Question

4£®Ä³Ñо¿ÐÔѧϰС×é³ÆÈ¡º¬NaClµÄNa₂CO₃¹ÌÌå25g£¬½«ÆäÅäÖƳÉÈÜÒº£¬ÔÙÏòÆäÖÐÖðµÎ¼ÓÈë×ãÁ¿µÄ7.3%µÄÏ¡ÑÎËᣬʹÆøÌåÍêÈ«·Å³ö£¬¹²ÊÕ¼¯µ½ 8.8g CO₂ÆøÌ壮
£¨l£©¼ÆËãÔ­¹ÌÌåÖÐNa₂CO₃µÄÖÊÁ¿·ÖÊýºÍÏûºÄÏ¡ÑÎËáµÄ×ÜÖÊÁ¿£®£¨Ð´³ö¼ÆËã¹ý³Ì£©
£¨2£©Èç±íΪÑо¿ÐÔѧϰС×éÏòÉÏÊöÅäÖƵĻìºÏÒºÖзÖÈý´ÎÖðµÎ¼ÓÈ루±ß¼Ó±ßÕñµ´£©Ï¡ÑÎËáºóËù¼Ç¼µÄ²¿·ÖÊý¾Ý£®¾­²éÔÄ×ÊÁÏÖªNa₂CO₃ÓëHCl·´Ó¦¿ÉÒÔ·ÖÁ½²½½øÐУºNa₂CO₃+HCl¨TNaCl+NaHCO₃£»NaHCO₃+HCl¨TNaCl+H₂O+CO₂¡ü£®
¢ÙÇëÍê³É±í¸ñÖÐδÌîµÄ²¿·Ö£®

ʵÑé´ÎÊý	ÿ´Î²úÉúµÄCO2µÄÖÊÁ¿/g
µÚÒ»´ÎÏÈÖðµÎµÎ¼ÓÏ¡ÑÎËá100g	0
µÚ¶þ´ÎÔÙÖðµÎµÎ¼ÓÏ¡ÑÎËá100g	8.8
µÚÈý´ÎÔÙÖðµÎµÎ¼ÓÏ¡ÑÎËá100g	0

¢Ú¸ù¾Ý±íÖÐÊý¾ÝÔÚ×ø±êϵÖл­³öCO₂µÄÖÊÁ¿£¨×Ý×ø±ê£©ÓëËù¼ÓÈëÏ¡ÑÎËáµÄÖÊÁ¿£¨ºá×ø±ê£©µÄ¹Øϵͼ£®

Answer 1

·ÖÎö £¨1£©¸ù¾ÝÍêÈ«·´Ó¦Ê±ÆøÌåµÄÖÊÁ¿Îª8.8g£¬½«¶þÑõ»¯Ì¼µÄÖÊÁ¿´úÈë̼ËáÄÆÓëÑÎËá·´Ó¦µÄ»¯Ñ§·´Ó¦·½³Ìʽ½øÐмÆËãµÃµ½Ì¼ËáÄƼ°ÑÎËáÖÊÁ¿£»
£¨2£©¸ù¾ÝNa₂CO₃ÓëHCl·´Ó¦¿ÉÒÔ·ÖÁ½²½½øÐУºNa₂CO₃+HCl=NaCl+NaHCO₃£»NaHCO₃+HCl=NaCl+H₂O+CO₂¡ü£¬´Ó²Î¼Ó·´Ó¦µÄÎïÖʵÄÖÊÁ¿À´·ÖÎö·¢ÉúµÄ»¯Ñ§·´Ó¦£¬²¢ÅжÏÊÇ·ñ²úÉúÆøÌ壮

½â´ð ½â£º£¨1£©ÉèÔ­¹ÌÌåÖÐNa₂CO₃µÄÖÊÁ¿Îªx£¬ÏûºÄÏ¡ÑÎËáµÄ×ÜÖÊÁ¿Îªy£®
Na₂CO₃+2HCl¨T2NaCl+H₂O+CO₂¡ü
106          73                        44
  x         y¡Á7.3%                   8.8g
$\frac{106}{x}=\frac{73}{y¡Á73%}=\frac{44}{8.8g}$
½âµÃx=21.2 g£¬y=200 g
Na₂CO₃µÄÖÊÁ¿·ÖÊý$\frac{21.2g}{25g}$¡Á100%=84.8%
´ð£ºÔ­¹ÌÌåÖÐNa₂CO₃µÄÖÊÁ¿·ÖÊýΪ84.8%£¬ÏûºÄÏ¡ÑÎËáµÄ×ÜÖÊÁ¿Îª200 g£®
£¨2£©¸ù¾ÝNa₂CO₃+HCl=NaCl+NaHCO₃£»NaHCO₃+HCl=NaCl+H₂O+CO₂¡ü¿ÉÖª£¬¼ÓÈë100gÑÎËáʱֻ·¢ÉúµÚÒ»²½·´Ó¦£¬
ÔÚ·´Ó¦Na₂CO₃+HCl=NaCl+NaHCO₃£»
            106      36.5
           21.2g 100g¡Á7.3%
$\frac{106}{21.2g}=\frac{36.5}{100g¡Á7.3%}$£¬
´Ëʱ²»²úÉú¶þÑõ»¯Ì¼ÆøÌ壬¼´ÆøÌåµÄÖÊÁ¿Îª0£¬ÔÙ¼Ó100gÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¬µÃµ½8.8g¶þÑõ»¯Ì¼ÆøÌ壬ÔÙ¼ÓÈëÑÎËá·´Ó¦²»·¢Éú£¬¼´µÚÈý´Î¼ÓÑÎËáµÃµ½ÆøÌåÖÊÁ¿Îª0£¬
¹Ê´ð°¸Îª£º£¨2£©¢Ù0
¢Ú

µãÆÀ ±¾Ìâ½ÏÄÑ£¬¿¼²é»¯Ñ§·´Ó¦·½³ÌʽµÄ¼ÆË㣬ѧÉúӦעÒâ·´Ó¦µÄ·Ö²½½øÐУ¬²¢×¢Òâ·´Ó¦ÓëͼÏóµÄ¶ÔÓ¦¹Øϵ£®