ÌâÄ¿ÄÚÈÝ

(10·Ö)¹¤ÒµÉÏÓÃÂÁÍÁ¿ó(º¬Ñõ»¯ÂÁ¡¢Ñõ»¯Ìú)ÖÆÈ¡ÂÁµÄ¹ý³ÌÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)³ÁµíCµÄ»¯Ñ§Ê½Îª________£¬¸ÃÎïÖʳýÁËÓÃÓÚ½ðÊôÒ±Á¶ÒÔÍ⣬»¹¿ÉÓÃ×÷________¡£

(2)µç½âÈÛÈÚµÄÑõ»¯ÂÁʱ£¬ÈôµÃµ½±ê×¼×´¿öÏÂ22.4 L O2£¬ÔòͬʱÉú³É________gÂÁ¡£

(3)²Ù×÷¢ñ¡¢²Ù×÷¢òºÍ²Ù×÷¢ó¶¼ÊÇ________(Ìî²Ù×÷Ãû³Æ)£¬ÊµÑéÊÒҪϴµÓAl(OH)3³ÁµíÓ¦¸ÃÔÚ________×°ÖÃÖнøÐУ¬Ï´µÓ·½·¨ÊÇ______________________________________

________________________________________________________________________¡£

(4)Éú²ú¹ý³ÌÖУ¬³ýNaOH¡¢H2O¿ÉÒÔÑ­»·Ê¹ÓÃÍ⣬»¹¿ÉÒÔÑ­»·Ê¹ÓõÄÎïÖÊÓÐ________(Ìѧʽ)¡£Óô˷¨ÖÆÈ¡ÂÁµÄ¸±²úÆ·ÊÇ________(Ìѧʽ)¡£

(5)д³öNa2CO3ÈÜÒºÓëCaO·´Ó¦µÄÀë×Ó·½³Ìʽ£º________________________________¡£

(6)ÈôÂÁÍÁ¿óÖл¹º¬ÓжþÑõ»¯¹è£¬´ËÉú²ú¹ý³ÌÖеõ½µÄÑõ»¯ÂÁ½«»ìÓÐÔÓÖÊ£º________(Ìѧʽ)¡£

 

¡¾´ð°¸¡¿

(1)Fe2O3¡¡ÑÕÁÏ¡¡(2)36¡¡(3)¹ýÂË¡¡¹ýÂË¡¡Ïò©¶·ÖмÓÕôÁóË®ÖÁ½þû³Áµí£¬Ê¹Ë®×ÔÈ»Á÷Íê£¬ÖØ¸´²Ù×÷2¡«3´Î

(4)CaOºÍCO2¡¡Fe2O3ºÍO2

(5)CO32¡ª£«CaO£«H2O===CaCO3¡ý£«2OH£­

(6)SiO2

¡¾½âÎö¡¿(1)ÒòAl2O3£«2NaOH===2NaAlO2£«H2O£¬¶øFe2O3²»ÓëNaOHÈÜÒº·¢Éú·´Ó¦£¬Òò´Ë¹ýÂ˵õ½µÄ³ÁµíCΪFe2O3£¬³ýÀûÓÃFe2O3·¢ÉúÂÁÈÈ·´Ó¦ÖÆÈ¡FeÍ⣬»¹¿É×÷ÑÕÁÏ¡£

(2)2Al2O3(ÈÛÈÚ)4Al£«3O2¡ü£¬n(O2)£½22.4/22.4£½1 mol£¬Ôòm(Al)£½4/3mol¡Á27 g/mol£½36 g¡£

(3)²Ù×÷¢ñ¡¢¢ò¡¢¢ó¾ùÊdzýÈ¥²»ÈÜÓÚÒºÌåµÄ¹ÌÌ壬¼´¹ýÂË£¬Ï´µÓ·½·¨ÊÇÏò©¶·ÖмÓÕôÁóË®ÖÁ½þû³Áµí£¬Ê¹Ë®×ÔÈ»Á÷Íê£¬ÖØ¸´²Ù×÷2¡«3´Î¡£

(4)¾ÝÁ÷³Ìͼ¿É¿´³öÑ­»·Ê¹ÓõÄÎïÖʳýNaOH¡¢H2OÍ⣬»¹ÓÐCaCO3CaO£«CO2¡ü£¬CaOÓëCO2Ò²¿ÉÑ­»·Ê¹Óã»ÖÆÈ¡AlͬʱµÃµ½Fe2O3ºÍO2¡£

(5)·¢ÉúµÄ»¯Ñ§·´Ó¦ÎªNa2CO3£«CaO£«H2O===CaCO3¡ý£«2NaOH¡£

(6)¾ÝSiO2ÐÔÖÊ£¬SiO2£«2NaOH===Na2SiO3£«H2O£¬Na2SiO3£«CO2£«H2O===H2SiO3¡ý£«Na2CO3£¬H2SiO3SiO2£«H2O£¬Òò´ËAl2O3Öн«»ìÓÐSiO2¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø