ÌâÄ¿ÄÚÈÝ

Óû²â¶¨Ä³Ê¯»ÒʯÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£¨Ê¯»ÒʯÑùÆ·ÖеÄÔÓÖÊÊÜÈȲ»·Ö½â£¬ÇÒ²»ÓëËá·´Ó¦£©£¬¼×¡¢ÒÒÁ½Î»Í¬Ñ§·Ö±ðÉè¼ÆÁËÈçÏÂÁ½¸öʵÑé·½°¸£º
·½°¸Ò»£º¢Ù³Æȡʯ»ÒʯÑùÆ·ÖÊÁ¿8g£»¢ÚÓþƾ«µÆ¼ÓÈÈÑùÆ·£¬Ö±ÖÁÖÊÁ¿²»Ôٸı䣻¢Û½«¹ÌÌå²ÐÓàÎï·ÅÔÚ¸ÉÔïµÄÆ÷ÃóÖÐÀäÈ´ºó³ÆµÃÖÊÁ¿6.9 g£»¢Ü¼ÆË㣮
·½°¸¶þ£º¢Ù³Æȡʯ»ÒʯÑùÆ·ÖÊÁ¿8 g£»¢Ú¼ÓÈëÖÊÁ¿·ÖÊýΪ7.3%µÄÑÎËá100 g£¬Ê¹Ì¼Ëá¸ÆÍêÈ«·´Ó¦£»¢ÛÏò·´Ó¦ºóµÄÈÜÒºÖмÓÈ뺬ÈÜÖÊ3.2 gµÄÇâÑõ»¯ÄÆÈÜÒº£¬Ç¡ºÃÖкͶàÓàµÄÑÎË᣻¢Ü¼ÆË㣮
Çë»Ø´ðÏÂÁÐÓйØÎÊÌ⣺
£¨1£©100gÖÊÁ¿·ÖÊýΪ7.3%µÄÑÎËáÖУ¬ÈÜÖÊÂÈ»¯ÇâµÄÖÊÁ¿Îª
 
g£»
£¨2£©ÄãÈÏΪÉÏÊö·½°¸ÖУ¬ÇÐʵ¿ÉÐеÄÊÇ·½°¸
 
£¬²»Ñ¡ÓÃÁíÒ»·½°¸µÄÔ­ÒòÊÇ
 
£»
£¨3£©Çó·½°¸¶þÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£®
·ÖÎö£º£¨1£©¸ù¾ÝÈÜÖÊÖÊÁ¿=ÈÜÒºÖÊÁ¿¡ÁÈÜÖÊÖÊÁ¿·ÖÊý¼ÆËã¼´¿É£»
£¨2£©´ÓÁ½¸ö·½°¸¿ÉÒÔ¿´³ö£¬µÚÒ»¸ö·½°¸µÄ¢ÚÖÐÓþƾ«µÆ¼ÓÈÈ£¬Î¶ȴﲻµ½Ê¯»Òʯ·Ö½âµÄ¸ßÎÂÒªÇó£»·½°¸¶þÖÐ̼Ëá¸Æ¿ÉÒÔºÍÑÎËá·¢Éú·´Ó¦£»
£¨3£©¢Ù¸ù¾ÝÇâÑõ»¯ÄÆÈÜÒººÍÑÎËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ¼ÆËã³öÓëÇâÑõ»¯ÄÆ·´Ó¦µÄHClÖÊÁ¿£¬È»ºó¸ù¾ÝÑÎËáÖÐHClµÄÖÊÁ¿-ÇâÑõ»¯ÄÆ·´Ó¦µÄHClÖÊÁ¿=Óë̼Ëá¸Æ·´Ó¦µÄHClµÄÖÊÁ¿£»
¢Ú¸ù¾Ý̼Ëá¸ÆºÍÑÎËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ¼ÆËã³ö̼Ëá¸ÆµÄÖÊÁ¿£¬ÔÙ¸ù¾ÝÖÊÁ¿·ÖÊý¹«Ê½¼ÆËã¼´¿É£®
½â´ð£º½â£º£¨1£©100g¡Á7.3%=7.3g£®¹Ê´ð°¸Îª£º7.3g£®
£¨2£©´ÓÁ½¸ö·½°¸¿ÉÒÔ¿´³ö£¬µÚÒ»¸ö·½°¸µÄ¢ÚÖÐÓþƾ«µÆ¼ÓÈÈ£¬Î¶ȴﲻµ½Ê¯»Òʯ·Ö½âµÄ¸ßÎÂÒªÇ󣬷½°¸¶þÖÐ̼Ëá¸Æ¿ÉÒÔºÍÑÎËá·¢Éú·´Ó¦£®
¹Ê´ð°¸Îª£º¶þ£¬Ì¼Ëá¸ÆÒªÔÚ¸ßÎÂϲÅÄÜÍêÈ«·Ö½â£®
£¨3£©½â£º¢ÙÉèÓëÇâÑõ»¯ÄÆ·´Ó¦µÄHClÖÊÁ¿Îªx
HCl+NaOH¨TNaCl+H2O
36.5 40
x    3.2g
¡à
36.5
40
=
x
3.2g

½âÖ®µÃ£ºx=2.92g£¬
ÄÇô£¬Óë̼Ëá¸Æ·´Ó¦µÄÑÎËáÖÊÁ¿Îª£º7.3%¡Á100g-2.92g=4.38g£¬
¢ÚÉèʯ»ÒʯÖÐ̼Ëá¸ÆµÄÖÊÁ¿Îªy£¬
CaCO3+2HCl=CaCl2+CO2¡ü+H2O
100    73
y      4.38g
¡à
100
73
=
y
4.38g

½âÖ®µÃy=6g£¬
̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ£º
6g
8g
¡Á 100%
=75%£®
´ð£º·½°¸¶þÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ75%£®
µãÆÀ£º±¾ÌâÖ÷Òª¿¼²éѧÉúÔËÓû¯Ñ§·½³ÌʽºÍÖÊÁ¿·ÖÊý¹«Ê½½øÐмÆËãµÄÄÜÁ¦£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø