ÌâÄ¿ÄÚÈÝ

1£®Ä³»¯Ñ§ÐËȤС×齫27.8gº¬ÔÓÖʵĴ¿¼îÑùÆ·£¨ÔÓÖÊΪÂÈ»¯ÄÆ£©Óë181gÏ¡ÑÎËáÏà»ìºÏ£¬³ä·Ö·´Ó¦£¬²âµÃ·´Ó¦ÖÐÉú³ÉÆøÌåµÄÖÊÁ¿£¨m£©Ó뷴Ӧʱ¼ä£¨t£©µÄÊý¾ÝÈç±íËùʾ£º
·´Ó¦Ê±¼ät/st0t1t2t3t4t5t6
ÆøÌåÖÊÁ¿m/g01.763.525.287.048.88.8
¸ù¾ÝÌâĿҪÇ󣬻شðÏÂÁÐÎÊÌ⣺
£¨1£©Ì¼ËáÄÆÓëÏ¡ÑÎËá·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºNa2CO3+2HCl=2NaCl+CO2¡ü+H2O£¬Ì¼ËáÄÆÍêÈ«·´Ó¦ºó£¬Éú³ÉCO2µÄÖÊÁ¿Îª8.8g
£¨2£©ÇëÔÚÈçͼËùʾµÄ×ø±êͼÖУ¬»­³ö·´Ó¦Ê±Éú³ÉÆøÌåµÄÖÊÁ¿£¨m£©ËæÊ±¼ä£¨t£©±ä»¯µÄÇúÏߣ®

£¨3£©Çë¼ÆËãÍêÈ«·´Ó¦ºóÈÜÒºÖÐÂÈ»¯ÄƵÄÖÊÁ¿·ÖÊý£®

·ÖÎö £¨1£©ÓÉÓÚ̼ËáÄÆºÍÏ¡ÑÎËá·´Ó¦ÄܲúÉú¶þÑõ»¯Ì¼£¬¸ù¾Ýͼ±íÊý¾Ý·ÖÎöµÃ³öÉú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿£»
£¨2£©¸ù¾Ýͼ±íÊý¾ÝºÍ×ø±êͼÖкáÖáÓë×ÝÖáµÄ±íʾ£¬°´ÒªÇó»­³ö·´Ó¦Éú³ÉÆøÌåµÄÖÊÁ¿£¨m£©ËæÊ±¼ä£¨t£©±ä»¯µÄÇúÏߣ»
£¨3£©ÓÉÉú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿£¬¸ù¾Ý̼ËáÄÆÓëÏ¡ÑÎËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ¿ÉÒÔ¼ÆËã³öÑùÆ·ÖÐ̼ËáÄÆµÄÖÊÁ¿ºÍÉú³ÉÂÈ»¯ÄƵÄÖÊÁ¿£®½ø¶ø¼ÆËã³öÑùÆ·ÖÐÂÈ»¯ÄƵÄÖÊÁ¿£¬ÔÙ¼ÓÉÏÉú³ÉÂÈ»¯ÄƵÄÖÊÁ¿¾ÍÊÇ·´Ó¦ºóÈÜÒºÖеÄÈÜÖÊÖÊÁ¿£¬×îºó½áºÏËùµÃÈÜÒºµÄÖÊÁ¿¸ù¾ÝÈÜÖÊÖÊÁ¿·ÖÊýµÄ¼ÆË㹫ʽ¿ÉÒÔ¼ÆËã³ö·´Ó¦ºóµÄÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£®

½â´ð ½â£º£¨1£©Ì¼ËáÄÆÓëÏ¡ÑÎËá·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºNa2CO3+2HCl=2NaCl+CO2¡ü+H2O£»ÓÉͼ±íÊý¾Ý¿ÉÖª£¬Ì¼ËáÄÆÍêÈ«·´Ó¦ºó£¬Éú³ÉCO2µÄÖÊÁ¿Îª£º8.8g£» 
    £¨2£©¸ù¾Ýͼ±íÊý¾Ý£¬»­³ö·´Ó¦Éú³ÉÆøÌåµÄÖÊÁ¿£¨m£©ËæÊ±¼ä£¨t£©±ä»¯µÄÇúÏߣº

£¨3£©ÉèÑùÆ·ÖÐ̼ËáÄÆµÄÖÊÁ¿Îªx£¬·´Ó¦Éú³ÉNaClµÄÖÊÁ¿Îªy
Na2CO3+2HCl=2NaCl+CO2¡ü+H2O
106                    117     44
 x                        y      8.8g  
$\frac{106}{x}=\frac{117}{y}=\frac{44}{8.8g}$
½âÖ®µÃ£ºx=21.2g   y=23.4g
¡àÑùÆ·ÖÐÂÈ»¯ÄƵÄÖÊÁ¿Îª27.8g-21.2g=6.6g
·´Ó¦ºóÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿=23.4g+6.6g=30g
·´Ó¦ºóÈÜÒºµÄÖÊÁ¿=27.8g+181g-8.8g=200g
¡àÍêÈ«·´Ó¦ºóËùµÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ$\frac{30g}{200g}$¡Á100%=15%
¹Ê´ð°¸Îª£º£¨1£©Na2CO3+2HCl=2NaCl+CO2¡ü+H2O£»8.8£»
£¨2£©
£»
£¨3£©£ºÍêÈ«·´Ó¦ºóËùµÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ15%£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÓйØÑùÆ·»ìºÏÎïµÄ»¯Ñ§·½³ÌʽµÄ¼ÆËãºÍÓйØÈÜÖÊÖÊÁ¿·ÖÊýµÄ¼ÆË㣬ÄܶÍÁ¶Ñ§ÉúµÄÉ÷ÃÜ˼ά¼°½âÌâÄÜÁ¦£¬ÄѶȽϴó£®Ã÷È·Êý¾ÝµÄ·ÖÎöÊǽâ´ð±¾ÌâµÄ¹Ø¼ü£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø