£¨1£©¹ýÑõ»¯ÇâÔÚ¶þÑõ»¯Ã̵Ĵ߻¯×÷ÓÃÏÂÉú³ÉË®ºÍÑõÆø£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2H
2O
22H
2O+O
2¡ü£»¸Ã·´Ó¦·ûºÏ¡°Ò»±ä¶à¡±µÄÌØÕ÷£¬ÊôÓڷֽⷴӦ£®
£¨2£©´¿¼îÈÜÒºÓëʯ»ÒË®·´Ó¦Éú³É̼Ëá¸Æ³ÁµíºÍÇâÑõ»¯ÄÆ£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºNa
2CO
3+Ca£¨OH£©
2=CaCO
3¡ý+2NaOH£»Äܹ۲쵽µÄÏÖÏóÊÇÓа×É«³ÁµíÉú³É£®
£¨3£©ÈýÑõ»¯ÁòÓëË®·´Ó¦Éú³ÉÁòËᣬ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºH
2O+SO
3=H
2SO
4£»Éú³ÉµÄÁòËáÏÔËáÐÔ£¬pHСÓÚ7£®
£¨4£©¹¤ÒµÉÏÓÃÒ»Ñõ»¯Ì¼»¹ÔÑõ»¯ÌúÁ¶Ìú£¬Ö÷ÒªÊÇÀûÓÃCOµÄ»¹ÔÐÔ£¬ÔÚ¸ßÎÂϺÍÑõ»¯Ìú·´Ó¦Éú³ÉÌúºÍ¶þÑõ»¯Ì¼£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪFe
2O
3+3CO
2Fe+3CO
2£®·´Ó¦ºóµÄÆøÌåͨÈë³ÎÇåʯ»ÒË®Öлá¹Û²ìµ½³ÎÇåʯ»ÒË®±ä»ë×Ç£®
£¨5£©Ì¼ÔÚ¹ýÁ¿ÑõÆøÖгä·ÖȼÉÕÉú³É¶þÑõ»¯Ì¼£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºC+O
2CO
2£»¸Ã·´Ó¦·Å³ö´óÁ¿µÄÈÈ£¬ÊôÓÚ·ÅÈÈ·´Ó¦£®
¹Ê´ð°¸Îª£º£¨1£©2H
2O
22H
2O+O
2¡ü£»·Ö½â·´Ó¦£»£¨2£©Na
2CO
3+Ca£¨OH£©
2=CaCO
3¡ý+2NaOH£»Óа×É«³ÁµíÉú³É£»£¨3£©H
2O+SO
3=H
2SO
4£»Ð¡ÓÚ£»
£¨4£©Fe
2O
3+3CO
2Fe+3CO
2£»³ÎÇåʯ»ÒË®±ä»ë×Ç£»£¨5£©C+O
2CO
2£»·ÅÈÈ£®