£¨12·Ö£©Í¬Ñ§ÃÇÓÃNa2CO3ÈÜÒººÍŨHClÀ´Ñо¿¼òÒ×Ãð»ðÆ÷µÄ·´Ó¦Ô­Àíʱ£¬¶Ô·ÏÒºµÄ³É·Ö½øÐÐ̽¾¿¡£ÉÏÊöÁ½ÖÖÎïÖÊ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                  £¬ÓÉ´ËÍƲâ³ö·ÏÒºÖÐÒ»¶¨ÓÐNaCl£¬¿ÉÄÜÓÐNa2CO3»òÑÎËá¡£

[ʵÑé̽¾¿]

¢åÈ·¶¨·ÏÒºÖÐÊÇ·ñº¬ÓÐÑÎËá

¢ÅÑ¡ÔñÊÔ¼Á£º¸ù¾ÝÑÎËáµÄ»¯Ñ§ÐÔÖÊ£¬Í¬Ñ§ÃÇÑ¡ÓÃÁËÈçÓÒͼËùʾµÄÎåÖÖÎïÖÊ£¬ÎïÖÊxÊÇËá¼îָʾ¼ÁÖеĠ         ÈÜÒº¡£

¢ÆʵÑéÑéÖ¤£ºÄ³Í¬Ñ§Ïò·ÏÒºÖмÓÈëÉÙÁ¿µÄþ·Û£¬¹Û²ìµ½                   £¬È·¶¨·ÏÒºÖÐÒ»¶¨Ã»ÓÐÑÎËá¡£

(3) YÊÇÒ»ÖÖºì×ØÉ«µÄ¹ÌÌå·ÛÄ©,ÓëÑÎËá·´Ó¦ºóÈÜÒº³Ê»ÆÉ«£¬Ð´³öÆäÓëÑÎËá·´Ó¦µÄ·½³Ìʽ                          ¡£

¢æÈ·¶¨·ÏÒºÖÐÊÇ·ñº¬ÓÐNa2CO3

ijͬѧѡÓà              ²â³ö·ÏÒºµÄpH=l0£¬È·¶¨·ÏÒºÖÐÒ»¶¨º¬ÓÐNa2CO3¡£

¢ç´¦Àí·ÏÒº£¬»ØÊÕÀûÓÃ

Óû´Ó·ÏÒºÖеõ½´¿¾»µÄNaCl£¬ÇëÍê³ÉÈçÏÂʵÑé·½°¸Éè¼Æ¡£

·½°¸

¼ÓÈëÊÔ¼Á

·ÖÀë·½·¨

·½°¸ÆÀ¼Û

1

ÊÊÁ¿Ca(NO3)2ÈÜÒº

¡¢Õô·¢½á¾§

²»¿ÉÐУ¬ÀíÓÉÊÇ

2

¹ýÁ¿µÄ

Õô·¢½á¾§

¿ÉÐÐ

·½°¸1Öз¢Éú·´Ó¦µÄ·½³ÌʽΪ                                    ¡£

 

¡¾´ð°¸¡¿

Na2CO3+2HCl = 2NaCl+H2O+CO2¡ü      ʯÈï      ÎÞÆøÅݲúÉú(»òÎÞÏÖÏó»òÎޱ仯) Fe2O3 + 6HCl = 2FeCl3 +  H2 O

·½°¸

¼ÓÈëÊÔ¼Á

·ÖÀë·½·¨

·½°¸ÆÀ¼Û

Ò»

 

¹ýÂË

Òý½øÐÂÔÓÖÊ(»òÓÐNaNO3Éú³É»òÒý½øNO3¡ª¸ùÀë×Ó)

¶þ

ÑÎËá (»òÏ¡HCl»òHCl)

 

 

Na2CO3+ Ca(NO3)2= 2NaNO3+ Ca CO3¡ý

¡¾½âÎö¡¿Na2CO3ÈÜÒººÍŨHCl·´Ó¦·½³ÌʽΪ£ºNa2CO3+2HCl = 2NaCl+H2O+CO2¡ü

¢å¡¢¢Å¡¢ÄÜÅжÏÈÜÒºµÄËá¼îÐÔµÄָʾ¼Á³£ÓÃ×ÏɫʯÈïÊÔÒº£¬ÒòΪÓöËá±äºì£¬Óö¼î±äÀ¶¡£

¢Æ¡¢ÈÜÒºÖÐÈçÓÐÏ¡ÑÎËᣬϡÑÎËá»áÓëþ·´Ó¦²úÉúÆøÌå¡£Òò´Ë£¬ÎÞÆøÅݲúÉú²Å»á˵Ã÷ÎÞÏ¡ÑÎËá¡£(3)ºìºÖÉ«¹ÌÌåÊÇÑõ»¯Ìú£¬»ÆÉ«ÈÜÒºÊÇÂÈ»¯ÌúÈÜÒº£¬·´Ó¦µÄ·½³ÌʽΪ£ºFe2O3 + 6HCl = 2FeCl3 +  H2 O

¢æ¡¢²â¶¨ÈÜÒºµÄPHÖµ£¬ÊµÑéÊÒ³£ÓÃPHÊÔÖ½¡£

¢ç¡¢±¾ÌâÖ÷ÒªÊdzýÈ¥NaClÖеÄNa2CO3¡£¼ÓµÄÊÔ¼ÁÓë̼ËáÄÆ·´Ó¦£¬Í¬Ê±²»ÄܲúÉúеÄÔÓÖÊ

·½°¸

¼ÓÈëÊÔ¼Á

·ÖÀë·½·¨

·½°¸ÆÀ¼Û

Ò»

 

¹ýÂË

Òý½øÐÂÔÓÖÊ(»òÓÐNaNO3Éú³É»òÒý½øNO3¡ª¸ùÀë×Ó)

¶þ

ÑÎËá (»òÏ¡HCl»òHCl)

 

 

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2011?¼ªÁÖ£©Í¬Ñ§ÃÇÓÃNa2CO3ÈÜÒººÍŨHClÀ´Ñо¿¼òÒ×Ãð»ðÆ÷µÄ·´Ó¦Ô­Àíʱ£¬¶Ô·ÏÒºµÄ³É·Ö½øÐÐ̽¾¿£®
[ÍÆÀí¼ÙÉè]
ÉÏÊöÁ½ÖÖÎïÖÊ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Na2CO3+2HCl=2NaCl+H2O+CO2¡ü
Na2CO3+2HCl=2NaCl+H2O+CO2¡ü
£¬ÓÉ´ËÍƲâ³ö·ÏÒºÖÐÒ»¶¨ÓÐNaCl£¬¿ÉÄÜÓÐNa2CO3»òÑÎËᣮ
[ʵÑé̽¾¿]
£¨Ò»£©È·¶¨·ÏÒºÖÐÊÇ·ñº¬ÓÐÑÎË᣺
£¨1£©Ñ¡ÔñÊÔ¼Á£º¸ù¾ÝÑÎËáµÄ»¯Ñ§ÐÔÖÊ£¬Í¬Ñ§ÃÇÑ¡ÓÃÁËÈçͼËùʾµÄÎåÖÖÎïÖÊ£¬ÆäÖÐÎïÖÊxÊÇËá¼îָʾ¼ÁÖеÄ
ʯÈï
ʯÈï
ÈÜÒº£®
£¨2£©ÊµÑéÑéÖ¤£ºÄ³Í¬Ñ§Ïò·ÏÒºÖмÓÈëÉÙÁ¿µÄþ·Û£¬¹Û²ìµ½È·¶¨·ÏÒºÖÐÒ»¶¨Ã»ÓÐÑÎËᣮ
£¨¶þ£©È·¶¨·ÏÒºÖÐÊÇ·ñº¬ÓÐNa2CO3ijͬѧѡÓÃ
pHÊÔÖ½£¨»òpH¼Æ£©
pHÊÔÖ½£¨»òpH¼Æ£©
²â³ö·ÏÒºµÄpH=l0£¬È·¶¨·ÏÒºÖÐÒ»¶¨º¬ÓÐNa2CO3£®
£¨Èý£©´¦Àí·ÏÒº£¬»ØÊÕÀûÓÃÓû´Ó·ÏÒºÖеõ½´¿¾»µÄNaCl£¬ÇëÍê³ÉÈçÏÂʵÑé·½°¸Éè¼Æ£®
·½°¸¼ÓÈëÊÔ¼Á·ÖÀë·½·¨·½°¸ÆÀ¼Û
ÊÊÁ¿Ca£¨NO3£©2ÈÜÒº¹ýÂË¡¢Õô·¢½á¾§²»¿ÉÐУ¬ÀíÓÉÊÇ
Òý½øÐÂÔÓÖÊ£¨»òÓÐNaNO3£¬Éú³É»òÒý½øNO¸ùÀë×Ó£©
Òý½øÐÂÔÓÖÊ£¨»òÓÐNaNO3£¬Éú³É»òÒý½øNO¸ùÀë×Ó£©
¹ýÁ¿µÄ
ÑÎËᣨ»òÏ¡HCl»òHCl£©
ÑÎËᣨ»òÏ¡HCl»òHCl£©
Õô·¢½á¾§¿ÉÐÐ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ͬѧÃÇÓÃNa2CO3ÈÜÒººÍŨÑÎËáÑо¿¼òÒ×Ãð»ðÆ÷µÄ·´Ó¦Ô­Àíʱ£¬¶Ô·ÏÒº³É·Ö½øÐÐ̽¾¿£®
¡¾ÍÆÀí¼ÙÉè¡¿ÓÉ´ËÍƲâ³ö·ÏÒºÖÐÒ»¶¨ÓÐNaCl£¬¿ÉÄÜÓÐNa2CO3»òÑÎËᣮ
¡¾ÊµÑé̽¾¿¡¿£¨1£©È·¶¨·ÏÒºÖÐÊÇ·ñº¬ÓÐÑÎËᣮ
¢ÙÑ¡ÔñÊÔ¼Á£º¸ù¾ÝÑÎËáµÄ»¯Ñ§ÐÔÖÊ£¬Í¬Ñ§ÃÇÑ¡ÓÃÁËÈçͼËùʾµÄÎåÖÖÎïÖÊ£¬ÆäÖÐÎïÖÊXÊÇËá¼îָʾ¼ÁÖеÄ
ʯÈï
ʯÈï
ÈÜÒº£®
¢ÚʵÑéÑéÖ¤£ºÄ³Í¬Ñ§Ïò·ÏÒºÖмÓÈëÉÙÁ¿µÄþ·Û£¬¹Û²ìµ½
ÎÞÆøÅݲúÉú
ÎÞÆøÅݲúÉú
£¬È·¶¨·ÏÒºÖÐÒ»¶¨Ã»ÓÐÑÎËᣮ
£¨2£©È·¶¨·ÏÒºÖÐÊÇ·ñº¬ÓÐNa2CO3£®Ä³Í¬Ñ§Ñ¡ÓÃ
pHÊÔÖ½
pHÊÔÖ½
²â³ö·ÏÒºµÄpH=10£¬È·¶¨·ÏÒºÖÐÒ»¶¨º¬ÓÐNa2CO3£®
£¨3£©´¦Àí·ÏÒº£¬»ØÊÕÀûÓã®Óû´Ó·ÏÒºÖеõ½´¿¾»µÄNaCl£¬ÇëÍê³ÉÈçÏÂʵÑé·½°¸Éè¼Æ£®
·½°¸ ¼ÓÈëÊÔ¼Á ·ÖÀë·½·¨ ·½°¸ÆÀ¼Û
¢ñ ÊÊÁ¿Ca£¨NO3£©2ÈÜÒº ¹ýÂË¡¢Õô·¢½á¾§ ²»¿ÉÐУ¬ÀíÓÉÊÇ
ÒýÈëÁËÐÂÔÓÖÊÏõËáÄÆ
ÒýÈëÁËÐÂÔÓÖÊÏõËáÄÆ
¢ò ¹ýÁ¿µÄ
Ï¡ÑÎËá
Ï¡ÑÎËá
Õô·¢½á¾§ ¿ÉÐÐ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2012?°ÍÖУ©Í¬Ñ§ÃÇÓÃNa2CO3ÈÜÒººÍŨHClÀ´Ñо¿¼òÒ×Ãð»ðÆ÷µÄ·´Ó¦Ô­Àíʱ£¬¶Ô·ÏÒºµÄ³É·Ö½øÐÐÁË̽¾¿£®
¡¾ÍÆÀí¼ÙÉè¡¿ÉÏÊöÁ½ÖÖ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Na2CO3+2HCl=2NaCl+H2O+CO2¡ü
Na2CO3+2HCl=2NaCl+H2O+CO2¡ü
£®ÓÉ´ËÍƲâ³ö·ÏÒºÖÐÒ»¶¨ÓÐNaCl£¬¿ÉÄÜÓÐNa2CO3»òÑÎËᣮ
¡¾ÊµÑé̽¾¿¡¿
£¨Ò»£©È·¶¨·ÏÒºÖÐÊÇ·ñº¬ÓÐÑÎËá
£¨1£©Ñ¡ÔñÊÔ¼Á£º¸ù¾ÝÑÎËáµÄ»¯Ñ§ÐÔÖÊ£¬Í¬Ñ§ÃÇÑ¡ÓÃÁËÈçͼËùʾµÄÎåÖÖÎïÖÊ£¬ÆäÖÐÎïÖÊXÊÇËá¼îָʾ¼ÁÖеÄ
ʯÈï
ʯÈï
ÈÜÒº£®
£¨2£©ÊµÑéÑéÖ¤£ºÄ³Í¬Ñ§Ïò·ÏÒºÖмÓÈëÉÙÁ¿µÄþ·Û£¬¹Û²ìµ½
ÎÞÃ÷ÏÔÏÖÏó
ÎÞÃ÷ÏÔÏÖÏó
£¬È·¶¨·ÏÒºÖÐÒ»¶¨Ã»ÓÐÑÎËᣮ
£¨¶þ£©È·¶¨·ÏÒºÖÐÊÇ·ñº¬ÓÐNa2CO3
ijͬѧѡÓÃ
pHÊÔÖ½
pHÊÔÖ½
²â³ö·ÏÒºµÄpH=10£¬È·¶¨·ÏÒºÖÐÒ»¶¨º¬ÓÐNa2CO3
£¨Èý£©´¦Àí·ÏÒº£¬»ØÊÕÀûÓÃ
Óû´Ó·ÏÒºÖеõ½´¿¾»µÄNaCl£¬Éè¼ÆÁËÈçÏÂʵÑé·½°¸£¬ÇëÍê³É·½°¸ÆÀ¼Û£®
·½°¸ ¼ÓÈëÊÔ¼Á ·ÖÀë·½·¨ ·½°¸ÆÀ¼Û
Ò» ÊÊÁ¿µÄCa£¨NO3£©2ÈÜÒº ¹ýÂË¡¢Õô·¢½á¾§
²»¿ÉÐÐ
²»¿ÉÐÐ
£¨Ñ¡Ìî¡°¿ÉÐС±»ò¡°²»¿ÉÐС±£¬ÏÂͬ£©
¶þ ÊÊÁ¿µÄÏ¡ÑÎËá Õô·¢½á¾§
¿ÉÐÐ
¿ÉÐÐ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?ºÍƽÇøÄ£Ä⣩ͬѧÃÇÓÃNa2CO3ÈÜÒººÍŨHClÀ´Ñо¿¼òÒ×Ãð»ðÆ÷µÄ·´Ó¦Ô­Àíʱ£¬¶Ô·ÏÒºµÄ³É·Ö½øÐÐ̽¾¿£®
ÍÆÀí¼ÙÉè
ÉÏÊöÁ½ÖÖÎïÖÊ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Na2CO3+2HCl=2NaCl+H2O+CO2¡ü
Na2CO3+2HCl=2NaCl+H2O+CO2¡ü
£¬ÓÉ´ËÍƲâ³ö·ÏÒºÖÐÒ»¶¨ÓÐNaCl£¬¿ÉÄÜÓÐNa2CO3»òHCl
ʵÑé̽¾¿
£¨Ò»£©È·¶¨·ÏÒºÖÐÊÇ·ñº¬ÓÐÑÎËá
£¨1£©Ñ¡ÔñÊÔ¼Á£º¸ù¾ÝÑÎËáµÄ»¯Ñ§ÐÔÖÊ£¬Í¬Ñ§ÃÇ¿ÉÑ¡ÓÃʯÈï¡¢Mg¡¢
CuO
CuO
¡¢
CaCO3
CaCO3
¡¢
Cu£¨OH£©2
Cu£¨OH£©2
£¨²»Í¬Àà±ðµÄ»¯ºÏÎ£®
£¨2£©ÊµÑéÑéÖ¤£ºÄ³Í¬Ñ§Ïò·ÏÒºÖмÓÈëÉÙÁ¿µÄþ·Û£¬¹Û²ìµ½·ÏÒºÖÐ
ÎÞÆøÅݲúÉú
ÎÞÆøÅݲúÉú
£¬È·¶¨Ò»¶¨Ã»ÓÐÑÎËᣮ
£¨¶þ£©È·¶¨·ÏÒºÖÐÊÇ·ñº¬ÓÐNa2CO3
ijͬѧѡÓÃ
pHÊÔÖ½
pHÊÔÖ½
²â³ö·ÏÒºµÄpH=l0£¬È·¶¨·ÏÒºÖÐÒ»¶¨º¬ÓÐNa2CO3£®
£¨Èý£©´¦Àí·ÏÒº£¬»ØÊÕÀûÓÃ
Óû´Ó·ÏÒºÖеõ½´¿¾»µÄNaCl£¬ÇëÍê³É¶ÔÏÂÁÐʵÑé·½°¸Éè¼ÆµÄÆÀ¼Û£®
·½°¸ ¼ÓÈëÊÔ¼Á ·ÖÀë·½·¨ ·½°¸ÆÀ¼Û
£¨Ðлò²»ÐУ¬²¢ËµÃ÷ÀíÓÉ£©
Ò» ÊÊÁ¿Ca£¨NO3£©2ÈÜÒº ¹ýÂË¡¢Õô·¢½á¾§
²»ÐУ¬Òý½øÐÂÔÓÖÊÏõËáÄÆ
²»ÐУ¬Òý½øÐÂÔÓÖÊÏõËáÄÆ
¶þ ¹ýÁ¿µÄHClÈÜÒº Õô·¢½á¾§
¿ÉÐУ¬HCl¾ßÓлӷ¢ÐÔ
¿ÉÐУ¬HCl¾ßÓлӷ¢ÐÔ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

ͬѧÃÇÓÃNa2CO3ÈÜÒººÍŨHClÀ´Ñо¿¼òÒ×Ãð»ðÆ÷µÄ·´Ó¦Ô­Àíʱ£¬¶Ô·ÏÒºµÄ³É·Ö½øÐÐ̽¾¿£®
ÍÆÀí¼ÙÉè
ÉÏÊöÁ½ÖÖÎïÖÊ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______£¬ÓÉ´ËÍƲâ³ö·ÏÒºÖÐÒ»¶¨ÓÐNaCl£¬¿ÉÄÜÓÐNa2CO3»òHCl
ʵÑé̽¾¿
£¨Ò»£©È·¶¨·ÏÒºÖÐÊÇ·ñº¬ÓÐÑÎËá
£¨1£©Ñ¡ÔñÊÔ¼Á£º¸ù¾ÝÑÎËáµÄ»¯Ñ§ÐÔÖÊ£¬Í¬Ñ§ÃÇ¿ÉÑ¡ÓÃʯÈï¡¢Mg¡¢______¡¢______¡¢______£¨²»Í¬Àà±ðµÄ»¯ºÏÎ£®
£¨2£©ÊµÑéÑéÖ¤£ºÄ³Í¬Ñ§Ïò·ÏÒºÖмÓÈëÉÙÁ¿µÄþ·Û£¬¹Û²ìµ½·ÏÒºÖÐ______£¬È·¶¨Ò»¶¨Ã»ÓÐÑÎËᣮ
£¨¶þ£©È·¶¨·ÏÒºÖÐÊÇ·ñº¬ÓÐNa2CO3
ijͬѧѡÓÃ______²â³ö·ÏÒºµÄpH=l0£¬È·¶¨·ÏÒºÖÐÒ»¶¨º¬ÓÐNa2CO3£®
£¨Èý£©´¦Àí·ÏÒº£¬»ØÊÕÀûÓÃ
Óû´Ó·ÏÒºÖеõ½´¿¾»µÄNaCl£¬ÇëÍê³É¶ÔÏÂÁÐʵÑé·½°¸Éè¼ÆµÄÆÀ¼Û£®
·½°¸¼ÓÈëÊÔ¼Á·ÖÀë·½·¨·½°¸ÆÀ¼Û
£¨Ðлò²»ÐУ¬²¢ËµÃ÷ÀíÓÉ£©
Ò»ÊÊÁ¿Ca£¨NO3£©2ÈÜÒº¹ýÂË¡¢Õô·¢½á¾§
______
¶þ¹ýÁ¿µÄHClÈÜÒºÕô·¢½á¾§
______

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸