ÌâÄ¿ÄÚÈÝ

£¨£±£³·Ö£©ÏÂͼÊÇÒ»¸ö»¯Ñ§¹ý³ÌµÄʾÒâͼ¡£ÒÑÖª¼×³ØµÄ×Ü·´Ó¦Ê½Îª£º

2CH3OH+3O2+4KOH    2K2CO3+6H2O

£¨1£©Çë»Ø´ðͼÖмס¢ÒÒÁ½³ØµÄÃû³Æ¡£¼×µç³ØÊÇ           ×°Öã¬ÒÒ³ØÊÇ           ×°Öá£

£¨2£©Çë»Ø´ðÏÂÁе缫µÄÃû³Æ£ºÍ¨ÈëCH3OHµÄµç¼«Ãû³ÆÊÇ        £¬B£¨Ê¯Ä«£©µç¼«µÄÃû³ÆÊÇ        ¡£

£¨3£©Ð´³öµç¼«·´Ó¦Ê½£º ͨÈëO2µÄµç¼«µÄµç¼«·´Ó¦Ê½ÊÇ                            ¡£

A£¨Fe£©µç¼«µÄµç¼«·´Ó¦Ê½Îª                                  £¬

£¨4£©ÒÒ³ØÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                        ¡£

£¨5£©µ±ÒÒ³ØÖÐA£¨Fe£©¼«µÄÖÊÁ¿Ôö¼Ó5.40gʱ£¬¼×³ØÖÐÀíÂÛÉÏÏûºÄO­2    mL£¨±ê¿öÏ£©¡£

 

¡¾´ð°¸¡¿

£¨£±£³·Ö£©

£¨1£©Ô­µç³Ø  µç½â³Ø  £¨Ã¿¿Õ1·Ö£©

£¨2£©¸º¼«   Ñô¼«£¨Ã¿¿Õ1·Ö£©

£¨3£©   O2£«2H2O£«4e£­£½4OH£­       Ag + £«  e£­  =  Ag     £¨Ã¿¿Õ2·Ö£©

£¨4£©

£¨5£©  280   £¨£²·Ö£©

¡¾½âÎö¡¿

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø