ÌâÄ¿ÄÚÈÝ

10£®Ä³Ñ§Ð£¿ÆÑ§ÐËȤС×éΪÁË̽¾¿ÊµÑéÊÒÖоÃÖõÄNaOHµÄ±äÖʳ̶ȣ¬¾ßÌåÈçͼ£º
¡¾Ñо¿·½°¸¡¿ÏȳÆÈ¡13.3g µÄNaOHÑùÆ·£¨ÔÓÖÊΪNa2CO3£©£¬Åä³ÉÈÜÒº£¬È»ºóÏòÈÜÒºÖÐÖðµÎ¼ÓÈëÖÊÁ¿·ÖÊýΪ14.6%µÄÏ¡ÑÎËᣬ¸ù¾ÝÉú³ÉCO2µÄÖÊÁ¿²â¶¨Na2CO3µÄÖÊÁ¿£®´Ó¶ø½øÒ»²½È·¶¨ÑùÆ·ÖÐNaOHµÄ±äÖʳ̶ȣ®£¨±äÖʳ̶ÈÊÇÖ¸ÒѱäÖʵÄNaOHÔÚÔ­NaOH ÖеÄÖÊÁ¿·ÖÊý£©
£¨1£©ÊµÑé²âµÃ¼ÓÈëÏ¡ÑÎËáµÄÖÊÁ¿Óë²úÉúCO2ÆøÌåµÄÖÊÁ¿¹ØÏµÈçͼËùʾ£®¸ÃÑùÆ·ÖÐNa2CO3µÄÖÊÁ¿ÊǶàÉÙ£¿
£¨2£©ÇóʵÑé¹ý³ÌÖÐÓëNaOH·´Ó¦ËùÓõÄÑÎËáÈÜÖʵÄÖÊÁ¿£®
£¨3£©ÇóÑùÆ·ÖÐNaOHµÄ±äÖʳ̶ȣ®

·ÖÎö £¨1£©¸ù¾ÝÇâÑõ»¯ÄÆÄÜºÍ¿ÕÆøÖеĶþÑõ»¯Ì¼·´Ó¦Éú³É̼ËáÄÆºÍË®£¬ÄܺÍÏ¡ÑÎËá·´Ó¦Éú³ÉÂÈ»¯ÄƺÍË®£»Ì¼ËáÄÆºÍÏ¡ÑÎËá·´Ó¦Éú³ÉÂÈ»¯ÄÆ¡¢Ë®ºÍ¶þÑõ»¯Ì¼£¬ÀûÓöþÑõ»¯Ì¼µÄÖÊÁ¿Çó³ö̼ËáÄÆµÄÖÊÁ¿¼´¿É£»
£¨2£©¸ù¾ÝÇâÑõ»¯ÄƵÄÖÊÁ¿Çó³öʵÑé¹ý³ÌÖÐÓëNaOH·´Ó¦ËùÓõÄÑÎËáÈÜÖʵÄÖÊÁ¿¼´¿É£»
£¨3£©¸ù¾Ý̼ËáÄÆµÄÖÊÁ¿Çó³ö±äÖʵÄÇâÑõ»¯ÄƵÄÖÊÁ¿£¬½ø¶øÇó³öÑùÆ·ÖÐNaOHµÄ±äÖʳ̶ȼ´¿É£®

½â´ð ½â£º£¨1£©Éè̼ËáÄÆµÄÖÊÁ¿Îªx£¬
ÓÉͼÖÐÐÅÏ¢¿ÉÖª£¬²úÉú¶þÑõ»¯Ì¼µÄÖÊÁ¿ÊÇ2.2g£¬
Na2CO3+2HCl¨T2NaCl+H2O+CO2¡ü£¬
  106                  44
   x                  2.2g
$\frac{106}{x}=\frac{44}{2.2g}$
x=5.3g
£¨2£©ÉèºÍÇâÑõ»¯ÄÆ·´Ó¦µÄÂÈ»¯ÇâÖÊÁ¿Îªy£¬
ºÍÂÈ»¯Çâ·´Ó¦µÄÇâÑõ»¯ÄÆÖÊÁ¿Îª£º13.3g-5.3g=8g£¬
NaOH+HCl=NaCl+H2O£¬
 40  36.5
 8g   y
$\frac{40}{8g}=\frac{36.5}{y}$
y=7.3g£¬
£¨3£©Éè±äÖʵÄÇâÑõ»¯ÄÆÖÊÁ¿Îªz£¬
 2NaOH+CO2¨TNa2CO3+H2O£¬
  80         106
  z          5.3g
$\frac{80}{z}=\frac{106}{5.3g}$
z=4g
´ËNaOHµÄ±äÖʳ̶ÈΪ£º$\frac{4g}{13.3g-5.3g+4g}$¡Á100%=33.3%£¬
´ð£º£¨1£©¸ÃÑùÆ·ÖÐNa2CO3µÄÖÊÁ¿ÊÇ5.3g£»
£¨2£©ÊµÑé¹ý³ÌÖÐÓëNaOH·´Ó¦ËùÓõÄÑÎËáÈÜÖʵÄÖÊÁ¿Îª7.3g£»
£¨3£©´ËNaOHµÄ±äÖʳ̶ÈΪ33.3%£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éѧÉúÔËÓüÙÉè·¨ºÍ»¯Ñ§·½³Ìʽ½øÐмÆËãºÍÍÆ¶ÏµÄÄÜÁ¦£¬Í¬Ê±¿¼²éÁË·ÖÎöͼÖÐÊý¾ÝµÄÄÜÁ¦£¬¼ÆËãʱҪעÒâ¹æ·¶ÐÔºÍ׼ȷÐÔ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø