ÌâÄ¿ÄÚÈÝ

ij¹¤³§ÓÃÈíÃ̿󣨺¬MnO2Ô¼70%¼°Al2 O3£©ºÍÉÁп¿ó£¨º¬ZnSÔ¼80%¼°FeS£©¹²Í¬Éú²úMnO2£¬ºÍZn£¨¸Éµç³ØÔ­ÁÏ£©¡£Á÷³ÌÈçÏ£º

ÒÑÖª£º¢ÙAÊÇMnSO4¡¢ZnSO4¡¢Fe2(SO4)3£¬Al2(SO4)3µÄ»ìºÏÒº¡£
¢ÚIVÖеĵç½â·´Ó¦Ê½ÎªMnSO4+ZnSO4+2H2O    MnO2+Zn +2H2SO4¡£
£¨1£©AÖÐÊôÓÚ»¹Ô­²úÎïµÄÊÇ             ¡£
£¨2£©¼ÓÈëMnCO3¡¢Zn2(OH)2CO3µÄ×÷ÓÃÊÇ         £ºCµÄ»¯Ñ§Ê½ÊÇ           ¡£
£¨3£©¸ÃÉú²úÖгýµÃµ½Na2SO4¡¢SµÈ¸±²úÆ·Í⣬»¹¿ÉµÃµ½µÄ¸±²úÆ·ÊÇ            ¡£
£¨4£©¸±²úÆ·S¿ÉÓÃÓÚÖÆÁòËᣬת»¯¹ý³ÌÊÇ£ºS¡úSO2¡úSO3¡úH2SO4¡£Ð´³öµÚ¶þ²½×ª»¯µÄ»¯Ñ§·½³Ìʽ           ¡£
£¨5£©Òª´ÓNa2SO4ÈÜÒºÖеõ½Ã¢Ïõ£¨ Na2SO4£®10H2O£©£¬Ðè½øÐеIJÙ×÷ÓÐÕô·¢Å¨Ëõ¡¢          ¡¢
¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔïµÈ¡£
£¨6£©´ÓÉú²úMnO2ºÍZnµÄ½Ç¶È¼ÆË㣬ÈíÃÌ¿óºÍÉÁп¿óͶÁϵÄÖÊÁ¿±È´óÔ¼ÊÇ          ¡£
£¨1£©MnSO4
(2)Ôö´óÈÜÒºµÄp H,ʹFe3+ºÍAl3+Éú³É³Áµí  H2SO4
£¨3£©Fe2O3¡¢Al2O3
(4)   2SO2+O22SO3
(5) ÀäÈ´½á¾§
£¨6£©1£º1(»ò1.03:1)

ÊÔÌâ·ÖÎö£º±È½ÏÐÅÏ¢1¿ÉÖªMn»¯ºÏ¼Û½µµÍ£¬ÊôÓÚ»¹Ô­²úÎïµÄΪMnSO4£¬Óɹ¤ÒÕÁ÷³Ì·ÖÎö¼ÓÈëMnCO3¡¢Zn2(OH)2CO3µÄ×÷ÓÃÊÇÔö´óÈÜÒºµÄp H,ʹFe3+ºÍAl3+Éú³É³Áµí£»ÇÒCµÄ»¯Ñ§Ê½ÎªH2SO4£»²»ÄѵóöµÃµ½µÄ¸±²úÆ·»¹ÓÐFe2O3¡¢Al2O3£»¼ÙÉèÈíÃÌ¿óºÍÉÁп¿óµÄÖÊÁ¿·Ö±ðΪx ¡¢y£¬¸ù¾Ý¢Ú¿ÉÖªMnO2ºÍZnµÄÎïÖʵÄÁ¿Ö®±ÈΪ1:1£¬Ôò0.7x g/87g/mol:0.8y g/97g/mol=1:1,½âµÃx/y=1.03:1.
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨16·Ö£©ÁòËáǦ£¨PbSO4£©¹ã·ºÓ¦ÓÃÓÚÖÆÔìǦÐîµç³Ø¡¢°×É«ÑÕÁϵȡ£ÀûÓ÷½Ç¦¿ó¾«¿ó£¨PbS£©Ö±½ÓÖƱ¸ÁòËáǦ·ÛÄ©µÄÁ÷³ÌÈçÏ£º

ÒÑÖª£º£¨¢¡£©PbCl2£¨s£©+2Cl£­(aq)PbCl4-(aq) ¡÷H£¾0
£¨¢¢£©Ksp(PbSO4)=1.08¡Á10-8, Ksp(PbCl2)=1.6¡Á10-5
£¨¢££©Fe3£«¡¢Pb2£«ÒÔÇâÑõ»¯ÎïÐÎʽÍêÈ«³Áµíʱ£¬ÈÜÒºµÄPHÖµ·Ö±ðΪ3.2¡¢7.04
£¨1£©²½Öè¢ñÖÐÉú³ÉPbCl2ºÍSµÄÀë×Ó·½³Ìʽ                               £¬¼ÓÈëÑÎËáµÄÁíÒ»¸öÄ¿µÄÊÇΪÁË¿ØÖÆPHÖµÔÚ0.5¡«1.0£¬Ô­ÒòÊÇ                                     ¡£
£¨2£©Óû¯Ñ§Æ½ºâÒƶ¯µÄÔ­Àí½âÊͲ½Öè¢òÖÐʹÓñùˮԡµÄÔ­Òò                       ¡£
£¨3£©Ð´³öPbCl2¾§Ìåת»¯ÎªPbSO4³ÁµíµÄÀë×Ó·½³Ìʽ                           ¡£
£¨4£©ÇëÓÃÀë×Ó·½³Ìʽ½âÊÍÂËÒº2¼ÓÈëH2O2¿ÉÑ­»·ÀûÓõÄÔ­Òò                         £¬ÂËÒº3ÊÇ               ¡£
£¨5£©Ç¦Ðîµç³ØµÄµç½âÒºÊÇÁòËᣬ³äµçºóÁ½¸öµç¼«ÉϳÁ»ýµÄ  PbSO4·Ö±ðת»¯ÎªPbO2ºÍPb£¬³äµçʱÒõ¼«µÄµç¼«·´Ó¦Ê½Îª                      ¡£
°´ÒªÇóÌî¿Õ
£¨Ò»£©¡¢X¡¢YºÍZ¾ùΪ¶ÌÖÜÆÚÔªËØ£¬Ô­×ÓÐòÊýÒÀ´ÎÔö´ó£¬XµÄµ¥ÖÊΪÃܶÈ×îСµÄÆøÌ壬YµÄÒ»ÖÖµ¥ÖʾßÓÐÌØÊâ³ô棬ZÓëXÔ­×Ó×î´¦²ãµç×ÓÊýÏàͬ¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÉÉÏÊöÔªËØ×é³ÉµÄ»¯ºÏÎïÖУ¬¼Èº¬Óм«ÐÔ¹²¼Û¼üÓÖº¬ÓÐÀë×Ó¼üµÄ»¯ºÏÎïµÄµç×Óʽ                        £»
£¨2£©XºÍY×é³ÉµÄ»¯ºÏÎïÖУ¬ÓÐÒ»ÖּȺ¬Óм«ÐÔ¹²¼Û¼üÓÖº¬ÓзǼ«ÐÔ¹²¼Û¼ü¡£´Ë»¯ºÏÎï¿É½«¼îÐÔ¹¤Òµ·ÏË®ÖеÄCN¡¥Ñõ»¯ÎªÌ¼ËáÑκͰ±£¬ÏàÓ¦µÄÀë×Ó·½³ÌʽΪ           
£¨¶þ£©¡¢ÔÚÒ»¶¨Ìõ¼þÏ£¬RO3n¡¥ºÍI¡¥·¢Éú·´Ó¦£¬Àë×Ó·½³ÌʽΪ£º RO3n¡¥+6I¡¥+6H+==R¡¥+3I2+3H2
RO3n¡¥-ÖÐRÔªËصĻ¯ºÏ¼ÛΪ       £¬RÔªËصÄÔ­×Ó×îÍâ²ãµç×ÓÓР      ¸ö¡£
£¨Èý£©¡¢Na2SxÔÚ¼îÐÔÈÜÒºÖпɱ»NaClOÑõ»¯ÎªNa2SO4£¬¶øNaClO±»»¹Ô­ÎªNaCl£¬Èô·´Ó¦ÖÐNa2SxÓëNaClOµÄÎïÖʵÄÁ¿Ö®±ÈΪ1©U16£¬ÔòxÖµÊÇ           
£¨ËÄ£©¡¢ÒÑÖªM2On2¡¥¿ÉÓëR2¡¥×÷Óã¬R2¡¥±»Ñõ»¯ÎªRµÄµ¥ÖÊ£¬M2On2¡¥µÄ»¹Ô­²úÎïÖУ¬MΪ+3¼Û£¬ÓÖÖªc(M2On2¡¥)=0£®3mol/LµÄÈÜÒº100mL¿ÉÓëc(R2¡¥)=0£®6mol/LµÄÈÜÒº150mLÇ¡ºÃÍêÈ«·´Ó¦£¬ÔònֵΪ            ¡£

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

¾«Ó¢¼Ò½ÌÍø