Question

ÈçͼÖÐ,AΪ¿ÕÆøÖг£¼ûµÄÆøÌåµ¥ÖÊ¡¢B¡¢C¡¢EÊǽðÊôµ¥ÖÊ,DΪ·Ç½ðÊôµ¥ÖÊ¡£ÒÑÖª:¢ÙIÊÇÒ»ÖÖ³£¼ûµÄÎÂÊÒÆøÌå,EÔ­×ÓºËÄÚÓÐ12¸öÖÊ×Ó¢ÚGΪºìÉ«ÎïÖÊ,·´Ó¦C+GB+HÄܷųö´óÁ¿µÄÈÈ,¸Ã·´Ó¦ÔøÓ¦ÓÃÓÚÌú¹ìµÄº¸½Ó;»Ø´ðÏÂÁÐÎÊÌâ:

(1)·Ö±ðд³öF¡¢G¡¢H¡¢IµÄ»¯Ñ§Ê½:

F________¡¡G________¡¡H________¡¡I_______

(2)ÊéдÏÂÁл¯Ñ§·½³Ìʽ:

C+GB+H___________________________________________________;

2E+I2F+D__________________________________________________¡£

(3)CÓëNaOHÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________________,·´Ó¦ºóÈÜÒºÓë¹ýÁ¿»¯ºÏÎïI·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________________________¡£

(4)1.6 g GÈÜÓÚÑÎËá,µÃµ½µÄÈÜÒºÓëÍ­·ÛÍêÈ«·´Ó¦,¼ÆËãÖÁÉÙËùÐèµÄÍ­·ÛµÄÖÊÁ¿Îª:________¡£

Answer 1

.¡¾½âÎö¡¿ÓÉÌâ¸ÉÒÑÖª¿ÉµÃ:IΪCO₂,EΪMg,GΪFe₂O₃,HΪAl₂O₃,CΪAl,BΪFe,ËùÒÔAΪO₂,FΪMgO,DΪC¡£

m(Fe₂O₃)=1.6 g/160 g¡¤mol^-1=0.01 mol

Fe₂O₃+¡¡6HCl====¡¡2FeCl₃¡¡¡¡+¡¡¡¡3H₂O

0.01 mol¡¡¡¡¡¡¡¡¡¡0.02 mol

2FeCl₃¡¡+¡¡Cu====2FeCl₂+CuCl₂

0.02 mol ¡¡0.01 mol

m(Cu)=0.01 mol¡Á64 g¡¤mol^-1=0.64 g¡£

´ð°¸:(1)MgO¡¡Fe₂O₃¡¡Al₂O₃¡¡CO₂

(2)2Al+Fe₂O₃2Fe+Al₂O₃

2Mg+CO₂2MgO+C

(3)2Al+2NaOH+6H₂O====2Na[Al(OH)₄]+3H₂¡ü

Na[Al(OH)₄]+CO₂====Al(OH)₃¡ý+NaHCO₃

(4)0.64 g