Question

3£®½«5.08gÓÉNa₂CO₃ºÍNaHCO₃×é³ÉµÄ¹ÌÌå»ìºÏÎïÍêÈ«ÈÜÓÚË®£¬ÖƳÉÈÜÒº£¬È»ºóÏò¸ÃÈÜÒºÖÐÖðµÎ¼ÓÈë2mol?L^-1µÄÑÎËᣬËù¼ÓÈëÑÎËáµÄÌå»ýÓë²úÉúCO₂Ìå»ý£¨±ê×¼×´¿ö£©µÄ¹ØϵÈçͼËùʾ£®
£¨1£©Ð´³öOA¶ÎËù·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£ºCO₃^2-+H⁺¨THCO₃^-£®
£¨2£©µ±¼ÓÈë35mLÑÎËáʱ£¬²úÉúµÄCO₂ÔÚ±ê×¼×´¿öϵÄÌå»ýΪ672mL£®
£¨3£©Ô­»ìºÏÎïÖеÄNa₂CO₃ºÍNaHCO₃µÄÎïÖʵÄÁ¿Ö®±ÈΪ4£º1£®
£¨4£©NaHCO₃¹ÌÌåÊÜÈÈ·Ö½âµÄ»¯Ñ§·½³ÌʽΪ2NaHCO₃$\frac{\underline{\;\;¡÷\;\;}}{\;}$Na₂CO₃+H₂O+CO₂¡ü£®

Answer 1

·ÖÎö £¨1£©Na₂CO₃ºÍNaHCO₃×é³ÉµÄ»ìºÏÎïÖмÓÈëÑÎËᣬÏÈ·¢Éú̼ËáÄÆÓëÑÎËá·´Ó¦Éú³É̼ËáÇâÄÆÓëÂÈ»¯ÄÆ£¬È»ºó̼ËáÇâÄÆÓëÑÎËá·´Ó¦Éú³É¶þÑõ»¯Ì¼ÆøÌ壻
£¨2£©½áºÏͼÏó¿ÉÖª£¬µ±¼ÓÈë35mLÑÎËáʱ£¬20¡«35mL·¢Éú̼ËáÇâÄÆÓëÑÎËáµÄ·´Ó¦Éú³É¶þÑõ»¯Ì¼ÆøÌ壻
£¨3£©¸ù¾ÝOA¶ÎµÄ·´Ó¦¼°ÏûºÄÑÎËáµÄÁ¿¼ÆËã³ö̼ËáÄƵÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾Ý×ÜÖÊÁ¿¼ÆËã³ö̼ËáÇâÄƵÄÎïÖʵÄÁ¿£¬×îºó¼ÆËã³ö¶þÕßµÄÎïÖʵÄÁ¿Ö®±È£»
£¨4£©Ì¼ËáÇâÄÆÊÜÈÈ·Ö½âΪ̼ËáÄÆ¡¢¶þÑõ»¯Ì¼ÓëË®£®

½â´ð ½â£º£¨1£©¼ÓÈëÑÎËáºó̼ËáÄÆÏÈÓëÑÎËá·´Ó¦Éú³É̼ËáÇâÄÆ£ºNa₂CO₃+HCl¨TNaHCO₃+NaCl£¬ËùÒÔOA¶ÎµÄÀë×Ó·´Ó¦·½³ÌʽΪ£ºCO₃^2-+H⁺¨THCO₃^-£¬
¹Ê´ð°¸Îª£ºCO₃^2-+H⁺¨THCO₃^-£»
£¨2£©½áºÏͼÏó¿ÉÖª£¬µ±¼ÓÈë35mLÑÎËáʱ£¬25¡«35mL·¢Éú̼ËáÇâÄÆÓëÑÎËáµÄ·´Ó¦Éú³É¶þÑõ»¯Ì¼ÆøÌ壬
ÉèÉú³É¶þÑõ»¯Ì¼µÄÎïÖʵÄÁ¿Îªn£¬Éú³ÉÆøÌåʱÏûºÄµÄËáµÄÎïÖʵÄÁ¿Îª£¨35-20£©¡Á10^-3L¡Á2mol•L^-1=0.03mol£¬Ôò
HCO₃^-+H⁺¨TCO₂¡ü+H₂O
     1     1
  0.03mol  n
¹Ên=0.03mol£¬±ê×¼×´¿öÏÂÆäÌå»ýΪ0.03mol¡Á22.4L/mol=0.672L£¬¼´Îª672mL£¬
¹Ê´ð°¸Îª£º672£»
£¨3£©¸ù¾ÝOA¶ÎµÄ·´Ó¦£ºNa₂CO₃+HCl¨TNaHCO₃+NaCl£¬¿ÉÖªn£¨Na₂CO₃£©=n£¨HCl£©=2mol/L¡Á0.02L=0.04mol£¬
¸ù¾Ý×ÜÖÊÁ¿¿ÉµÃ£ºn£¨NaHCO₃£©¡Á84g/mol+106g/mol¡Á0.04mol=5.08g£¬½âµÃ£ºn£¨NaHCO₃£©=0.01mol£¬
¹ÊÔ­»ì»áÎïÖÐNa₂CO₃ºÍNaHCO₃µÄÎïÖʵÄÁ¿Ö®±ÈΪ0.04mol£º0.01mol=4£º1£¬
¹Ê´ð°¸Îª£º4£º1£»
£¨4£©NaHCO₃¹ÌÌåÊÜÈÈ·Ö½âµÄ»¯Ñ§·½³ÌʽΪ£º2NaHCO₃$\frac{\underline{\;\;¡÷\;\;}}{\;}$Na₂CO₃+H₂O+CO₂¡ü£¬
¹Ê´ð°¸Îª£º2NaHCO₃$\frac{\underline{\;\;¡÷\;\;}}{\;}$Na₂CO₃+H₂O+CO₂¡ü£®

µãÆÀ ±¾Ì⿼²é»ìºÏÎï·´Ó¦µÄ¼ÆË㣬ÌâÄ¿ÄѶÈÖеȣ¬Ã÷È··´Ó¦µÄÏȺó˳Ðò¼°Í¼ÏóÖÐÿ¶ÎͼÏó¶ÔÓ¦µÄ»¯Ñ§·´Ó¦Êǽâ´ð±¾ÌâµÄ¹Ø¼ü£¬ÊÔÌâÅàÑøÁËѧÉúµÄ·ÖÎöÄÜÁ¦¼°»¯Ñ§¼ÆËãÄÜÁ¦£®