ÌâÄ¿ÄÚÈÝ

ÌúÊÇÈÕ³£Éú»îÖÐÓÃ;×î¹ã¡¢ÓÃÁ¿×î´óµÄ½ðÊô²ÄÁÏ¡£

£¨1£©³£ÎÂÏ£¬¿ÉÓÃÌúÖÊÈÝÆ÷ʢװŨÁòËáµÄÔ­ÒòÊÇ         ¡£

£¨2£©Ä³ÊµÑéС×éÀûÓÃÏÂͼװÖÃÑéÖ¤ÌúÓëË®ÕôÆøµÄ·´Ó¦¡£

¢ÙʪÃÞ»¨µÄ×÷ÓÃÊÇ      £¬ÊÔ¹ÜÖз´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ      ¡£

¢ÚʵÑé½áÊøºó£¬È¡³öÉÙÁ¿·´Ó¦ºóµÄ¹ÌÌåÓÚÊÔ¹ÜÖУ¬¼ÓÈë¹ýÁ¿ÑÎËᣬ

¹ÌÌåÍêÈ«Èܽ⣬ËùµÃÈÜÒºÖдæÔÚµÄÑôÀë×ÓÊÇ_____                   £¨ÌîÐòºÅ£©¡£

a£®Ò»¶¨ÓÐFe2+¡¢H+ºÍFe3+             b£®Ò»¶¨ÓÐFe2+¡¢H+£¬¿ÉÄÜÓÐFe3+   

c£®Ò»¶¨ÓÐFe2+¡¢Fe3+£¬¿ÉÄÜÓÐ H+      d£®Ò»¶¨ÓÐFe3+¡¢H+£¬¿ÉÄÜÓÐFe2+

£¨3£©Áí³ÆÈ¡Ò»¶¨Á¿µÄÌú¶¤·ÅÈëÊÊÁ¿µÄŨÁòËáÖУ¬¼ÓÈÈ£¬³ä·Ö·´Ó¦ºóÊÕ¼¯ÆøÌå¡£¾­²â¶¨ÆøÌåÖк¬ÓÐSO2¡¢CO2ºÍH2¡£

¢Ù ÌúÓëŨÁòËá·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ        ¡£

¢Ú ÆøÌåÖлìÓÐCO2µÄÔ­ÒòÊÇ£¨Óû¯Ñ§·½³Ìʽ±íʾ£©        ¡£

¢Û ½«672 mL£¨±ê×¼×´¿ö£©ÊÕ¼¯µ½µÄÆøÌåͨÈë×ãÁ¿äåË®ÖУ¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ                               £¬È»ºó¼ÓÈë×ãÁ¿BaCl2ÈÜÒº£¬¾­Ï´µÓ¡¢¸ÉÔïµÃµ½¹ÌÌå4.66 g¡£ÓÉ´ËÍÆÖªÊÕ¼¯µ½µÄÆøÌåÖÐSO2µÄÌå»ý·ÖÊýÊÇ        ¡£

 

¡¾´ð°¸¡¿

£¨1£© ŨÁòËáʹÌú±íÃæÐγÉÒ»²ãÖÂÃÜÎȶ¨µÄÑõ»¯Ä¤

£¨2£©¢ÙÌṩˮÕôÆø  3Fe+ 4H2O Fe3O4 + 4H2   ¢Ú b

£¨3£©¢Ù 2Fe + 6H2SO4(Ũ)Fe2(SO4)3 + 3SO2 + 6H2O

 ¢Ú C +2 H2SO4(Ũ)CO2 +2SO2 +2H2

¢Û SO2+Br2+2H2O==2Br- + SO42-+ 4H+    2/3

¡¾½âÎö¡¿£¨1£©³£ÎÂÏ£¬ÌúÔÚŨÁòËáÖз¢Éú¶Û»¯£¬Å¨ÁòËáʹÌú±íÃæÐγÉÒ»²ãÖÂÃÜÎȶ¨µÄÑõ»¯Ä¤£¬ËùÒÔ¿ÉÓÃÌúÖÊÈÝÆ÷ʢװŨÁòËá¡£

£¨2£©¢Ù¸ßÎÂÏ£¬ÌúºÍË®ÕôÆø·´Ó¦Éú³ÉÇâÆø£¬ËùÒÔʪÃÞ»¨µÄ×÷ÓÃÊÇÌṩˮÕôÆø£¬·´Ó¦µÄ·½³ÌʽÊÇ3Fe+ 4H2O Fe3O4 + 4H2 ¡£

¢ÚÓÉÓÚÔÚ·´Ó¦ÖÐÌúÊǹýÁ¿µÄ£¬ËùÒÔ·´Ó¦ºóµÄ¹ÌÌå»ìºÏÎïÖУ¬¼ÈÓÐËÄÑõ»¯ÈýÌú£¬Ò²ÓÐÌú¡£ÓÉÓÚÑÎËáÊǹýÁ¿µÄ£¬ËùÒÔÈÜÒºÖÐÒ»¶¨ÓÐFe2+¡¢H+£¬¿ÉÄÜÓÐFe3+£¬´ð°¸Ñ¡b¡£

£¨3£©¢ÙŨÁòËá¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬ËùÒÔÌúÓëŨÁòËá·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ

2Fe + 6H2SO4(Ũ)Fe2(SO4)3 + 3SO2 + 6H2O

¢ÚÓÉÓÚÌúÖл¹Ô­ÔÓÖÊ̼£¬¶ø̼ҲÄܺÍŨÁòËá·´Ó¦£¬·½³ÌʽÊÇC +2 H2SO4(Ũ)CO2 +2SO2 +2H2O¡£

¢ÛSO2ÄÜ°ÑäåË®Ñõ»¯£¬·´Ó¦µÄ·½³ÌʽÊÇ SO2+Br2+2H2O==2Br- + SO42-+ 4H+¡£4.66g°×É«³ÁµíÊÇÁòËá±µ£¬ËùÒÔÁòËá±µµÄÎïÖʵÄÁ¿ÊÇ0.02mol£¬ÔòSO2µÄÎïÖʵÄÁ¿Ò²ÊÇ0.02mol£¬»ìºÏÆøÊÇ0.03mol£¬Òò´ËSO2µÄÌå»ý·ÖÊýÊÇ2/3¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÌúÊÇÈÕ³£Éú»îÖÐÓÃ;×î¹ã¡¢ÓÃÁ¿×î´óµÄ½ðÊô²ÄÁÏ£®
£¨1£©³£ÎÂÏ£¬¿ÉÓÃÌúÖÊÈÝÆ÷ʢװŨÁòËáµÄÔ­ÒòÊÇ
ŨÁòËáʹÌú±íÃæÐγÉÒ»²ãÖÂÃÜÎȶ¨µÄÑõ»¯Ä¤
ŨÁòËáʹÌú±íÃæÐγÉÒ»²ãÖÂÃÜÎȶ¨µÄÑõ»¯Ä¤
£®
£¨2£©Ä³ÊµÑéС×éÀûÓÃÓÒͼװÖÃÑéÖ¤ÌúÓëË®ÕôÆøµÄ·´Ó¦£®
¢ÙʪÃÞ»¨µÄ×÷ÓÃÊÇ
ÌṩˮÕôÆø
ÌṩˮÕôÆø
£¬ÊÔ¹ÜÖз´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
3Fe+4H2O
 ¸ßΠ
.
 
Fe3O4+4H2
3Fe+4H2O
 ¸ßΠ
.
 
Fe3O4+4H2
£®
¢ÚʵÑé½áÊøºó£¬È¡³öÉÙÁ¿·´Ó¦ºóµÄ¹ÌÌåÓÚÊÔ¹ÜÖУ¬¼ÓÈë¹ýÁ¿ÑÎËᣬ¹ÌÌåÍêÈ«Èܽ⣬ËùµÃÈÜÒºÖдæÔÚµÄÑôÀë×ÓÊÇ
b
b
£¨ÌîÐòºÅ£©£®
a£®Ò»¶¨ÓÐFe2+¡¢H+ºÍFe3+             b£®Ò»¶¨ÓÐFe2+¡¢H+£¬¿ÉÄÜÓÐFe3+
c£®Ò»¶¨ÓÐFe2+¡¢Fe3+£¬¿ÉÄÜÓРH+      d£®Ò»¶¨ÓÐFe3+¡¢H+£¬¿ÉÄÜÓÐFe2+
£¨3£©Áí³ÆÈ¡Ò»¶¨Á¿µÄÌú¶¤·ÅÈëÊÊÁ¿µÄŨÁòËáÖУ¬¼ÓÈÈ£¬³ä·Ö·´Ó¦ºóÊÕ¼¯ÆøÌ壮¾­²â¶¨ÆøÌåÖк¬ÓÐSO2¡¢CO2ºÍH2£®
¢ÙÌúÓëŨÁòËá·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
2Fe+6H2SO4£¨Å¨£©
  ¡÷  
.
 
Fe2£¨SO4£©3+3SO2¡ü+6H2O
2Fe+6H2SO4£¨Å¨£©
  ¡÷  
.
 
Fe2£¨SO4£©3+3SO2¡ü+6H2O
£®
¢ÚÆøÌåÖлìÓÐCO2µÄÔ­ÒòÊÇ£¨Óû¯Ñ§·½³Ìʽ±íʾ£©
C+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CO2+2SO2+2H2O
C+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CO2+2SO2+2H2O
£®
¢Û½«672mL£¨±ê×¼×´¿ö£©ÊÕ¼¯µ½µÄÆøÌåͨÈë×ãÁ¿äåË®ÖУ¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ
SO2+Br2+2H2O¨T2Br-+SO42-+4H+
SO2+Br2+2H2O¨T2Br-+SO42-+4H+
£¬È»ºó¼ÓÈë×ãÁ¿BaCl2ÈÜÒº£¬¾­Ï´µÓ¡¢¸ÉÔïµÃµ½¹ÌÌå4.66g£®ÓÉ´ËÍÆÖªÊÕ¼¯µ½µÄÆøÌåÖÐSO2µÄÌå»ý·ÖÊýÊÇ
2
3
2
3
£®

ÌúÊÇÈÕ³£Éú»îÖÐÓÃ;×î¹ã¡¢ÓÃÁ¿×î´óµÄ½ðÊô²ÄÁÏ¡£

£¨1£©³£ÎÂÏ£¬¿ÉÓÃÌúÖÊÈÝÆ÷ʢװŨÁòËáµÄÔ­ÒòÊÇ        ¡£

£¨2£©Ä³ÊµÑéС×éÀûÓÃÓÒͼװÖÃÑéÖ¤ÌúÓëË®ÕôÆøµÄ·´Ó¦¡£

¢ÙʪÃÞ»¨µÄ×÷ÓÃÊÇ      £¬ÊÔ¹ÜÖз´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ      ¡£

¢ÚʵÑé½áÊøºó£¬È¡³öÉÙÁ¿·´Ó¦ºóµÄ¹ÌÌåÓÚÊÔ¹ÜÖУ¬¼ÓÈë¹ýÁ¿ÑÎËᣬ¹ÌÌåÍêÈ«Èܽ⣬ËùµÃÈÜÒºÖдæÔÚµÄÑôÀë×ÓÊÇ_____£¨ÌîÐòºÅ£©¡£

a£®Ò»¶¨ÓÐFe2+¡¢H+ºÍFe3+             b£®Ò»¶¨ÓÐFe2+¡¢H+£¬¿ÉÄÜÓÐFe3+   

c£®Ò»¶¨ÓÐFe2+¡¢Fe3+£¬¿ÉÄÜÓÐ H+       d£®Ò»¶¨ÓÐFe3+¡¢H+£¬¿ÉÄÜÓÐFe2+

£¨3£©Áí³ÆÈ¡Ò»¶¨Á¿µÄÌú¶¤·ÅÈëÊÊÁ¿µÄŨÁòËáÖУ¬¼ÓÈÈ£¬³ä·Ö·´Ó¦ºóÊÕ¼¯ÆøÌå¡£¾­²â¶¨ÆøÌåÖк¬ÓÐSO2¡¢CO2ºÍH2¡£

¢Ù ÌúÓëŨÁòËá·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ        ¡£

¢Ú ÆøÌåÖлìÓÐCO2µÄÔ­ÒòÊÇ£¨Óû¯Ñ§·½³Ìʽ±íʾ£©        ¡£

¢Û ½«672 mL£¨±ê×¼×´¿ö£©ÊÕ¼¯µ½µÄÆøÌåͨÈë×ãÁ¿äåË®ÖУ¬·¢Éú·´Ó¦£º

SO2 + Br2 + 2H2O = 2HBr+ H2SO4£¬È»ºó¼ÓÈë×ãÁ¿BaCl2ÈÜÒº£¬¾­Ï´µÓ¡¢¸ÉÔïµÃµ½¹ÌÌå4.66 g¡£ÓÉ´ËÍÆÖªÊÕ¼¯µ½µÄÆøÌåÖÐSO2µÄÌå»ý·ÖÊýÊÇ       ¡£

 

ÌúÊÇÈÕ³£Éú»îÖÐÓÃ;×î¹ã¡¢ÓÃÁ¿×î´óµÄ½ðÊô²ÄÁÏ¡£
£¨1£©³£ÎÂÏ£¬¿ÉÓÃÌúÖÊÈÝÆ÷ʢװŨÁòËáµÄÔ­ÒòÊÇ        ¡£
£¨2£©Ä³ÊµÑéС×éÀûÓÃÏÂͼװÖÃÑéÖ¤ÌúÓëË®ÕôÆøµÄ·´Ó¦¡£

¢ÙʪÃÞ»¨µÄ×÷ÓÃÊÇ     £¬ÊÔ¹ÜÖз´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ     ¡£
¢ÚʵÑé½áÊøºó£¬È¡³öÉÙÁ¿·´Ó¦ºóµÄ¹ÌÌåÓÚÊÔ¹ÜÖУ¬¼ÓÈë¹ýÁ¿ÑÎËᣬ
¹ÌÌåÍêÈ«Èܽ⣬ËùµÃÈÜÒºÖдæÔÚµÄÑôÀë×ÓÊÇ_____                  £¨ÌîÐòºÅ£©¡£
a£®Ò»¶¨ÓÐFe2+¡¢H+ºÍFe3+             b£®Ò»¶¨ÓÐFe2+¡¢H+£¬¿ÉÄÜÓÐFe3+   
c£®Ò»¶¨ÓÐFe2+¡¢Fe3+£¬¿ÉÄÜÓÐ H+      d£®Ò»¶¨ÓÐFe3+¡¢H+£¬¿ÉÄÜÓÐFe2+
£¨3£©Áí³ÆÈ¡Ò»¶¨Á¿µÄÌú¶¤·ÅÈëÊÊÁ¿µÄŨÁòËáÖУ¬¼ÓÈÈ£¬³ä·Ö·´Ó¦ºóÊÕ¼¯ÆøÌå¡£¾­²â¶¨ÆøÌåÖк¬ÓÐSO2¡¢CO2ºÍH2¡£
¢ÙÌúÓëŨÁòËá·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ       ¡£
¢ÚÆøÌåÖлìÓÐCO2µÄÔ­ÒòÊÇ£¨Óû¯Ñ§·½³Ìʽ±íʾ£©       ¡£
¢Û½«672 mL£¨±ê×¼×´¿ö£©ÊÕ¼¯µ½µÄÆøÌåͨÈë×ãÁ¿äåË®ÖУ¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ                              £¬È»ºó¼ÓÈë×ãÁ¿BaCl2ÈÜÒº£¬¾­Ï´µÓ¡¢¸ÉÔïµÃµ½¹ÌÌå4.66 g¡£ÓÉ´ËÍÆÖªÊÕ¼¯µ½µÄÆøÌåÖÐSO2µÄÌå»ý·ÖÊýÊÇ       ¡£

ÌúÊÇÈÕ³£Éú»îÖÐÓÃ;×î¹ã¡¢ÓÃÁ¿×î´óµÄ½ðÊô²ÄÁÏ¡£

(1)³£ÎÂÏ£¬¿ÉÓÃÌúÖÊÈÝÆ÷ʢװŨÁòËáµÄÔ­ÒòÊÇ                                   ¡£

£¨2£©Ä³ÊµÑéС×éÀûÓÃÓÒͼװÖÃÑéÖ¤ÌúÓëË®ÕôÆøµÄ·´Ó¦¡£

¢ÙʪÃÞ»¨µÄ×÷ÓÃÊÇ      £¬ÊÔ¹ÜÖз´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                                  ¡£

¢ÚʵÑé½áÊøºó£¬È¡³öÉÙÁ¿·´Ó¦ºóµÄ¹ÌÌåÓÚÊÔ¹ÜÖУ¬¼ÓÈë¹ýÁ¿ÑÎËᣬ¹ÌÌåÍêÈ«Èܽ⣬ËùµÃÈÜÒºÖдæÔÚµÄÑôÀë×ÓÊÇ            £¨ÌîÐòºÅ£©¡£

a£®Ò»¶¨ÓÐFe2+¡¢H+ºÍFe3+          b£®Ò»¶¨ÓÐFe2+¡¢H+£¬¿ÉÄÜÓÐFe3+   

c£®Ò»¶¨ÓÐFe2+¡¢Fe3+£¬¿ÉÄÜÓÐ H+    d£®Ò»¶¨ÓÐFe3+¡¢H+£¬¿ÉÄÜÓÐFe2+

£¨3£©Áí³ÆÈ¡Ò»¶¨Á¿µÄÌú¶¤·ÅÈëÊÊÁ¿µÄŨÁòËáÖУ¬¼ÓÈÈ£¬³ä·Ö·´Ó¦ºóÊÕ¼¯ÆøÌå¡£¾­²â¶¨ÆøÌåÖк¬ÓÐSO2¡¢CO2ºÍH2¡£

¢Ù ÌúÓëŨÁòËá·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                                              ¡£

¢Ú ÆøÌåÖлìÓÐCO2µÄÔ­ÒòÊÇ£¨Óû¯Ñ§·½³Ìʽ±íʾ£©                                  ¡£

¢Û ½«672 mL£¨±ê×¼×´¿ö£©ÊÕ¼¯µ½µÄÆøÌåͨÈë×ãÁ¿äåË®ÖУ¬·¢Éú·´Ó¦£º

SO2 + Br2 + 2H2O = 2HBr + H2SO4£¬È»ºó¼ÓÈë×ãÁ¿BaCl2ÈÜÒº£¬¾­Ï´µÓ¡¢¸ÉÔïµÃµ½¹ÌÌå4.66 g¡£ÓÉ´ËÍÆÖªÊÕ¼¯µ½µÄÆøÌåÖÐSO2µÄÌå»ý·ÖÊýÊÇ        ¡£

 

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

¾«Ó¢¼Ò½ÌÍø