Question

½«27.4g Na₂CO₃ºÍNaHCO₃µÄ»ìºÏÎïƽ¾ù·Ö³ÉµÈÁ¿µÄÁ½·Ý£¬Ò»·ÝÈÜÓÚË®ºó¼ÓÈë×ãÁ¿Ä³Å¨¶ÈµÄÑÎËᣬÊÕ¼¯µ½CO₂ÆøÌåV L£¬ÏûºÄÑÎËá100mL£®ÁíÒ»·ÝÖ±½Ó¼ÓÈÈÖÁºãÖØ£¬Éú³ÉCO₂ÆøÌå1.12L£¨ËùÓÐÆøÌåÌå»ý¾ùÔÚ±ê×¼×´¿öϲⶨ£©£®ÊÔ¼ÆË㣺
£¨1£©Ô­»ìºÏ¹ÌÌåÖÐNa₂CO₃ºÍNaHCO₃µÄÎïÖʵÄÁ¿Ö®±È£ºn£¨Na₂CO₃£©£ºn£¨NaHCO₃£©=

 

£»
£¨2£©V=

 

L£¬ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶Èc£¨HCl£©=

 

 mol?L^-1£®

Answer 1

¿¼µã£ºÓйػìºÏÎï·´Ó¦µÄ¼ÆËã

רÌ⣺

·ÖÎö£º£¨1£©»ìºÏÎïÖ±½Ó¼ÓÈÈʱֻÓÐNaHCO₃·Ö½â£¬·¢Éú·´Ó¦2NaHCO₃

¡÷   .

Na₂CO₃+CO₂¡ü+H₂O£¬¸ù¾Ý·½³Ìʽ֪£¬n£¨NaHCO₃£©=2n£¨CO₂£©=2¡Á

1.12L

22.4L/mol

=0.1mol£¬m£¨NaHCO₃£©=0.1mol¡Á84g/mol=8.4g£¬Ôòm£¨Na₂CO₃£©=

27.4g

2

-8.4g=5.3g£¬n£¨Na₂CO₃£©=

5.3g

106g/mol

=0.05mol£»
£¨2£©»ìºÏÎïºÍÏ¡ÑÎËá·´Ó¦Éú³ÉCO₂£¬¸ù¾ÝCÔ­×ÓÊغãµÃn£¨Na₂CO₃£©+n£¨NaHCO₃£©=n£¨CO₂£©=

V

22.4

mol£¬¾Ý´Ë¼ÆËã¶þÑõ»¯Ì¼Ìå»ý£»·´Ó¦ºóÈÜÒºÖеÄÈÜÖÊΪNaCl£¬¸ù¾ÝClÔ­×ÓÊغã¼ÆËãHClµÄÎïÖʵÄÁ¿Å¨¶È£®

½â´ð£º ½â£º£¨1£©»ìºÏÎïÖ±½Ó¼ÓÈÈʱֻÓÐNaHCO₃·Ö½â£¬·¢Éú·´Ó¦2NaHCO₃

¡÷   .

Na₂CO₃+CO₂¡ü+H₂O£¬¸ù¾Ý·½³Ìʽ֪£¬n£¨NaHCO₃£©=2n£¨CO₂£©=2¡Á

1.12L

22.4L/mol

=0.1mol£¬m£¨NaHCO₃£©=0.1mol¡Á84g/mol=8.4g£¬Ôòm£¨Na₂CO₃£©=

27.4g

2

-8.4g=5.3g£¬n£¨Na₂CO₃£©=

5.3g

106g/mol

=0.05mol£¬ËùÒÔn£¨Na₂CO₃£©£ºn£¨NaHCO₃£©=0.05mol£º0.1mol=1£º2£¬¹Ê´ð°¸Îª£º1£º2£»
£¨2£©»ìºÏÎïºÍÏ¡ÑÎËá·´Ó¦Éú³ÉCO₂£¬¸ù¾ÝCÔ­×ÓÊغãµÃn£¨Na₂CO₃£©+n£¨NaHCO₃£©=n£¨CO₂£©=

V

22.4

mol=0.05mol+0.1mol£¬ËùÒÔV=0.15mol¡Á22.4L/mol=3.36L£¬
·´Ó¦ºóÈÜÒºÖеÄÈÜÖÊΪNaCl£¬¸ù¾ÝNaÔ­×ÓÊغãµÃn£¨NaCl£©=2n£¨Na₂CO₃£©+n£¨NaHCO₃£©=0.1mol+0.1mol=0.2mol£¬¸ù¾ÝClÔ­×ÓÊغãµÃn£¨HCl£©=n£¨NaCl£©=0.2mol£¬ÔòC£¨HCl£©=

0.2mol

0.1L

=2.0mol/L£¬¹Ê´ð°¸Îª£º3.36£»2.0£®

µãÆÀ£º±¾Ì⿼²é»ìºÏÎï·´Ó¦µÄÓйؼÆË㣬²àÖØ¿¼²é·ÖÎö¡¢¼ÆËãÄÜÁ¦£¬ÀûÓÃÔ­×ÓÊغã½øÐмÆË㣬֪µÀÎïÖÊÖ®¼äµÄ·´Ó¦£¬ÌâÄ¿ÄѶȲ»´ó£®