Question

1£®Ð¿¼°ÆäÑõ»¯ÎZnO£¬°×É«£¬ÄÑÈÜÓÚË®£¬ÈÜÓÚÇ¿Ëᣩ¾ßÓй㷺µÄÓ¦Óã®
£¨Ò»£©Ð¿µÄ¹ã·ºÓ¦ÓÃ
£¨1£©Ð¿±»³Æ֮Ϊ¡°ÉúÃüÔªËØ¡±£®¶ùͯ¡¢ÇàÉÙÄêÈç¹ûȱпÑÏÖØ£¬½«»áµ¼Ö¡°ÙªÈå Ö¢¡±ºÍÖÇÁ¦·¢Óý²»Á¼£®¾­³£³Ô±´¿ÇÀຣ²úÆ·¡¢ºìÉ«ÈâÀàµÈº¬Ð¿µÄʳÎïÓÐÀû²¹¡°Ð¿¡±£®ÕâÀïµÄ¡°Ð¿¡±ÊÇÖ¸B£®
A£®µ¥ÖÊ           B£®ÔªËØ          C£®Ô­×Ó         D£®·Ö×Ó
£¨2£©ÊµÑéÊÒÓÃпÓëÏ¡ÁòËáÖÆÈ¡ÇâÆø£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪZn+H₂SO₄¨TZnSO₄+H₂¡ü£®
£¨3£©Ð¿ÓëÍ­ÈۺϿÉÐγɡ°»ÆÍ­¡±£¬ËüÊôÓÚA£¨Ñ¡ÌîÐòºÅ£©£®
A£®½ðÊô²ÄÁÏ      B£®ÎÞ»ú·Ç½ðÊô²ÄÁÏ      C£®ºÏ³É²ÄÁÏ      D£®¸´ºÏ²ÄÁÏ
£¨¶þ£©»îÐÔZnOÖÆÈ¡µÄ̽¾¿
¡¾ÊµÑé·½°¸¡¿¹¤ÒµÉÏÓôÖÑõ»¯Ð¿£¨º¬ÉÙÁ¿FeO£©ÖÆÈ¡»îÐÔÑõ»¯Ð¿£¬ÆäÁ÷³ÌÈçÏ£º

³ÁµíÎï	Fe£¨OH£©3	Zn£¨OH£©2	Fe£¨OH£©2
¿ªÊ¼³ÁµípH	1.5	6.2	6.3
ÍêÈ«³ÁµípH	3.2	8.0	9.7

¡¾²éÔÄ×ÊÁÏ¡¿Ò»Ð©ÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽ¿ªÊ¼³Áµí¡¢ÍêÈ«³ÁµíʱÈÜÒºµÄpH¼ûÉÏ±í£®
¡¾ÎÊÌâ̽¾¿¡¿
£¨4£©¡°Èܽ⡱ǰ½«´ÖÑõ»¯Ð¿·ÛËé³Éϸ¿ÅÁ££¬Ä¿µÄÊÇÔö´ó½Ó´¥Ãæ»ý£¬Ê¹Æä³ä·Ö·´Ó¦£®
£¨5£©¡°Èܽ⡱ºóµÃµ½µÄËáÐÔÈÜÒºÖк¬ÓÐZnSO₄¡¢H₂SO₄¡¢FeSO₄£®³ýÌú³ØÖмÓÈëÊÊÁ¿H₂O₂£¬Ê¹Fe²⁺ת»¯ÎªFe³⁺£»ÎªÊ¹ÈÜÒºÖÐFe³⁺È«²¿×ª»¯ÎªFe£¨OH£©₃£¬¶øZn²⁺²»ÐγÉZn£¨OH£©₂£¬ÔòÓ¦¼ÓÈ백ˮ¿ØÖÆÈÜÒºµÄpHµÄ·¶Î§Îª1.5¡«3.2£®
£¨6£©¡°³Áµí³Ø¡±ÖеijÁµí¾­¹ýÂË¡¢Ï´µÓµÈ²Ù×÷µÃ´¿¾»¹ÌÌåM£¬Æä×é³ÉΪ£ºaZnCO₃•bZn£¨OH£©₂•cH₂O£®
È·ÈϹÌÌåMÒѾ­Ï´µÓ¸É¾»µÄ²Ù×÷ÊÇ£ºÈ¡×îºóÒ»´ÎÏ´µÓÒº£¬×îºÃÑ¡ÔñÏÂÁÐC£¬
ÏòÆäÖеÎÈëÎÞÏÖÏó£¬ÔòÒѾ­Ï´¸É¾»£®
A£®×ÏɫʯÈïÊÔÒº     B£®ÉÙÁ¿Ï¡ÑÎËá    C£®ÂÈ»¯±µÈÜÒº»òÏõËá±µÈÜÒº¿ÕÆø
×é³É²â¶¨¹ÌÌåBµÄ×é³É»áÓ°ÏìÖƵõÄZnOµÄ»îÐÔ£®ÎªÈ·¶¨aZnCO₃•bZn£¨OH£©₂•cH₂OµÄ×é³É£¬½øÐÐÈçÏÂʵÑ飨¼ÙÉèÿ²½·´Ó¦¡¢ÎüÊÕ¾ùÍêÈ«£©£º

¡¾²éÔÄ×ÊÁÏ¡¿aZnCO₃•bZn£¨OH£©₂•cH₂OÊÜÈÈ·Ö½âÉú³ÉZnO¡¢H₂O¡¢CO₂ÈýÖÖ²úÎ
£¨7£©Ð´³ö×°ÖÃBÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽaZnCO₃•bZn£¨OH£©₂•cH₂O$\frac{\underline{\;\;¡÷\;\;}}{\;}$£¨a+b£©ZnO+£¨b+c£©H₂O+aCO₂¡ü£®
ZHENÏÖ³ÆÈ¡35.9g aZnCO₃•bZn£¨OH£©₂•cH₂OÔÚB×°ÖÃÖнøÐÐÍêÈ«ìÑÉÕ£¬²âµÃ×°ÖÃCºÍDµÄÖÊÁ¿·Ö±ðÔöÖØ7.2gºÍ4.4g£®
¡¾ÊµÑé·ÖÎö¼°Êý¾Ý´¦Àí¡¿
£¨8£©×°ÖÃAµÄ×÷ÓÃÊdzýÈ¥¿ÕÆøÖеÄH₂OºÍCO₂£®
£¨9£©¸ù¾ÝÉÏÊöÊý¾Ý£¬ÔòÉú³ÉZnOµÄÖÊÁ¿Îª24.3g£»a£ºb£ºc=1£º2£º2£®
£¨10£©ÈôûÓÐE×°Öã¬Ôò²âµÃµÄaֵƫ´ó£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°²»±ä¡±£©£®

Answer 1

·ÖÎö £¨1£©ÈËÌåÈç¹ûȱ·¦Ä³ÖÖÔªËØ£¬»á»¼¼²²¡£¬ÀýÈçȱ·¦¸ÆÔªËØʱÈÝÒ×»¼ØþÙͲ¡»ò¹ÇÖÊÊèËÉÖ¢£¬È±·¦µâÔªËØʱÈÝÒ×»¼¼××´ÏÙÖ×´óµÈ£»
ʳƷº¬Óеĸơ¢Ð¿¡¢Ìú¡¢µâµÈͨ³£ÊÇÖ¸ÔªËØ£»
£¨2£©¸ù¾Ý·´Ó¦Îï¡¢Éú³ÉÎïÊéд»¯Ñ§·½³Ìʽ£»
£¨3£©¸ù¾ÝÎïÖʵķÖÀà½øÐзÖÎö£»
£¨4£©´ÖÑõ»¯Ð¿·ÛËé³Éϸ¿ÅÁ£ÊÇÔö´ó½Ó´¥Ãæ»ý£¬¼Ó¿ì·´Ó¦ËÙÂÊ£¬Ìá¸ßÉú²úЧÂÊ£»
£¨5£©¸ù¾ÝÑõ»¯Ð¿ºÍÏ¡ÁòËá·´Ó¦Éú³ÉÁòËáпºÍË®½øÐзÖÎö£»
£¨6£©¸ù¾Ý̼Ëá¸ùÀë×ӵļìÑé·ÖÎö£»
£¨7£©¸ù¾Ý·´Ó¦Îï¡¢Éú³ÉÎïÊéд»¯Ñ§·½³Ìʽ£»
£¨8£©¸ù¾ÝAÄÜÎüÊÕ¿ÕÆøÖеÄË®ºÍ¶þÑõ»¯Ì¼½â´ð£»
£¨9£©¸ù¾ÝCÎüÊÕµÄÊÇË®£¬DÎüÊÕµÄÊǶþÑõ»¯Ì¼£¬¼îʽ̼ËáпÑùÆ·35.9g£¬×°ÖÃCÔö¼ÓµÄÖÊÁ¿Îª7.2g£¬¹ÊÉú³ÉË®µÄÖÊÁ¿Îª7.2g£»×°ÖÃDÔö¼ÓµÄÖÊÁ¿Îª4.4g£¬Éú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿Îª4.4g£¬¹ÊÑõ»¯Ð¿µÄÖÊÁ¿Îª35.9g-7.2g-4.4g=24.3g£¬¼ÆËã¸÷ÎïÖʵÄÎïÖʵÄÁ¿£¬ÀûÓÃÔªËØÊغã¼ÆËã¼îʽ̼ËáпÖÐa¡¢b¡¢c£»
£¨10£©¸ù¾ÝEÄÜÎüÊÕ¿ÕÆøÖеÄË®ºÍ¶þÑõ»¯Ì¼½â´ð£®

½â´ð ½â£º£¨1£©ÈËÌåȱÉÙпԪËػᵼÖ¶ùͯ·¢Óý³Ù»º£¬ÖÇÁ¦µÍÏ£¬µ¼ÖÂÙªÈåÖ¢£»Ê³Æ·º¬Óеĸơ¢Ð¿¡¢Ìú¡¢µâµÈͨ³£ÊÇÖ¸ÔªËØ£¬¹ÊÌ٪È壻B£»
£¨2£©ÊµÑéÊÒÓÃпÓëÏ¡ÁòËáÖÆÈ¡ÇâÆø£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºZn+H₂SO₄¨TZnSO₄+H₂¡ü£»¹ÊÌZn+H₂SO₄¨TZnSO₄+H₂¡ü£»
£¨3£©¸ù¾ÝÎïÖʵķÖÀ࣬ºÏ½ðÊôÓÚ½ðÊô²ÄÁÏ£¬¹ÊÑ¡A£»
£¨4£©½«´ÖÑõ»¯Ð¿·ÛËé³Éϸ¿ÅÁ£ÊÇÔö´ó½Ó´¥Ãæ»ý£¬¼Ó¿ì·´Ó¦ËÙÂÊ£¬Ìá¸ßÉú²úЧÂÊ£»¹Ê´ð°¸£ºÔö´ó½Ó´¥Ãæ»ý£¬Ê¹Æä³ä·Ö·´Ó¦£»
£¨5£©Ñõ»¯Ð¿ºÍÏ¡ÁòËá·´Ó¦Éú³ÉÁòËáпºÍË®£¬FeOÓëÁòËá·´Ó¦Éú³ÉÁòËáÑÇÌú£¬ËùÒÔ¡°Èܽ⡱ºóµÃµ½µÄËáÐÔÈÜÒºÖк¬Óз´Ó¦Éú³ÉµÄÁòËáп¡¢ÁòËáÑÇÌúºÍ¹ýÁ¿µÄÁòËᣬ´Ó±íÖпÉÒÔ¿´³ö£¬ÔÚpH¿ØÖÆÔÚ1.5¡«3.2µÄ·¶Î§ÄÚ£¬Ö»Éú³ÉÇâÑõ»¯Ìú³Áµí£¬¹ÊÌH₂SO₄£»1.5¡«3.2£»
£¨6£©È·ÈϹÌÌåMÒѾ­Ï´µÓ¸É¾»µÄ²Ù×÷ÊÇ£ºÈ¡×îºóÒ»´ÎÏ´µÓÒº£¬ÏòÆäÖеÎÈëÂÈ»¯±µÈÜÒº»òÏõËá±µÈÜÒºÎÞÏÖÏó£¬ÔòÒѾ­Ï´¸É¾»£¬¹ÊÑ¡C£»
£¨7£©aZnCO₃•bZn£¨OH£©₂•cH₂OÊÜÈÈ·Ö½âÉú³ÉZnO¡¢H₂O¡¢CO₂ÈýÖÖ²úÎﻯѧ·½³Ìʽ£ºaZnCO₃•bZn£¨OH£©₂•cH₂O$\frac{\underline{\;\;¡÷\;\;}}{\;}$£¨a+b£©ZnO+£¨b+c£© H₂O+aCO₂¡ü£»
£¨8£©×°ÖÃAÄÜÎüÊÕ¿ÕÆøÖеÄË®ºÍ¶þÑõ»¯Ì¼£¬Åųý¿ÕÆøÖеÄË®ºÍ¶þÑõ»¯Ì¼µÄ¸ÉÈÅ£»¹ÊÌÎüÊÕ¿ÕÆøÖеÄË®ºÍ¶þÑõ»¯Ì¼£»
£¨9£©CÎüÊÕµÄÊÇË®£¬DÎüÊÕµÄÊǶþÑõ»¯Ì¼£¬¼îʽ̼ËáпÑùÆ·35.9g£¬×°ÖÃCÔö¼ÓµÄÖÊÁ¿Îª7.2g£¬¹ÊÉú³ÉË®µÄÖÊÁ¿Îª7.2g£»×°ÖÃDÔö¼ÓµÄÖÊÁ¿Îª4.4g£¬Éú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿Îª4.4g£¬¹ÊÑõ»¯Ð¿µÄÖÊÁ¿Îª35.9g-7.2g-4.4g=24.3g£¬¸ù¾Ý»¯Ñ§·½³ÌʽaZnCO₃•bZn£¨OH£©₂•cH₂O$\frac{\underline{\;\;¡÷\;\;}}{\;}$£¨a+b£©ZnO+£¨b+c£© H₂O+aCO₂¡ü£»
n£¨H₂O£©=$\frac{7.2g}{18g/mol}$=0.4mol£¬n£¨CO₂£©=$\frac{4.4g}{44g/mol}$=0.1mol£¬n£¨ZnO£©=$\frac{24.3g}{81g/mol}$=0.3mol£¬£¨a+b£©£º£¨b+c£©£ºa=0.3mol£º0.4mol£º0.1mol=3£º4£º1£¬Ôòa£ºb£ºc=1£º2£º2£»¹ÊÌ24.3£»1£º2£º2£»
£¨10£©E´¦Ó¦¸ÃÌí¼ÓÒ»¸öÊ¢ÓеĸÉÔï¹Ü£¬¼îʯ»ÒµÄ×÷ÓÃÊÇ·ÀÖ¹¿ÕÆøÖеĶþÑõ»¯Ì¼ºÍË®ÕôÆø½øÈë×°ÖÃD£¬ÄÇôװÖÃDÎüÊÕÁËÒ»²¿·Ö¿ÕÆøÖеĶþÑõ»¯Ì¼£¬ÔöÖØ´óÓÚʵ¼Ê·Ö½âµÃµ½µÄ¶þÑõ»¯Ì¼£¬Ê¹ÊµÑé½á¹ûÆ«´ó£»¹ÊÌƫ´ó£®
¹Ê´ð°¸Îª£º£¨1£©ÙªÈ壬B£»
£¨2£©Zn+H₂SO₄¨TZnSO₄+H₂¡ü£»
£¨3£©A£»
£¨4£©Ôö´ó½Ó´¥Ãæ»ý£¬Ê¹Æä³ä·Ö·´Ó¦£»
£¨5£©H₂SO₄£¬1.5¡«3.2£»
£¨6£©C£»
£¨7£©aZnCO₃•bZn£¨OH£©₂•cH₂O$\frac{\underline{\;\;¡÷\;\;}}{\;}$£¨a+b£©ZnO+£¨b+c£© H₂O+aCO₂¡ü£»
£¨8£©³ýÈ¥¿ÕÆøÖеÄH₂OºÍCO₂£»
£¨9£©24.3£»1£º2£º2£»
£¨10£©Æ«´ó£®

µãÆÀ ±¾Ì⿼²éѧÉú¶Ô¹¤ÒÕÁ÷³ÌµÄÀí½â¡¢ÔĶÁ»ñÈ¡ÐÅÏ¢ÄÜÁ¦¡¢¶Ô²Ù×÷²½ÖèµÄ·ÖÎöÆÀ¼Û¡¢ÎïÖʵķÖÀëÌá´¿¡¢Ñõ»¯Ñ§¼ÆËãµÈ£¬ÊǶÔËùѧ֪ʶµÄ×ÛºÏÔËÓÃÓëÄÜÁ¦µÄ¿¼²é£¬ÐèҪѧÉú¾ß±¸ÔúʵµÄ»ù´¡ÖªÊ¶Óë×ÛºÏÔËÓÃ֪ʶ¡¢ÐÅÏ¢½øÐнâ¾öÎÊÌâµÄÄÜÁ¦£®