Question

1£®Ê¹ÓÃÃܶÈС¡¢Ç¿¶È´óµÄþºÏ½ðÄܼõÇáÆû³µ×ÔÖØ£¬´Ó¶ø¼õÉÙÆûÓÍÏûºÄºÍ·ÏÆøÅÅ·Å£®
£¨1£©Ã¾ÔªËØÔÚ×ÔÈ»½çÖÐÊÇÒÔ»¯ºÏÎÌî¡°µ¥ÖÊ¡±»ò¡°»¯ºÏÎ£©ÐÎʽ´æÔڵģ®
£¨2£©¹¤ÒµÖÆþµÄÒ»ÖÖÔ­ÀíÊÇ2MgO+Si+2CaO$\underline{\underline{1600¡ãC}}$2Mg¡ü+Ca₂SiO₄£¬³é×ßÈÝÆ÷ÖеĿÕÆø¶Ô·´Ó¦Óдٽø×÷Óã¬ÇÒÄÜ·ÀÖ¹¿ÕÆøÖеÄÎïÖÊÓëMg·´Ó¦Ê¹²úÆ·»ìÓÐMgO£¨Ìѧʽ£©£®
£¨3£©ÖÆþµÄÔ­ÁÏMgO¿É´Óº£Ë®ÖлñµÃ£®Ð¡»ªÀûÓÃþÓëÑÎËá·´Ó¦ºóµÄ·ÏÒº£¬Ä£Äâ´Óº£Ë®ÖлñÈ¡MgOµÄ¹ý³Ì£¬ÊµÑéÈçÏ£º
²½Öè1£ºÏòÉÏÊö·ÏÒºÖУ¬±ß½Á°è±ß·ÖÅú¼ÓÈëCaO£¬ÖÁMgCl₂ÍêÈ«³ÁµíΪֹ£¬¹ýÂ˵ÃMg£¨OH£©₂¹ÌÌ壮¹²ÏûºÄ8.4g CaO£®
²½Öè2£º½«Mg£¨OH£©₂¹ÌÌå¼ÓÈÈ·Ö½âΪMgOºÍË®£¬ËùµÃMgOµÄÖÊÁ¿Îª4.0g£®
¢Ù²½Öè2ÖÐMg£¨OH£©₂·Ö½âµÄ»¯Ñ§·½³ÌʽΪMg£¨OH£©₂$\frac{\underline{\;\;¡÷\;\;}}{\;}$MgO+H₂O£®
¢Úͨ¹ýËùµÃMgOµÄÖÊÁ¿¼ÆËãÉÏÊö·ÏÒºÖк¬MgCl₂µÄÖÊÁ¿m=9.5g£»
¢Û·ÖÎöʵÑéÊý¾Ý£¬¿ÉÖª²½Öè1Öз¢ÉúµÄ»¯Ñ§·´Ó¦ÓУº
CaO+H₂O=Ca£¨OH£©₂£»
Ca£¨OH£©₂+MgCl₂=CaCl₂+Mg£¨OH£©₂¡ý£»Ca£¨OH£©₂+2HCl=CaCl₂+2H₂O£¨»òCaO+2HCl=CaCl₂+H₂O£©£®

Answer 1

·ÖÎö £¨1£©¸ù¾ÝÖ»Óнð¡¢ÒøµÈÉÙÊýµÄ»¯Ñ§ÐÔÖʷdz£Îȶ¨µÄ½ðÊô£¬ÔÚ×ÔÈ»½çÖÐÖ÷ÒªÊÇÒÔµ¥ÖÊÐÎʽ´æÔÚ½øÐнâ´ð£»
£¨2£©Ã¾¿ÉÒÔÓëÑõÆø·´Ó¦Éú³ÉÑõ»¯Ã¾£»
£¨3£©Mg£¨OH£©₂¹ÌÌå¼ÓÈÈ·Ö½âΪMgOºÍË®£»¸ù¾Ý·½³ÌʽµÄ·´Ó¦¹Øϵ½øÐÐÏà¹ØµÄ¼ÆËã¼°Åжϣ®

½â´ð ½â£º£¨1£©Ã¾µÄ»¯Ñ§ÐÔÖʽϻîÆã¬Ã¾ÔªËØÔÚ×ÔÈ»½çÖÐÊÇÒÔ»¯ºÏÎïÐÎʽ´æÔÚ£»
£¨2£©Ã¾¿ÉÒÔÓëÑõÆø·´Ó¦Éú³ÉÑõ»¯Ã¾£¬²úÆ·»ìÓÐMgO£»
£¨3£©¢ÙMg£¨OH£©₂¹ÌÌå¼ÓÈÈ·Ö½âΪMgOºÍË®£¬·½³ÌʽΪ£ºMg£¨OH£©₂$\frac{\underline{\;\;¡÷\;\;}}{\;}$MgO+H₂O£»
¢ÚÓÉCaO+H₂O¨TCa£¨OH£©₂£»Ca£¨OH£©₂+MgCl₂¨TCaCl₂+Mg£¨OH£©₂¡ý£»Mg£¨OH£©₂$\frac{\underline{\;\;¡÷\;\;}}{\;}$MgO+H₂O  ¿ÉµÃMgOÓëMgCl₂¹ØϵʽΪ£ºMgCl₂¡«MgO
Éè·ÏÒºÖк¬MgCl₂µÄÖÊÁ¿Îªx
MgCl₂¡«MgO
95     40
x      4.0g
$\frac{95}{x}$=$\frac{40}{4.0g}$
x=9.5g
¢ÛÓÉCaO+H₂O¨TCa£¨OH£©₂£»Ca£¨OH£©₂+MgCl₂¨TCaCl₂+Mg£¨OH£©₂¡ý£»¿ÉµÃCaOÓëMgCl₂¹ØϵʽΪ£ºMgCl₂¡«CaO
ÉèÏûºÄÑõ»¯¸ÆÖÊÁ¿Îªy£®
MgCl₂¡«CaO
95    56
9.5g   y
$\frac{95}{9.5g}$=$\frac{56}{y}$
y=5.6g
ʵ¼ÊÏûºÄ8.4g CaO£¬ËµÃ÷·ÏÒºÖÐÓйýÁ¿µÄÑÎËᣬ·¢ÉúÁËCaO+2HCl¨TCaCl₂+H₂O£®
¹Ê´ð°¸Îª£º£¨1£©»¯ºÏÎï¡¡£¨2£©MgO  £¨3£©¢ÙMg£¨OH£©₂$\frac{\underline{\;\;¡÷\;\;}}{\;}$MgO+H₂O  ¢Ú9.5¡¡¢ÛCa£¨OH£©₂+2HCl=CaCl₂+2H₂O£¨»òCaO+2HCl=CaCl₂+H₂O£©

µãÆÀ ±¾ÌâÄѶȽϴ󣬿¼²éº£Ë®×ÊÔ´µÄÀûÓã¬Ñ§ÉúÓ¦ÔÚÎïÖÊת»¯ÖÐӦѧ»áÀûÓû¯Ñ§·½³ÌʽµÄ¼ÆËãÀ´·ÖÎöÎÊÌ⣮