A+B¡úX+Y+H2O£¨Î´Åäƽ£¬·´Ó¦Ìõ¼þÂÔÈ¥£©ÊÇÖÐѧ³£¼û·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¬ÆäÖÐA¡¢BµÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º4¡£Çë»Ø´ð£º
£¨1£©ÈôYÊÇ»ÆÂÌÉ«ÆøÌ壬Ôò¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ _____________________________£»½«YÓëµÈÎïµÄÁ¿µÄSO2³ä·Ö»ìºÏºóͨÈëÆ·ºìÈÜÒº£¬Î´¼ûÆäÍÊÉ«£¬Ô­ÒòÊÇ__________________________________________£¨ÇëÓû¯Ñ§·½³Ìʽ½âÊÍ˵Ã÷£©¡£
£¨2£©ÈôAΪ·Ç½ðÊôµ¥ÖÊ£¬¹¹³ÉËüµÄÔ­×ÓºËÍâ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄ2±¶£¬BµÄÈÜ
ҺΪijŨËᣬÔò·´Ó¦ËùµÃµÄÑõ»¯²úÎïµÄ»¯Ñ§Ê½Îª           ¡£
£¨3£©ÈôAΪ½ðÊôµ¥ÖÊ£¬³£ÎÂÏÂAÔÚBµÄŨÈÜÒºÖС°¶Û»¯¡±£¬¶øÇÒA¿ÉÈÜÓÚXÈÜÒºÖУº
¢Ù AÔªËصÄÔªËØ·ûºÅΪ            £¬ÔòAÓëBµÄ·´Ó¦ÖÐÑõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈÊÇ         ¡£
¢Ú Èôº¬a mol XµÄÈÜÒºÈܽâÁËÒ»¶¨Á¿Aºó£¬´ËʱÈÜÒºÖÐÁ½ÖÖ½ðÊôÑôÀë×ÓµÄÎïÖʵÄÁ¿Ç¡ºÃÏàµÈ£¬Ôò±»»¹Ô­µÄXµÄÎïÖʵÄÁ¿Îª         mol£¨Óú¬a´úÊýʽ±íʾ£©£»Îª±£Ö¤AÓëB³ä·Ö·´Ó¦ºó×îÖÕËùµÃµÄÈÜÒºÖÐͬʱº¬ÓÐÉÏÊöÁ½ÖÖ½ðÊôÑôÀë×Ó£¬Ôò´ËʱBÓëAµÄÖÊÁ¿±ÈÓ¦Âú×ãµÄÈ¡Öµ·¶Î§ÊÇ             ¡£


£¨1£©MnO2+4Cl£­+2H£«Mn2£«+Cl2¡ü+2H2O £¨2·Ö£©  SO2 + Cl2 + 2H2O = H2SO4 + 2HCl £¨2·Ö£©
£¨2£©CO2£¨1·Ö£©
£¨3£©¢ÙFe£¨1·Ö£©  1£º1 £¨1·Ö£©   ¢Ú0.4a£¨1·Ö £© 3/1 < m(B)/m(A) < 9/2 £¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©ÖÐѧ»¯Ñ§ÖлÆÂÌÉ«ÆøÌå¾ÍÊÇÖ¸ÂÈÆø¡£MnO2+4Cl£­+2H£«Mn2£«+Cl2¡ü+2H2O£»½«ÂÈÆøÓëµÈÎïµÄÁ¿µÄSO2³ä·Ö»ìºÏºóͨÈëÆ·ºìÈÜÒºÈÜÒº²»±äÉ«£¬ÊÇÒòΪ¶þÕß·´Ó¦Éú³ÉÁòËáºÍÑÎËᣬʧȥƯ°×ÐÔ¡£  
SO2 + Cl2 + 2H2O = H2SO4 + 2HCl £¨2·Ö£©
£¨2£©Ô­×ÓºËÍâ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄ2±¶£¬±íÃ÷AÊÇC£¬A¡¢BµÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º4£¬±íÃ÷BÒ»¶¨ÊÇŨÏõËá¡£Òò´Ë·´Ó¦ËùµÃµÄÑõ»¯²úÎïµÄ»¯Ñ§Ê½Îª£ºCO2£¨1·Ö£©
£¨3£©¢Ù½ðÊôµ¥Öʳ£ÎÂÏ·¢Éú¡°¶Û»¯¡±µÄÖ»ÓÐÌúºÍÂÁ£¬µ«ÊÇÖ»ÓÐÌúÄÜÈÜÓÚÈý¼ÛÌúÀë×ÓµÄÈÜÒº£¬Òò´ËAÔªËصÄÔªËØ·ûºÅΪ£ºFe£»Fe£«4HNO3£½Fe(NO3)3£«NO¡ü£«2H2O £»AÓëBµÄ·´Ó¦ÖÐÑõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈÊÇ1£º1
¢Ú     Fe  £«   2Fe3£«  £½   3Fe2£«
¿ªÊ¼£º           amol       0mol
ת»¯£º           bmol      1.5bmol
×îÖÕ£º         (a£­b)mol    1.5bmol
ÒÀÌâÒâÓУº(a£­b)mol£½1.5bmol
½âÖ®µÃ£ºb£½0.4a
Fe£«4HNO3(Ï¡)£½Fe(NO3)3£«NO¡ü£«2H2O£»3Fe£«8HNO3(Ï¡)£½3Fe(NO3)2£«2NO¡ü£«4H2O
56  4¡Á63                          3¡Á56   8¡Á63
ËùÒÔBÓëAµÄÖÊÁ¿±ÈÓ¦Âú×ãµÄÈ¡Öµ·¶Î§ÊÇ£º3/1 < m(B)/m(A) < 9/2 £¨2·Ö£©
¿¼µã£ºÑõ»¯»¹Ô­·´Ó¦µÄÓйØ֪ʶ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

A¡¢B¡¢C¡¢D¡¢E¡¢F¡¢X´æÔÚÈçͼËùʾת»»¹Øϵ£¬ÆäÖУ¬AÊÇÒ»ÖÖÕýÑΣ¬BÊÇÆø̬Ç⻯ÎCÊǵ¥ÖÊ£¬FÊÇÇ¿Ëá¡£X¿ÉÄÜÊÇÇ¿ËᣬҲ¿ÉÄÜÊÇÇ¿¼î¡£

£¨1£©AµÄ»¯Ñ§Ê½ÊÇ___________________¡£
£¨2£©ÈôXÊÇÇ¿Ëᣬ½«DÓëCl2ͬʱͨÈëË®Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________________¡£
£¨3£©ÈôXÊÇÇ¿¼î£¬¹ýÁ¿µÄB¸úCl2·´Ó¦³ýÉú³ÉCÍ⣬ÁíÒ»²úÎïÊÇÂÈ»¯Îï¡£
¢Ù¹ýÁ¿µÄB¸úCl2·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________________¡£
¢Ú¹¤ÒµÉú²úÖÐB¡úDµÄ»¯Ñ§·½³ÌʽΪ_______________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÖÐѧ³£¼û·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£ºA+B¡úX+Y+H2O£¨Î´Åäƽ£¬·´Ó¦Ìõ¼þÂÔÈ¥£©£¬ÆäÖÐA¡¢BµÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º4¡£Çë»Ø´ð£º
£¨1£©ÈôYÊÇ»ÆÂÌÉ«ÆøÌ壬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                                   ¡£
£¨2£©ÈôAΪ³£¼ûµÄ·Ç½ðÊôµ¥ÖÊ£¬BµÄÈÜҺΪijŨËᣬÆäÑõ»¯²úÎïµÄ½á¹¹Ê½Îª  ______      
£¨3£©ÈôAΪ½ðÊôµ¥ÖÊ£¬³£ÎÂÏÂAÔÚBµÄŨÈÜÒºÖС°¶Û»¯¡±£¬ÇÒA¿ÉÈÜÓÚXÈÜÒºÖС£º¬ a mol XµÄÈÜÒºÈܽâÁËÒ»¶¨Á¿Aºó£¬ÈôÈÜÒºÖÐÁ½ÖÖ½ðÊôÑôÀë×ÓµÄÎïÖʵÄÁ¿Ç¡ºÃÏàµÈ£¬Ôò±»»¹Ô­µÄXÊÇ   mol¡£
£¨4£©ÈôA¡¢B¡¢X¡¢Y¾ùΪ»¯ºÏÎï¡£ÏòAÈÜÒºÖмÓÈëÏõËáËữµÄAgNO3ÈÜÒº£¬²úÉú°×É«³Áµí£»BµÄÑæɫΪ»ÆÉ«¡£ÔòAÓëB°´ÎïÖʵÄÁ¿Ö®±È1£º4·´Ó¦ºó£¬ÈÜÒºÖÐÈÜÖʵĻ¯Ñ§Ê½Îª               ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÔÚÏÂÁÐÎïÖÊת»¯ÖУ¬AÊÇÒ»ÖÖÕýÑΣ¬DµÄÏà¶Ô·Ö×ÓÖÊÁ¿±ÈCµÄÏà¶Ô·Ö×ÓÖÊÁ¿´ó16£¬EÊÇË᣻µ±XÎÞÂÛÊÇÇ¿ËỹÊÇÇ¿¼îʱ£¬¶¼ÓÐÈçϵÄת»¯¹Øϵ£º

µ±XÊÇÇ¿ËáʱA¡¢B¡¢C¡¢D¡¢E¾ùº¬Í¬Ò»ÖÖÔªËØ£»µ±XÊÇÇ¿¼îʱ£¬A¡¢B¡¢C¡¢D¡¢E¾ùº¬ÁíÍâͬһÖÖÔªËØ¡££¨ÒÑÖªH2CO3¡¢H2S¡¢H2SiO3½ÔΪ¶þÔªÈõËᣩ
Çë»Ø´ð£º
£¨1£©AÊÇ________£¬YÊÇ________£¬ZÊÇ________¡£ £¨Ìѧʽ£¬ÏÂͬ£©
£¨2£©µ±XÊÇÇ¿Ëáʱ£¬EÊÇ________£¬Ð´³öBÉú³ÉCµÄ»¯Ñ§·½³Ìʽ£º___________________
£¨3£©µ±XÊÇÇ¿¼îʱ£¬EÊÇ________£¬Ð´³öBÉú³ÉCµÄ»¯Ñ§·½³Ìʽ£º__________________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ϱíΪԪËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬Çë²ÎÕÕÔªËØ¢ÙÒ»¢àÔÚ±íÖеÄλÖã¬Óû¯Ñ§ÓÃÓï»Ø´ðÏÂÁÐÎÊÌ⣺

¢ñ.£¨1£©Çë»­³öÔªËØ¢àµÄÒõÀë×ӽṹʾÒâͼ               ¡£
£¨2£©¢Ü¡¢¢Ý¡¢¢ßµÄÔ­×Ӱ뾶ÓÉСµ½´óµÄ˳ÐòΪ                 ¡£
£¨3£©¢ÝºÍ¢ÞµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ¼îÐÔÇ¿ÈõΪ         >       ¡£
£¨4£©¢Ü¡¢¢ÝÁ½ÖÖÔªËصÄÔ­×Ó°´1:1×é³ÉµÄ³£¼û»¯ºÏÎïµÄµç×ÓʽΪ            ¡£
¢ò.ÓɱíÖТÙÒ»¢àÖеÄÒ»ÖÖ»ò¼¸ÖÖÔªËØÐγɵij£¼ûÎïÖÊA¡¢B¡¢C¿É·¢ÉúÒÔÏ·´Ó¦£¨¸±²úÎïÒÑÂÔÈ¥£©£¬ÊԻشð£º

£¨1£©ÈôXÊÇÒ»ÖÖ³£¼û¹ý¶É½ðÊôµ¥ÖÊ£¬ÏòCµÄË®ÈÜÒºÖеμÓAgNO3ÈÜÒº£¬²úÉú²»ÈÜÓÚÏ¡HNO3µÄ°×É«³Áµí£¬¼ìÑé´ËCÈÜÒºÖнðÊôÀë×ӵķ½·¨ÊÇ                 £»ÓÖÖªÔÚËáÐÔÈÜÒºÖиýðÊôÀë×ÓÄܱ»Ë«ÑõË®Ñõ»¯£¬Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ                                      ¡£
£¨2£©ÈôA¡¢B¡¢CΪº¬ÓÐͬһ½ðÊôÔªËصÄÎÞ»ú»¯ºÏÎXΪǿµç½âÖÊ£¬AÈÜÒºÓëCÈÜÒº·´Ó¦Éú³ÉB£¬ÔòBµÄ»¯Ñ§Ê½Îª            £¬Ð´³öAÓëCÁ½ÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ____                                  ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

½«¼îÈÜÒºA¡¢ÑÎÈÜÒºB°´ÈçϳÌÐò½øÐÐʵÑ飬¸ù¾ÝÏÂÊöÏÖÏóÅжϣº

£¨1£©AµÄ»¯Ñ§Ê½                  BµÄ»¯Ñ§Ê½              ¡£  
£¨2£©Íê³ÉÏÂÁÐת»¯µÄ»¯Ñ§·½³Ìʽ£¬²¢Óá°µ¥ÏßÇÅ·¨¡±±êÃ÷µç×ÓתÒƵķ½ÏòºÍÊýÄ¿£º
D+H2£½H+P£º                                                    
£¨3£© д³öÏÂÁз´Ó¦µÄÀë×Ó·½³Ìʽ£º
A£«B£º                                                    
F£«ÑÎË᣺                                                    
£¨4£©ÑÎÈÜÒºBÖÐÒõÀë×ӵļìÑé·½·¨ÊÇ                                          

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Öû»·´Ó¦¿ÉÓÃÏÂͼ±íʾ£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Èô¼×Ϊ»ÆÂÌÉ«ÆøÌ壬µ¥ÖÊÒÒÄÜʹµí·ÛÈÜÒº±äÀ¶É«£¬Ôò¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ
                                                                               ¡£
£¨2£©Èô¹ýÁ¿µ¥Öʼ×Ó뻯ºÏÎïA·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2Al+Fe2O3Al2O3+2Fe£¬ ³ýÈ¥·´Ó¦ºó»ìºÏÎïÖÐÊ£ÓàµÄÂÁ·ÛÓëÉú³ÉµÄAl2O3ËùÓõÄÊÔ¼ÁÊÇ                      £¬Ð´³öÂÁ·ÛÓëËùÓÃÊÔ¼Á·´Ó¦µÄÀë×Ó·½³Ìʽ                                                        ¡£
£¨3£©¹¤ÒµÉÏÀûÓÃÖû»·´Ó¦Ô­ÀíÖƱ¸Ò»ÖÖ°ëµ¼Ìå²ÄÁÏ£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ
                                                                        ¡£
£¨4£©Èôµ¥ÖÊÒÒÊǺÚÉ«·Ç½ðÊô¹ÌÌåµ¥ÖÊ£¬»¯ºÏÎïBÊÇÓÅÖʵÄÄ͸ßβÄÁÏ£¬Ôò¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                                                               ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(13·Ö)ÒÑÖª£ºÕýÑÎAÇ¿Èȿɵõ½B¡¢C¡¢D¡¢EËÄÖÖÎïÖÊ£¬Bͨ³£Çé¿öÏÂΪÎÞÉ«ÎÞζҺÌ壬E¡¢F ÊÇ¿ÕÆøÖ÷Òª³É·Ö£¬DÄܲúÉúËáÓ꣬IΪºì×ØÉ«ÆøÌ壬CÓëJ·´Ó¦¿ÉµÃA£¬J¡¢KΪÁ½ÖÖ³£¼ûµÄËá¡£ÎïÖÊÖ®¼äµÄת»¯¹ØϵÈçͼËùʾ(ͼÖв¿·Ö·´Ó¦Îï»òÉú³ÉÎï¼°·´Ó¦Ìõ¼þδÁгö£©¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)EÎïÖʵĵç×ÓʽÊÇ________¡£
(2)¼ìÑéCµÄÊÔÖ½ÊÇ________£¬¼ìÑéDµÄÊÔ¼ÁÊÇ________(ÌîÊÔÖ½¡¢ÊÔ¼ÁÃû³Æ)¡£
(3)д³öAÇ¿ÈÈ·Ö½âÉú³ÉB¡¢C¡¢D¡¢EµÄ»¯Ñ§·½³Ìʽ________¡£
(4)д³öDͨÈËFeCl3ÈÜҺʱ£¬·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ_____ ¡£
(5) ¡ª¶¨Å¨¶ÈJ¡¢K»ìºÏºóµÄÏ¡ÈÜÒº200mL£¬Æ½¾ù·Ö³ÉÁ½·Ý¡£ÏòÆäÖÐÒ»·ÝÖÐÖð½¥¼ÓÈËÍ­·Û£¬×î¶àÄÜÈܽâa g(²úÉúÆøÌåֻΪG)¡£ÏòÁíÒ»·ÝÖÐÖð½¥¼ÓÈËÌú·Û£¬²úÉúÆøÌåµÄÁ¿ËæÌú·ÛÖÊÁ¿Ôö¼ÓµÄ±ä»¯ÈçͼËùʾ¡£Ôò¢Ùa£½________g£¬¢ÚÆøÌåG±ê×¼×´¿öÏÂÌå»ýΪ________£¬¢ÛJµÄÎïÖʵÄÁ¿Å¨¶ÈΪ______¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÓйØʵÑé²Ù×÷¡¢ÏÖÏóºÍ½âÊÍ»ò½áÂÛ¶¼ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

Ñ¡Ïî
ʵÑé²Ù×÷
ÏÖÏó
½âÊÍ»ò½áÂÛ
A
¹ýÁ¿µÄÌú·Û¼ÓÈëÏ¡ÏõËáÖУ¬³ä·Ö·´Ó¦ºó£¬µÎÈëKSCNÈÜÒº
ÈÜÒº³ÊºìÉ«
Ï¡ÏõËὫFeÑõ»¯ÎªFe2+
B
AlCl3ÈÜÒºÖеμӹýÁ¿µÄ°±Ë®ÈÜÒº
ÏȳöÏÖ°×É«³Áµí£¬ºó³ÁµíÓÖÖð½¥Èܽâ
ÇâÑõ»¯ÂÁÄÜÈÜÓÚ°±Ë®
C
ÂÁ²­²åÈëÏ¡ÏõËáÖÐ
ÎÞÃ÷ÏÔÏÖÏó
ÂÁ²­±íÃ汻ϡÏõËáÑõ»¯£¬ÐγÉÖÂÃܵÄÑõ»¯Ä¤
D
Óò£Á§°ôպȡŨÁòËáµãµ½pHÊÔÖ½ÉÏ
ÊÔÖ½±äºÚÉ«
ŨÁòËá¾ßÓÐÍÑË®ÐÔ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸