ÌâÄ¿ÄÚÈÝ
£¨8·Ö£©³ÇÊÐʹÓõÄȼÁÏ£¬ÏÖ´ó¶àÓÃúÆø¡¢Òº»¯Ê¯ÓÍÆø¡£ÃºÆøµÄÖ÷Òª³É·ÖÊÇCOºÍH2µÄ»ìºÏÆøÌ壬ËüÓÉú̿ºÍË®ÕôÆø·´Ó¦ÖƵ㬹ÊÓÖ³ÆˮúÆø¡£
£¨1£©ÊÔд³öÖÆȡˮúÆøµÄÖ÷Òª»¯Ñ§·½³Ìʽ____________________________.
£¨2£©Òº»¯Ê¯ÓÍÆøµÄÖ÷Òª³É·ÖÊDZûÍ飬±ûÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ
C3H8(g) +5O2(g) == 3CO2(g) + 4H2O(l)£»¡÷H=¨C2220.0 kJ¡¤
,
ÓÖÖªCOÆøÌåȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ£º
CO(g) +1/2O2(g) == CO2(g) £»¡÷H=¨C282.57kJ¡¤
,
ÊԱȽÏͬÖÊÁ¿µÄC3H8ºÍCOȼÉÕ£¬²úÉúµÄÈÈÁ¿±ÈֵԼΪ_________£º1¡£
£¨3£©ÒÑÖªÇâÆøȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ2H2(g) + O2(g) == 2H2O(l)£»¡÷H=¨C571.6 kJ¡¤
,ÊԱȽÏͬÖÊÁ¿µÄH2ºÍC3H8ȼÉÕ£¬²úÉúµÄÈÈÁ¿±ÈֵԼΪ____________£º1¡£
£¨4£©ÇâÆøÊÇδÀ´µÄÄÜÔ´£¬³ýÀ´Ô´·á¸»Ö®Í⣬»¹¾ßÓеÄÓŵãÊÇ____________________
£¨1£©ÊÔд³öÖÆȡˮúÆøµÄÖ÷Òª»¯Ñ§·½³Ìʽ____________________________.
£¨2£©Òº»¯Ê¯ÓÍÆøµÄÖ÷Òª³É·ÖÊDZûÍ飬±ûÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ
C3H8(g) +5O2(g) == 3CO2(g) + 4H2O(l)£»¡÷H=¨C2220.0 kJ¡¤

ÓÖÖªCOÆøÌåȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ£º
CO(g) +1/2O2(g) == CO2(g) £»¡÷H=¨C282.57kJ¡¤

ÊԱȽÏͬÖÊÁ¿µÄC3H8ºÍCOȼÉÕ£¬²úÉúµÄÈÈÁ¿±ÈֵԼΪ_________£º1¡£
£¨3£©ÒÑÖªÇâÆøȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ2H2(g) + O2(g) == 2H2O(l)£»¡÷H=¨C571.6 kJ¡¤

£¨4£©ÇâÆøÊÇδÀ´µÄÄÜÔ´£¬³ýÀ´Ô´·á¸»Ö®Í⣬»¹¾ßÓеÄÓŵãÊÇ____________________
£¨1£©C+H2O(g)== CO+H2 £¨2£©5.0 £¨3£©2.8
£¨4£©µÈÖÊÁ¿Ê±·ÅÈȶࣻ²úÎïÎÞÎÛȾ
£¨4£©µÈÖÊÁ¿Ê±·ÅÈȶࣻ²úÎïÎÞÎÛȾ
ÂÔ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿