ÌâÄ¿ÄÚÈÝ

ÍùÒ»½ðÊôÂÁµÄÒ×À­¹ÞÄÚ³äÂúCO2£¬È»ºóÍù¹ÞÄÚ×¢Èë×ãÁ¿µÄNaOHÈÜÒº£¬Á¢¼´Óýº²¼·â¹Þ¿Ú¡£¾­¹ýÒ»¶Îʱ¼äºó£¬¹Þ±ÚÄÚ°¼¶ø±ñ¡£ÔÙ¹ýÒ»¶Îʱ¼äºó£¬±ñÁ˵ĹޱÚÖØÐÂ¹ÄÆðÀ´¡£½âÊÍÏÂÊö±ä»¯µÄʵÑéÏÖÏó£¬Å䯽Óйػ¯Ñ§·½³Ìʽ¡£

£¨1£©¹Þ±ÚÄÚ°¼¶ø±ñµÄÔ­Òò£º________________________________________________£¬

·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º__________________________________________________¡£

£¨2£©¹ÞÖØÐÂ¹ÄÆðµÄÔ­Òò£º__________________________________________________£¬

·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_______________________________________________________¡£

£¨3£©ÊµÑéʱ£¬½«¡°³äÂúCO2¡±¸ÄΪ£º¡°·ÅÈëÒ»Ò©³×°×É«¹ÌÌ塱£¬ÆäËû²»±ä£¬Ò²ÄÜÈ¡µÃÏàͬµÄʵÑéЧ¹û£¬Ôò¸Ã°×É«¹ÌÌåÊÇ________________£¨Ð´Ãû³Æ£©¡£

£¨1£©NaOHÎüÊÕÁËCO2ÆøÌ壬ʹ¹ÞÄÚÆøÑ¹Ð¡ÓÚ¹ÞÍâÆøÑ¹£¬ËùÒÔ¹ÞÄÚ°¼¶ø±ñ

2NaOH+CO2¨T¨TNa2CO3+H2O

£¨2£©¹ýÁ¿NaOHºÍAl·´Ó¦²úÉúH2£¬Ê¹¹ÞÄÚÆøÑ¹Ôö´ó£¬ËùÒÔ¹ÞÔÙ´Î¹ÄÆð

2Al+2NaOH+2H2O¨T¨T2NaAlO2+3H2¡ü

£¨3£©¸É±ù


½âÎö:

NaOHÄÜÎüÊÕCO2ʹ¹ÞÄÚÆøÑ¹¼õС£»Ò²ÄÜÓëAl·´Ó¦·Å³öH2£¬Ê¹¹ÞÄÚÆøÑ¹Ôö´ó£»±¾ÌâÕýÊǸù¾ÝÕâÁ½¸ö·´Ó¦Éè¼ÆµÄ¡£·ÅÈëÒ»Ò©³×°×É«¹ÌÌ壬ÄÜÓëCO2¾ßÓÐÏàͬµÄʵÑéЧ¹û£¬°×É«¹ÌÌåÓ¦¸ÃÊǸɱù¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø