Question

20£®Îª²âÁ¿Na₂CO₃ºÍNaCl×é³ÉµÄÑùÆ·ÖÐNaClµÄÖÊÁ¿·ÖÊý£¬È¡15g¸Ã¹ÌÌåÑùÆ·£¬È«²¿ÈÜÓÚ200mLË®ÖУ¬ÏòËùµÃµ½µÄ»ìºÏÈÜÒºÖÐÖðµÎµÎÈëÈÜÖÊÖÊÁ¿·ÖÊýΪ7.3%µÄÏ¡ÑÎËᣬ¼Ç¼ÁËÈçͼËùʾµÄÇúÏß¹Øϵ£®£¨Ë®µÄÃܶÈΪ1g/mL£¬ÖÊÁ¿·ÖÊý¼ÆËã½á¹û¾«È·µ½0.1%£©
£¨1£©µ±Na₂CO₃ÓëÏ¡ÑÎËáÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ÏûºÄ7.3%µÄÏ¡ÑÎËáµÄÖÊÁ¿ÊǶàÉÙ£¿
£¨2£©ÑùÆ·ÖÐNaClµÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿
£¨3£©¼ÆËãNa₂CO₃ÓëÏ¡ÑÎËáÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ËùµÃÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊý£®

Answer 1

·ÖÎö £¨1£©¸ù¾Ýͼʾ£¬ÕÒ³öÔÚºÎʱ·´Ó¦ÒÑÍêÈ«£¬¼´¿ÉÖªµÀµ±ÑÎËáÓë̼ËáÄÆÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ÏûºÄÑÎËáµÄÖÊÁ¿£»
£¨2£©ÒòΪ̼ËáÄÆÓëÑÎËá·´Ó¦Éú³É¶þÑõ»¯Ì¼ÆøÌ壮¸ù¾Ý·´Ó¦µÄ»¯Ñ§·½³ÌʽºÍÉú³ÉÆøÌåµÄÖÊÁ¿£¬Áгö±ÈÀýʽ£¬¾Í¿É¼ÆËã³öÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿£»È»ºó¸ù¾Ý¡°ÑùÆ·ÖÊÁ¿-̼ËáÄƵÄÖÊÁ¿=»ìºÏÎïÖÐÂÈ»¯ÄƵÄÖÊÁ¿¡±¼ÆËã¼´¿É£»
£¨3£©µ±ÑÎËáÓë̼ËáÄÆÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ËùµÃÈÜҺΪÂÈ»¯ÄÆÈÜÒº£¬ÆäÖÐÈÜÖÊÂÈ»¯ÄÆÖÊÁ¿ÎªÑùÆ·ÖÐÂÈ»¯ÄÆÓë·´Ó¦Éú³ÉÂÈ»¯ÄƵÄÖÊÁ¿ºÍ£¬ÀûÓÃÖÊÁ¿Êغ㶨ÂÉ£¬ÇóµÃ·´Ó¦ºóÈÜÒºµÄÖÊÁ¿£¬Ê¹ÓÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý¼ÆË㹫ʽ¼ÆËãËùµÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£®

½â´ð ½â£º£¨1£©ÓÉͼʾ¿ÉÖª£¬¼ÓÈëÑÎËáµ½100gʱ£¬ÆøÌå²»ÔÙÉú³É£¬ËµÃ÷·´Ó¦ÒÑÍêÈ«£»
¹Ê´ð°¸Îª£º100g£»
£¨2£©ÉèÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿Îªx£¬Éú³ÉÂÈ»¯ÄƵÄÖÊÁ¿Îªy£¬
Na₂CO₃+2HCl¨T2NaCl+CO₂¡ü+H₂O
 106         117   44
  x          y     4.4g
$\frac{106}{x}=\frac{117}{y}=\frac{44}{4.4g}$
x=10.6 g     y=11.7g
ÑùÆ·ÖÐNaClµÄÖÊÁ¿·ÖÊý=$\frac{15g-10.6}{15g}¡Á$100%=29.3%
£¨3£©Na₂CO₃ÓëÏ¡ÑÎËáÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ËùµÃÈÜҺΪÂÈ»¯ÄÆÈÜÒº£¬ÈÜÖÊÖÊÁ¿Îª11.7g+15g-10.6g=16.1g£¬ÈÜÒºµÄÖÊÁ¿=15g+200g+100g-4.4g=310.6g
ÈÜÒºÖÐNaCl%=$\frac{16.1g}{310.6g}$¡Á100%=5.2%£®
´ð°¸£º
£¨1£©100g£»
£¨2£©ÑùÆ·ÖÐNaClµÄÖÊÁ¿·ÖÊýÊÇ29.3%£»
£¨3£©ËùµÃÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊýΪ5.2%£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éѧÉú¶ÔÍêÈ«·´Ó¦µÄÈÏʶ£¬ÒÔ¼°ÔËÓû¯Ñ§·½³ÌʽºÍÈÜÖÊÖÊÁ¿·ÖÊý¹«Ê½½øÐмÆËãµÄÄÜÁ¦£®Ôö¼ÓÁËѧÉú·ÖÎöÎÊÌâµÄ˼ά¿ç¶È£¬Ç¿µ÷ÁËѧÉúÕûºÏ֪ʶµÄÄÜÁ¦£®