄÈÜÒº£¬Ç¡ºÃÍêÈ«·´Ó¦£¬Çó¸ÃÁòËáÌú淋Ļ¯Ñ§Ê½_________¡£
£¨4£©ÏÖ½«Ò»¿é·óÓÐͲµÄ¾øÔµ°å½þÈë800mL 3mol/LµÄFeCl3ÈÜÒºÖУ¬Ò»¶Îʱ¼äºó£¬½«¸ÃÏß·°åÈ¡³ö£¬ÏòÈÜÒºÖмÓÈëÌú·Û56.0 g£¬³ä·Ö·´Ó¦ºóÊ£Óà¹ÌÌå51.2 g£¬ÇóËùµÃÈÜÒºÖÐÈÜÖʵÄÎïÖʵÄÁ¿Å¨¶È_________£¨ºöÂÔ·´Ó¦Ç°ºóÈÜÒºÌå»ýµÄ±ä»¯£©¡£
¡¾´ð°¸¡¿33% Ba(HCO3)2 1:1 Fe2£¨SO4£©32£¨NH4£©2SO42H2O FeCl2 4.25 mol/L£»CuCl2 0.25 mol/L
¡¾½âÎö¡¿
£¨1£©ÉèFe2O3ºÍFe3O4µÄ»ìºÏÎïµÄÎïÖʵÄÁ¿·Ö±ðÊÇx¡¢y£¬Ôòn(Fe):n(O)=(2x+3y):(3x+4y)=1:1.375£¬Ôòx:y=1:2£¬Ôò¸ÃÑùÆ·ÖÐFe2O3µÄÎïÖʵÄÁ¿·ÖÊýΪ=33%£¬¹Ê´ð°¸Îª£º33%£»
£¨2£©100mL1.2mol/LBa(OH)2µÄÎïÖʵÄÁ¿Îª0.12mol£¬Éú³É15.76g¼´0.08mol°×É«³Áµí̼Ëá±µ£¬Èô¶þÑõ»¯Ì¼²»×㣬¸ù¾Ý̼ԪËØÊغ㣬²úÉúµÄ¶þÑõ»¯Ì¼ÊÇ0.08mol£¬´Ëʱ¸ù¾Ý»ìºÏ×ÜÖÊÁ¿¼ÆË㣬Éú³ÉµÄ¶þÑõ»¯Ì¼¿Ï¶¨´óÓÚ0.08mol£¬¹Ê¶þÑõ»¯Ì¼¹ýÁ¿£¬´ËʱÈÜÒºÖÐÈÜÖÊΪBa(HCO3)2£»ÓÉBa(OH)2+ CO2 = BaCO3¡ý+H2O¡¢BaCO3+ CO2+H2O = Ba(HCO3)2µÃ£¬¸ù¾Ý̼Ô×ÓÊغãµÃn(CO2)=0.08mol+(0.012mol-0.08mol)¡Á2=0.16mol£¬Áî»ìºÏÎïÖÐCuO¡¢Fe2O3µÄÎïÖʵÄÁ¿·Ö±ðΪxmol¡¢ymol£¬Ôò£ºx+3y=0.16£¬80x+160y=9.6£¬½âµÃx=0.04mol£¬y=0.04mol£¬ËùÒÔ»ìºÏÎïÖÐCuOºÍFe2O3µÄÎïÖʵÄÁ¿Ö®±ÈΪ1:1£¬¹Ê´ð°¸Îª£ºBa(HCO3)2£»1:1£»
£¨3£©³ÆÈ¡7.00gÑùÆ·£¬½«ÆäÈÜÓÚË®ÅäÖóÉ100mLÈÜÒº,·Ö³ÉÁ½µÈ·Ý£¬ÏòÆäÖÐÒ»·ÝÖмÓÈë×ãÁ¿NaOHÈÜÒº£¬¹ýÂËÏ´µÓµÃµ½1.07g³Áµí£¬Ó¦ÎªFe(OH)3£¬n(Fe(OH)3)==0.01mol£¬ÏòÁíÒ»·ÝÈÜÒºÖмÓÈë0.025molBa(NO3)2ÈÜÒº£¬Ç¡ºÃÍêÈ«·´Ó¦£¬Ôòn(SO42)=0.025mol£¬ËùÒÔ7.00gÑùÆ·Öк¬ÓÐFe2(SO4)30.01mol£¬ÆäÖꬵÄn(SO42)Ϊ0.05mol£¬Ôò(NH4)2SO4Ϊ0.05mol0.01¡Á3mol=0.02mol£¬Ôòm(H2O)=7.00g0.01mol¡Á400g/mol0.02mol¡Á132g/mol=0.72g£¬n(H2O)=
=0.04mol£¬n(Fe2(SO4)3):n((NH4)2SO4):n(H2O)=0.01:0.02:0.02=1:2:2£¬ËùÒÔ»¯Ñ§Ê½ÎªFe2£¨SO4£©32£¨NH4£©2SO42H2O£¬¹Ê´ð°¸Îª£ºFe2£¨SO4£©32£¨NH4£©2SO42H2O£»
£¨4£©ÂÈ»¯ÌúÈܽâÍÉú³ÉÂÈ»¯ÑÇÌúºÍÂÈ»¯Í£¬·´Ó¦ºóÈÜÒºÈÜÒºÖмÓÈëÌú·Û£¬Ê£Óà¹ÌÌåÖÊÁ¿Ð¡ÓÚFe·ÛµÄÖÊÁ¿£¬ÔòÈܽâCuºóÈÜÒºÖл¹ÓÐÊ£ÓàµÄFeCl3£¬¼ÓÈëÌú·ÛÏÈ·¢Éú£º2Fe3++Fe=3Fe2+£¬È»ºó·¢Éú£ºFe+Cu2+=Fe2++Cu£¬×îÖÕÓйÌÌåÊ£Ó࣬ÔòFe3+ÍêÈ«·´Ó¦£¬ÈÜÒºÖк¬ÓÐÈÜÖÊFeCl2£¬¼ÙÉè·´Ó¦ºóÈÜÒºÖÐûÓÐCuCl2£¬¸ù¾ÝÂÈÀë×ÓÊغã¿ÉÖªÈÜÒºÖÐFeCl2Ϊ=3.6mol£¬Ôò¼ÓÈëµÄFeӦΪ3.6mol0.8L¡Á3mol/L=1.2mol£¬¶øʵ¼Ê¼ÓÈëFeΪ
=1mol<1.2mol£¬¹ÊÈÜÒºÖÐÒ»¶¨º¬ÓÐCuCl2£¬¸ù¾ÝFeÔ×ÓÊغ㣬ÈÜÒºn(FeCl2)=n(FeCl3)+n(Fe)=0.8L¡Á3mol/L+1mol=3.4mol£¬ÔòÈÜÒºc(FeCl2)=
=4.25mol/L£¬¸ù¾ÝÂÈÀë×ÓÊغ㣬n(CuCl2)=
¡Á(0.8L¡Á3mol/L¡Á33.4mol¡Á2)=0.2mol£¬ÔòÈÜÒºc(CuCl2)=
=0.25mol/L£¬¹Ê´ð°¸Îª£ºFeCl2 4.25 mol/L£»CuCl2 0.25 mol/L¡£
![](http://thumb2018.1010pic.com/images/loading.gif)