A 10.0g chunk of sodium metal (Na) is dropped into a mixture of 50.0g of water
and 50.0g of ice at 0'C. The reaction is as follows:
2Na(s)+ 2H2O(l)-->2NaOH(aq)+H2(g) △H=-368KJ
The △H(fus) of ice is 6.0 kj/mole. Assuming that the final mixture has a
heat capacity of 4.18J/g'c, calculate the final temperature.
麻煩高手囉
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 61.224.46.207
> -------------------------------------------------------------------------- <
作者: fierybear () 看板: TransBioChem
標題: Re: [普化] 80中山
時間: Tue Jun 21 20:19:54 2005
※ 引述《ymme (迷)》之銘言：
: A 10.0g chunk of sodium metal (Na) is dropped into a mixture of 50.0g of water
: and 50.0g of ice at 0'C. The reaction is as follows:
: 2Na(s)+ 2H2O(l)-->2NaOH(aq)+H2(g) △H=-368KJ
: The △H(fus) of ice is 6.0 kj/mole. Assuming that the final mixture has a
: heat capacity of 4.18J/g'c, calculate the final temperature.
: 麻煩高手囉
10.0g Na 10.0/23 = 0.435mole
50.0g H2O(l) 50.0/18 = 2.78 mole
50.0g H2O(s) 50.0/18 = 2.78 mole
2Na(s)+ 2H2O(l)-->2NaOH(aq)+H2(g) △H=-368KJ
Na為限制試劑，得放熱
368*0.435/2 = 80.0 KJ
2.7 mole H2O(s) △H(fus) = 6.0 kj/mole
冰溶化耗熱2.78*6.0
剩餘熱量 80.0-2.78*6.0 = 63 KJ
heat capacity of 4.18J/g'c 總重 = 110.0g
63/(110.0*4.18) = 0.14 ℃
有請高手確認指正
話說...那麼舊的考古題哪來的??
好奇~~
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 218.171.210.158
※ 編輯: fierybear 來自: 218.171.210.158 (06/21 20:20)