ÌâÄ¿ÄÚÈÝ
ÔÚÎÒ¹úÇຣºþµØÇøÓÐÒ»ÖÖ˵·¨£¬¶¬ÌìÀ̼ÏÄÌìɹÑΣ®ÕâÀïµÄ¼îÊÇÖ¸Na2CO3£¬ÑÎÊÇÖ¸NaCl£®ÈËÃÇ´ÓÑκþÖÐÀ̵õļî»áº¬ÓÐÉÙÁ¿µÄNaCl£®Ä³Ñо¿ÐÔѧϰС×é³ÆÈ¡º¬NaClµÄNa2CO3£¬¹ÌÌå25.0g£¬½«ÆäÅäÖÆ³ÉÈÜÒº£¬ÔÙÏòÆäÖÐÖðµÎ¼ÓÈë×ãÁ¿µÄÈÜÖÊÖÊÁ¿·ÖÊýΪ7.3%µÄÏ¡ÑÎËá£¬Ê¹ÆøÌåÍêÈ«·Å³ö£¬¹²ÊÕ¼¯µ½8.8g¶þÑõ»¯Ì¼ÆøÌ壮ÊÔ¼ÆË㣺£¨1£©Ô¹ÌÌåÖÐNa2CO3£¬µÄÖÊÁ¿·ÖÊý£»£¨2£©·´Ó¦ÖÐÏûºÄµÄÑÎËáµÄ×ÜÖÊÁ¿£®
½â£º£¨1£©Éè̼ËáÄÆµÄÖÊÁ¿Îªx£¬ÑÎËáÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿Îªy£®
Na2CO3+2HCl=2NaCl+H2O+CO2¡ü
106 73 44
x y 8.8g

x=21.2g

y=14.6g
%=84.8%
´ð£ºÔ¹ÌÌåÖÐ̼ËáÄÆµÄÖÊÁ¿·ÖÊýΪ84.8%£®
£¨2£©·´Ó¦ÖÐÏûºÄµÄÑÎËáµÄ×ÜÖÊÁ¿Îª£º
=200g ´ð£º·´Ó¦ÖÐÏûºÄÑÎËáµÄ×ÜÖÊÁ¿Îª200g£®
·ÖÎö£º£¨1£©¿É¸ù¾Ý¶þÑõ»¯Ì¼µÄÖÊÁ¿Çó³ö̼ËáÄÆµÄÖÊÁ¿£¬ÔÙ¸ù¾Ý
%Çó³öÔ¹ÌÌåÖÐ̼ËáÄÆµÄÖÊÁ¿·ÖÊý£®
£¨2£©¸ù¾Ý¶þÑõ»¯Ì¼µÄÖÊÁ¿Çó³öÑÎËáÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿£¬ÔÙ¸ù¾Ý
Çó³öÑÎËáµÄ×ÜÖÊÁ¿£®
µãÆÀ£º´¿¼î²»ÊǼ¶øÊÇÑΣ¬ÊÇ̼ËáÄÆ£¬Ö»²»¹ý̼ËáÄÆµÄË®ÈÜÒºÏÔ¼îÐÔ£®
Na2CO3+2HCl=2NaCl+H2O+CO2¡ü
106 73 44
x y 8.8g
x=21.2g
y=14.6g
´ð£ºÔ¹ÌÌåÖÐ̼ËáÄÆµÄÖÊÁ¿·ÖÊýΪ84.8%£®
£¨2£©·´Ó¦ÖÐÏûºÄµÄÑÎËáµÄ×ÜÖÊÁ¿Îª£º
·ÖÎö£º£¨1£©¿É¸ù¾Ý¶þÑõ»¯Ì¼µÄÖÊÁ¿Çó³ö̼ËáÄÆµÄÖÊÁ¿£¬ÔÙ¸ù¾Ý
£¨2£©¸ù¾Ý¶þÑõ»¯Ì¼µÄÖÊÁ¿Çó³öÑÎËáÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿£¬ÔÙ¸ù¾Ý
µãÆÀ£º´¿¼î²»ÊǼ¶øÊÇÑΣ¬ÊÇ̼ËáÄÆ£¬Ö»²»¹ý̼ËáÄÆµÄË®ÈÜÒºÏÔ¼îÐÔ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿