ÌâÄ¿ÄÚÈÝ
ÒÔ¸»º¬ÁòËáÑÇÌúµÄ¹¤Òµ·ÏҺΪÔÁÏÉú²úÑõ»¯ÌúµÄ¹¤ÒÕÈçÏÂ(²¿·Ö²Ù×÷ºÍÌõ¼þÂÔ)£º
¢ñ.´Ó·ÏÒºÖÐÌá´¿²¢½á¾§³öFeSO47H2O¡£
¢ò.½«FeSO47H2OÅäÖƳÉÈÜÒº¡£
¢ó.FeSO4ÈÜÒºÓëÉÔ¹ýÁ¿µÄNH4HCO3ÈÜÒº»ìºÏ£¬µÃµ½º¬FeCO3µÄ×ÇÒº¡£
¢ô.½«×ÇÒº¹ýÂË£¬ÓÃ90¡æÈÈˮϴµÓ³Áµí£¬¸ÉÔïºóµÃµ½FeCO3¹ÌÌå¡£
¢õ.ìÑÉÕFeCO3£¬µÃµ½Fe2O3¹ÌÌå¡£
ÒÑÖª£ºNH4HCO3ÔÚÈÈË®Öзֽ⡣
(1)¢ñÖУ¬¼Ó×ãÁ¿µÄÌúм³ýÈ¥·ÏÒºÖеÄFe3+£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_________________¡£
(2)¢òÖУ¬Ðè¼ÓÒ»¶¨Á¿ÁòËá¡£ÔËÓû¯Ñ§Æ½ºâÔÀí¼òÊöÁòËáµÄ×÷ÓÃ_____________¡£
(3)¢óÖУ¬Éú³ÉFeCO3µÄÀë×Ó·½³ÌʽÊÇ_____________¡£ÈôFeCO3×ÇÒº³¤Ê±¼ä±©Â¶ÔÚ¿ÕÆøÖУ¬»áÓв¿·Ö¹ÌÌå±íÃæ±äΪºìºÖÉ«£¬¸Ã±ä»¯µÄ»¯Ñ§·½³ÌʽÊÇ_____________¡£
(4)¢ôÖУ¬Í¨¹ý¼ìÑéSO42-À´ÅжϳÁµíÊÇ·ñÏ´µÓ¸É¾»¡£¼ìÑéSO42-µÄ²Ù×÷ÊÇ_____________¡£
(5)ÒÑÖªìÑÉÕFeCO3µÄ»¯Ñ§·½³ÌʽÊÇ4FeCO3+O22Fe2O3+4CO2¡£ÏÖìÑÉÕ464.0kgµÄFeCO3,µÃµ½316.8kg²úÆ·¡£Èô²úÆ·ÖÐÔÓÖÊÖ»ÓÐFeO£¬Ôò¸Ã²úÆ·ÖÐFe2O3µÄÖÊÁ¿ÊÇ_________kg¡£(Ħ¶ûÖÊÁ¿/gmol-1:FeCO3116Fe2O3160FeO72)