ÌâÄ¿ÄÚÈÝ

¹¤ÒµÉÏ¡°ºîÊÏ¡±ÖÆ¼î·¨ÖÆµÃµÄ´¿¼î²úÆ·Öг£º¬ÓÐÉÙÁ¿µÄNaCl¡£Îª²â¶¨´¿¼î²úÆ·ÖÐNa2CO3µÄº¬Á¿£¬È¡23gÑùÆ·ÖÃÓÚÉÕ±­ÖУ¬¼ÓË®½«ÆäÈܽ⣬ȻºóÖðµÎµÎÈëÏ¡ÑÎËᣬµ±µÎÈë200gÏ¡ÑÎËáʱ£¬Á½ÕßÇ¡ºÃÍêÈ«·´Ó¦£¬Éú³ÉµÄÆøÌåÈ«²¿Òݳö£¬¹²ÊÕ¼¯µ½8.8gCO2¡£

£¨1£©ËùÓÃÏ¡ÑÎËáÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ____________£»23gÑùÆ·ÖÐNa2CO3µÄÖÊÁ¿Îª_____

____g

£¨2£©·´Ó¦ºóÉÕ±­ÖеÄÈÜҺΪ²»±¥ºÍÈÜÒº£¬ÊÔͨ¹ý¼ÆËãÇó³ö¸ÃÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿¡£

£¨1£©7.3£¥           21.2

£¨2£©·´Ó¦ºóËùµÃÈÜÒºÖÐÈÜÖÊΪNaCl£¬ÉèÉú³ÉNaClµÄÖÊÁ¿Îªx¡£

    Na2CO3 +  2HCl  =  2NaCl  +  CO2¡ü  +  H2O

                         117      44

                          x      8.8g   

                        ½âµÃx=23.4g

      NaClµÄÖÊÁ¿=23.4g+£¨23g-21.2g£©=25.2g 

      ´ð£º£¨ÂÔ£©

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø