ÌâÄ¿ÄÚÈÝ
½ðÊô¸ÆÏßÊÇÁ¶ÖÆÓÅÖʸֲĵÄÍÑÑõÍÑÁ×¼Á£¬Ä³¸ÆÏßµÄÖ÷Òª³É·ÖΪ½ðÊôMºÍCa£¬²¢º¬ÓÐ3. 5%£¨ÖÊÁ¿·ÖÊý£©
CaO¡£
(1)CaÔªËØÔÚÖÜÆÚ±íÖÐλÖÃÊÇ___________¡£
(2)CaÓë×î»îÆÃµÄ·Ç½ðÊôÔªËØAÐγɻ¯ºÏÎïD£¬DµÄµç×ÓʽΪ______________£¬DµÄ·Ðµã±ÈAÓëSiÐγɵϝºÏÎïEµÄ·Ðµã___________¡£
(3)Å䯽ÓøÆÏßÍÑÑõÍÑÁ׵Ļ¯Ñ§·½³Ìʽ
___P+ ___FeO+___CaO
___Ca3(PO4)2+___Fe
(4)½«¸ÆÏßÊÔÑùÈÜÓÚÏ¡ÑÎËáºó£¬¼ÓÈë¹ýÁ¿NaOHÈÜÒº£¬Éú³É°×É«Ðõ×´³Áµí²¢Ñ¸ËÙ±ä³É»ÒÂÌÉ«£¬×îºó±ä³ÉºìºÖÉ«M(OH)n£¬Ôò¼ì²âMn+µÄÊÔ¼ÁÊÇ______________¡£
(5)È¡1.6 g¸ÆÏßÊÔÑù£¬ÓëË®³ä·Ö·´Ó¦£¬Éú³É224 mL H2£¨±ê×¼×´¿ö£©£¬ÔÙÏòÈÜÒºÖÐͨÈëÊÊÁ¿µÄCO2£¬×î¶àÄܵõ½CaCO3____g¡£
CaO¡£
(1)CaÔªËØÔÚÖÜÆÚ±íÖÐλÖÃÊÇ___________¡£
(2)CaÓë×î»îÆÃµÄ·Ç½ðÊôÔªËØAÐγɻ¯ºÏÎïD£¬DµÄµç×ÓʽΪ______________£¬DµÄ·Ðµã±ÈAÓëSiÐγɵϝºÏÎïEµÄ·Ðµã___________¡£
(3)Å䯽ÓøÆÏßÍÑÑõÍÑÁ׵Ļ¯Ñ§·½³Ìʽ
___P+ ___FeO+___CaO
(4)½«¸ÆÏßÊÔÑùÈÜÓÚÏ¡ÑÎËáºó£¬¼ÓÈë¹ýÁ¿NaOHÈÜÒº£¬Éú³É°×É«Ðõ×´³Áµí²¢Ñ¸ËÙ±ä³É»ÒÂÌÉ«£¬×îºó±ä³ÉºìºÖÉ«M(OH)n£¬Ôò¼ì²âMn+µÄÊÔ¼ÁÊÇ______________¡£
(5)È¡1.6 g¸ÆÏßÊÔÑù£¬ÓëË®³ä·Ö·´Ó¦£¬Éú³É224 mL H2£¨±ê×¼×´¿ö£©£¬ÔÙÏòÈÜÒºÖÐͨÈëÊÊÁ¿µÄCO2£¬×î¶àÄܵõ½CaCO3____g¡£
(1)µÚËÄÖÜÆÚ¢òA×å
(2)
£»¸ß
(3)2P+5FeO+3CaO
1Ca3(PO4)2+5Fe
(4)KSCN(»òNaOH»òÆäËûºÏÀí´ð°¸)
(5)1.1
(2)
(3)2P+5FeO+3CaO
(4)KSCN(»òNaOH»òÆäËûºÏÀí´ð°¸)
(5)1.1
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿