X g of ammonium chloride were boiled with excess sodium hydroxide solution. The ammonia evolved was absorbed in 50.0 cm^3 of 0.5M sulphuric acid. This solution was then made up to 250 cm^3 with deionized water and 25.0 cm^3 of it required 25.10 cm^3 of 0.1M sodium hydroxide for neutralization. Calculate the mass of ammonium chloride, X.
Thank you very much!
Chem question
2009-09-27 8:25 pm
回答 (2)
2009-09-27 9:41 pm
✔ 最佳答案
我是用中文讀chem,所以不太懂看英文,不知有否理解錯題目,但我都嘗試吧.
NH4Cl + NaOH → NaCl + NH3 + H2O
2NH3 + H2SO4 → (NH4)2SO4
然後我不太肯定題目說什麼,是指把reaction用剩的H2SO4稀釋至250cm3,再抽取其中25cm3與NaOH進行neutralization嗎?
(我就是看不懂這裡,那麼我就假設自己沒理解錯,繼續做吧)
H2SO4 + 2NaOH → Na2SO4 + 2H2O
NaOH = 0.1 * (25.1 / 1000)
= 0.00251mol
根據equation,H2SO4 : NaOH = 2 : 1
所以 H2SO4 = 0.00251 / 2
= 0.001255mol
注意,這裡的H2SO4是從250cm3抽取25cm3進行reaction,
也就是說,原本H2SO4的mole是大10倍:
250cm3的H2SO4 = 0.001255 * (250/25)
= 0.01255mol
該0.01255mol的H2SO4是在第2條equation用剩,
而第二條equation提到該H2SO4是50cm3 of 0.5M:
50cm3 of 0.5M 的H2SO4 = 0.5 * (50/1000)
= 0.025mol
也就表示第2條equation,共用了H2SO4:
0.025 - 0.01255
=0.01245mol
根據第2條equation,NH3 : H2SO4 = 2 : 1
NH3 = 0.01245 * 2
= 0.0249mol
根據第1條equation,NH3 : NH4Cl = 1 : 1
NH4Cl = 0.0249mol
mole = mass / molar mass
0.0249 * (14+4+35.5)
=1.33215g
所以X = 1.33215
2009-09-27 14:46:37 補充：
見到002的回答,也就證明了我沒理解錯題目吧^^
參考: 可能做錯,因為我唔識睇題目/.\
2009-09-27 10:01 pm
Consider the titration of the diluted excess acid with sodium hydroxide solution:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Mole ratio H2SO4 : NaOH = 1 : 2
No. of moles of NaOH used = 0.1 x (25.1/1000) = 0.00251 mol
No. of moles of diluted excess H2SO4 = 0.00251 x (1/2) = 0.001255 mol
Consider the reaction between H2SO4 and NH3:
H2SO4 + 2NH3 → (NH4)2SO4
Mole ration H2SO4 : NH3 = 1 : 2
Total no. of moles of H2SO4 added = 0.5 x (50/1000) = 0.025 mol
No. of moles of H2SO4 unreacted = 0.001255 x (250/50) = 0.01255 mol
No. of moles of H2SO4 reacted = 0.025 - 0.01255 = 0.01245 mol
No. of moles of NH3 reacted = 0.01245 x 2 = 0.0249 mol
Consider the reaction between NH3 and NaOH to produce NH3:
NH4Cl + NaOH → NaCl + NH3 + H2O
Mole ratio NH4Cl : NH3 = 1 : 1
No. of moles of NH3 formed = 0.0249 mol
No. of moles of NH4Cl used = 0.0249 mol
Molar mass of NH4Cl = 14 + 1x4 + 35.5 = 53.5 g/mol
Mass of NH4Cl used = 0.0249 x 53.5 = 1.33 g
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Mole ratio H2SO4 : NaOH = 1 : 2
No. of moles of NaOH used = 0.1 x (25.1/1000) = 0.00251 mol
No. of moles of diluted excess H2SO4 = 0.00251 x (1/2) = 0.001255 mol
Consider the reaction between H2SO4 and NH3:
H2SO4 + 2NH3 → (NH4)2SO4
Mole ration H2SO4 : NH3 = 1 : 2
Total no. of moles of H2SO4 added = 0.5 x (50/1000) = 0.025 mol
No. of moles of H2SO4 unreacted = 0.001255 x (250/50) = 0.01255 mol
No. of moles of H2SO4 reacted = 0.025 - 0.01255 = 0.01245 mol
No. of moles of NH3 reacted = 0.01245 x 2 = 0.0249 mol
Consider the reaction between NH3 and NaOH to produce NH3:
NH4Cl + NaOH → NaCl + NH3 + H2O
Mole ratio NH4Cl : NH3 = 1 : 1
No. of moles of NH3 formed = 0.0249 mol
No. of moles of NH4Cl used = 0.0249 mol
Molar mass of NH4Cl = 14 + 1x4 + 35.5 = 53.5 g/mol
Mass of NH4Cl used = 0.0249 x 53.5 = 1.33 g
收錄日期: 2021-04-13 16:51:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090927000051KK00606