ÌâÄ¿ÄÚÈÝ
£¨2013?ãÉÐÐÇøһģ£©¼¦µ°¿ÇµÄÖ÷Òª³É·ÖÊÇ̼Ëá¸Æ£¬Îª²â¶¨µ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£¬Ð¡ÀöºÍСæÃͬѧ½øÐÐÁËÈçÏÂʵÑ飮Çë»Ø´ðÏà¹ØÎÊÌ⣺
£¨1£©ÊµÑé¹ý³ÌºÍ²â¶¨µÄÏà¹ØʵÑéÊý¾ÝÈçÏÂËùʾ£º£¨Ìáʾ£º²ÐÔü²»ÓëÑÎËá·´Ó¦£©
¢Ù³ä·Ö·´Ó¦ºó£¬Ð¡Àö¸ù¾Ý¡°·´Ó¦¹ý³ÌÖмõÉÙµÄÖÊÁ¿¡±ÇóµÃÉú³ÉµÄCO2µÄÎïÖʵÄÁ¿ÊÇ
¢ÚСæøù¾Ý¡°µ°¿Ç²ÐÔüÖÊÁ¿Îª4g¡±£¬ÇóµÃµ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ
£¨2£©Ð¡ÀöºÍСæÃÇóµÃµÄµ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýÓвî±ð£¬ÇëÄã·ÖÎöÆäÖпÉÄܵÄÔÒò£®
£¨1£©ÊµÑé¹ý³ÌºÍ²â¶¨µÄÏà¹ØʵÑéÊý¾ÝÈçÏÂËùʾ£º£¨Ìáʾ£º²ÐÔü²»ÓëÑÎËá·´Ó¦£©
¢Ù³ä·Ö·´Ó¦ºó£¬Ð¡Àö¸ù¾Ý¡°·´Ó¦¹ý³ÌÖмõÉÙµÄÖÊÁ¿¡±ÇóµÃÉú³ÉµÄCO2µÄÎïÖʵÄÁ¿ÊÇ
0.1
0.1
mol£»È»ºó¸ù¾ÝCO2µÄÁ¿ÇóµÃµ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£¬ÇëÄãд³öСÀöµÄ¼ÆËã¹ý³ÌºÍ½á¹û£®£¨¾«È·µ½0.1%£¬ÏÂͬ£©¢ÚСæøù¾Ý¡°µ°¿Ç²ÐÔüÖÊÁ¿Îª4g¡±£¬ÇóµÃµ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ
66.7%
66.7%
£»£¨2£©Ð¡ÀöºÍСæÃÇóµÃµÄµ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýÓвî±ð£¬ÇëÄã·ÖÎöÆäÖпÉÄܵÄÔÒò£®
·ÖÎö£º£¨1£©¸ù¾Ý¼¦µ°¿ÇµÄÖ÷Òª³É·ÖÊÇ̼Ëá¸Æ£¬ÓëËá·´Ó¦ºóµ°¿Ç²ÐÔüÖÊÁ¿Îª4.3g£¬Òò̼Ëá¸ÆÄÜÓëËá·´Ó¦£¬Ôò¿É¼ÆËã̼Ëá¸ÆµÄÖÊÁ¿£¬ÔÙ¼ÆËãµ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£»
Ò²¿ÉÀûÓÃÖÊÁ¿Êغ㶨ÂÉ£¬ÀûÓ÷´Ó¦Ç°ºó»ìºÏÎïµÄÖÊÁ¿²îÀ´¼ÆËãÆøÌå¶þÑõ»¯Ì¼µÄÖÊÁ¿£¬ÀûÓû¯Ñ§·´Ó¦·½³ÌʽÀ´¼ÆËã̼Ëá¸ÆµÄÖÊÁ¿£¬½øÒ»²½¼ÆËãµ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£®
£¨2£©¸ù¾ÝʵÑéÖйÌÌåÖÊÁ¿µÄ³ÆÁ¿¼°ÆøÌåµÄÊÕ¼¯µÈ·½ÃæÀ´·ÖÎöÆä²î±ð£®
Ò²¿ÉÀûÓÃÖÊÁ¿Êغ㶨ÂÉ£¬ÀûÓ÷´Ó¦Ç°ºó»ìºÏÎïµÄÖÊÁ¿²îÀ´¼ÆËãÆøÌå¶þÑõ»¯Ì¼µÄÖÊÁ¿£¬ÀûÓû¯Ñ§·´Ó¦·½³ÌʽÀ´¼ÆËã̼Ëá¸ÆµÄÖÊÁ¿£¬½øÒ»²½¼ÆËãµ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£®
£¨2£©¸ù¾ÝʵÑéÖйÌÌåÖÊÁ¿µÄ³ÆÁ¿¼°ÆøÌåµÄÊÕ¼¯µÈ·½ÃæÀ´·ÖÎöÆä²î±ð£®
½â´ð£º½â£º£¨1£©¢Ù³ä·Ö·´Ó¦ºó£¬Ð¡Àö¸ù¾Ý¡°·´Ó¦¹ý³ÌÖмõÉÙµÄÖÊÁ¿¡±ÇóµÃÉú³ÉµÄCO2µÄÖÊÁ¿ÊÇ12.0g+100.0g-107.6g=4.4g£¬CO2µÄÎïÖʵÄÁ¿=
=0.1mol
Éè̼Ëá¸ÆµÄÖÊÁ¿Îªx£¬
CaCO3+2HCl=CaCl2+H2O+CO2¡ü
100 44
x 4.4g
=
x=10.0g
Ôò̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý=
¡Á100%=83.3%
´ð£ºÌ¼Ëá¸ÆµÄÖÊÁ¿·ÖÊý83.3%£®
¢Úµ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý=
¡Á100%=66.7%£»
£¨2£©Ð¡Àö¼ÆËãËùÓõĹÌÌå²ÐÔü¿ÉÄÜδ¸ÉÔʵ¼Ê¹ÌÌå²ÐÔüÖÊÁ¿Ð¡ÓÚ4g£¨»òСæüÆËãËùÓõÄÊý¾Ý¡°4.4g¡±Öаüº¬ÁË·´Ó¦¹ý³ÌÖÐÑÎËá»Ó·¢µÄÂÈ»¯ÇâÆøÌåµÄÖÊÁ¿£¬¼´Êµ¼ÊÉú³ÉµÄ¶þÑõ»¯Ì¼ÆøÌåÖÊÁ¿Ð¡ÓÚ4.4g£©£¬ËùÒÔСÀöºÍСæÃÇóµÃµÄµ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýÓвî±ð£®
¹Ê´ð°¸Îª£º£¨1£©¢Ù0.1mol£»83.3%£»¢Ú66.7%£®£¨2£©Ð¡Àö¼ÆËãËùÓõĹÌÌå²ÐÔü¿ÉÄÜδ¸ÉÔʵ¼Ê¹ÌÌå²ÐÔüÖÊÁ¿Ð¡ÓÚ4g£¨»òСæüÆËãËùÓõÄÊý¾Ý¡°4.4g¡±Öаüº¬ÁË·´Ó¦¹ý³ÌÖÐÑÎËá»Ó·¢µÄÂÈ»¯ÇâÆøÌåµÄÖÊÁ¿£¬¼´Êµ¼ÊÉú³ÉµÄ¶þÑõ»¯Ì¼ÆøÌåÖÊÁ¿Ð¡ÓÚ4.4g£©£®
4.4g |
44g/mol |
Éè̼Ëá¸ÆµÄÖÊÁ¿Îªx£¬
CaCO3+2HCl=CaCl2+H2O+CO2¡ü
100 44
x 4.4g
100 |
x |
44 |
4.4g |
x=10.0g
Ôò̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý=
10g |
12g |
´ð£ºÌ¼Ëá¸ÆµÄÖÊÁ¿·ÖÊý83.3%£®
¢Úµ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý=
12g-4g |
12g |
£¨2£©Ð¡Àö¼ÆËãËùÓõĹÌÌå²ÐÔü¿ÉÄÜδ¸ÉÔʵ¼Ê¹ÌÌå²ÐÔüÖÊÁ¿Ð¡ÓÚ4g£¨»òСæüÆËãËùÓõÄÊý¾Ý¡°4.4g¡±Öаüº¬ÁË·´Ó¦¹ý³ÌÖÐÑÎËá»Ó·¢µÄÂÈ»¯ÇâÆøÌåµÄÖÊÁ¿£¬¼´Êµ¼ÊÉú³ÉµÄ¶þÑõ»¯Ì¼ÆøÌåÖÊÁ¿Ð¡ÓÚ4.4g£©£¬ËùÒÔСÀöºÍСæÃÇóµÃµÄµ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýÓвî±ð£®
¹Ê´ð°¸Îª£º£¨1£©¢Ù0.1mol£»83.3%£»¢Ú66.7%£®£¨2£©Ð¡Àö¼ÆËãËùÓõĹÌÌå²ÐÔü¿ÉÄÜδ¸ÉÔʵ¼Ê¹ÌÌå²ÐÔüÖÊÁ¿Ð¡ÓÚ4g£¨»òСæüÆËãËùÓõÄÊý¾Ý¡°4.4g¡±Öаüº¬ÁË·´Ó¦¹ý³ÌÖÐÑÎËá»Ó·¢µÄÂÈ»¯ÇâÆøÌåµÄÖÊÁ¿£¬¼´Êµ¼ÊÉú³ÉµÄ¶þÑõ»¯Ì¼ÆøÌåÖÊÁ¿Ð¡ÓÚ4.4g£©£®
µãÆÀ£º±¾Ì⿼²éѧÉúÀûÓû¯Ñ§·´Ó¦·½³ÌʽµÄ¼ÆË㣬Ã÷È··´Ó¦ºó²ÐÔüµÄÖÊÁ¿¼°ÀûÓÃÖÊÁ¿ÊغãÀ´¼ÆËã¶þÑõ»¯Ì¼µÄÖÊÁ¿Êǽâ´ðµÄ¹Ø¼ü£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿