Question

6£®Ä³¹¤³§Éú²ú²úÆ·Na₂CO₃Öк¬ÓÐÉÙÁ¿NaHCO₃ÑùÆ·£¬ÎªÁ˲ⶨ²úÆ·ÖÐNa₂CO₃ÖÊÁ¿·ÖÊý£¬È¡20.0gÑùÆ·½øÐÐÈçͼʵÑ飬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
2NaHCO₃$\frac{\underline{\;¡÷\;}}{\;}$Na₂CO₃+CO₂¡ü+H₂O

ʵÑé²½ÖèÈçÏ£º¢Ù´ò¿ªK£¬»ºÂýͨÈë¸ÉÔïN₂Ò»¶Îʱ¼ä£»
¢Ú¹Ø±ÕK£¬¼ÓÈÈÖÁÒ»¶¨Î¶Èʹ֮·´Ó¦£®
¢ÛʵÑéºó£¬¹Û²ìµ½CÖеÄÈÜÒº²»±ä»ë×Ç£¬²âµÃ¸÷×°ÖÃÖÊÁ¿ÈçÏ£º

	ʵÑéÇ°	ʵÑéºó
²£Á§¹Ü¼°ÑùÆ·ÖÊÁ¿/g	45.5	42.4
B×°ÖÃÖÊÁ¿/g	100.5	103.6

ÎÊ£º£¨1£©BÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2NaOH+CO₂¨TNa₂CO₃+H₂O
£¨2£©D×°ÖõÄ×÷ÓÃÊÇ·ÀÖ¹¿ÕÆøÖеĶþÑõ»¯Ì¼ºÍË®ÕôÆø½øÈëC×°ÖÃ
£¨3£©ÊµÑé¹ý³ÌÖУ¬¹Û²ìµ½B×°ÖÃÖв»ÔÙ²úÉúÆøÅÝ£¬ËµÃ÷AÖÐÒÑ·´Ó¦ÍêÈ«£»Í£Ö¹¼ÓÈȺó£¬ÔÚÀäÈ´¹ý³ÌÖÐҪעÒâµÄÎÊÌâÍ£Ö¹¼ÓÈȺó¼ÌÐøͨµªÆøÖÁÀäÈ´£®
£¨4£©¸ù¾ÝʵÑéËù¸øÊý¾Ý£¬¿É²âµÃNa₂CO₃ÖÊÁ¿·ÖÊýΪ58%£®

Answer 1

·ÖÎö £¨1£©¸ù¾Ý¶þÑõ»¯Ì¼ºÍÇâÑõ»¯ÄÆ·´Ó¦Éú³É̼ËáÄƺÍË®½øÐзÖÎö£»
£¨2£©¸ù¾Ý¼îʯ»Ò»áÓë¶þÑõ»¯Ì¼¡¢Ë®·´Ó¦½øÐзÖÎö£»
£¨3£©¸ù¾Ý¶þÑõ»¯Ì¼ºÍÇâÑõ»¯¸Æ·´Ó¦»áÉú³É̼Ëá¸Æ³ÁµíºÍË®½øÐзÖÎö£»
£¨4£©¸ù¾Ý»¯Ñ§·½³ÌʽºÍ¹ÌÌå¼õÉÙµÄÖÊÁ¿½øÐмÆË㣮

½â´ð ½â£º£¨1£©ÇâÑõ»¯ÄÆÈÜÒºÄÜÎüÊÕ¿ÕÆøÖеĶþÑõ»¯Ì¼Éú³É̼ËáÄƺÍË®£¬ËùÒÔB×°ÖÃÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2NaOH+CO₂¨TNa₂CO₃+H₂O£»
£¨2£©¼îʯ»Ò»áÓë¶þÑõ»¯Ì¼¡¢Ë®·´Ó¦£¬ËùÒÔD×°ÖÃÄÜ·ÀÖ¹¿ÕÆøÖеĶþÑõ»¯Ì¼ºÍË®ÕôÆø½øÈëC×°Öã»
£¨3£©Ì¼ËáÇâÄÆÔÚ¼ÓÈȵÄÌõ¼þÏÂÉú³É̼ËáÄÆ¡¢Ë®ºÍ¶þÑõ»¯Ì¼£¬ËùÒÔʵÑé¹ý³ÌÖУ¬¹Û²ìµ½B×°ÖÃÖв»ÔÙ²úÉúÆøÅÝ£¬Í£Ö¹¼ÓÈȺó£¬ÔÚÀäÈ´¹ý³ÌÖÐҪעÒâµÄÎÊÌâÊÇ£ºÍ£Ö¹¼ÓÈȺó¼ÌÐøͨµªÆøÖÁÀäÈ´£»
£¨4£©¹ÌÌå¼õÉÙµÄÖÊÁ¿Îª£º45.5g-42.4g=3.1g£¬
ÉèÑùÆ·ÖÐ̼ËáÇâÄƵÄÖÊÁ¿Îªx
2NaHCO₃$\frac{\underline{\;¡÷\;}}{\;}$Na₂CO₃+CO₂¡ü+H₂O   ¹ÌÌå¼õÉÙ
  168                                                 62
    x                                                   3.1g
         $\frac{168}{x}$=$\frac{62}{3.1g}$
         x=8.4g
ËùÒÔNa₂CO₃ÖÊÁ¿·ÖÊýΪ$\frac{20g-8.4g}{20g}$¡Á100%=58%£®
¹Ê´ð°¸Îª£º£¨1£©2NaOH+CO₂¨TNa₂CO₃+H₂O£»
£¨2£©·ÀÖ¹¿ÕÆøÖеĶþÑõ»¯Ì¼ºÍË®ÕôÆø½øÈëC×°Öã»
£¨3£©B×°ÖÃÖв»ÔÙ²úÉúÆøÅÝ£¬Í£Ö¹¼ÓÈȺó¼ÌÐøͨµªÆøÖÁÀäÈ´£»
£¨4£©58%£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁË»¯Ñ§·½³ÌʽµÄÊéд¡¢¸ù¾Ý»¯Ñ§·½³Ìʽ¼ÆËã¡¢ÅжÏÎïÖʵÄ×é³ÉµÈ·½ÃæµÄÎÊÌ⣮