Question from (AQA,GCEAS, Unit 1, JAN,2005)
Ammonium sulphate reacts with aqueous sodium hydroxide as shown by the eqt. below:
(NH4)2 SO4 + 2NaOH --> 2NH3 + Na2SO4 + H2O
A sample of ammonium sulphate was heated with 100cm^3 of 0.500 ml dm^-3 aqueous sodium hydroxide. To ensure that all the ammonium sulphate reacted, an excess of sodium hydroxide was used.
Heating was continued until all of the ammonia had been driven off as a gas. The unreacted sodium hydroxide remaining in the solution required 27.3cm^-3 of 0.600mol dm^-3 hydrochloric acid for neutralization.
a) calculate the original number of moles of NaOH in 100cm^3 of 0.500mol dm^-3 aqueous sodium hydroxide.
b) Calculate the number of moles of HCl in 27.3cm^3 of 0.600mol dm^-3 hydrochloric acid.
c) Deduce the number of moles of the unreacted NaOH neutralized by the hydrochloric acid.
d) Use your answers from parts(a) and (c) to calculate the number of moles of NaOH which reacted with the ammonium sulphate.
Thank you for answering^^
chem (acid and base)
2011-02-06 5:21 am
回答 (3)
2011-02-06 8:29 am
✔ 最佳答案
a)The original no. of moles of NaOH = 0.500 x (100/1000) = 0.0500 mol
b)
No. of moles of HCl = 0.600 x (27.3/1000) = 0.0164 mol
c)
NaOH + HCl → NaCl + H2O
No. of moles of NaOH reacted with HCl = 0.0164 mol
d)
No. of moles of NaOH reacted with (NH4)2SO4 = 0.0500 - 0.0164 = 0.0336 mol
參考: micatkie
2011-02-07 1:34 am
a) calculate the original number of moles of NaOH in 100cm^3 of 0.500mol dm^-3 aqueous sodium hydroxide.
Molarity of NaOH = 0.5M
Volume of NaOH (in dm^3) = 100 / 1000 = 0.1dm^3
Molarity = no. of moles of solute / volume
Therefore, No. of moles of NaOH = 0.5 x 0.1 = 0.05 moles
b) Calculate the number of moles of HCl in 27.3cm^3 of 0.600mol dm^-3 hydrochloric acid.
Molarity of HCl = 0.6M
Volume of NaOH (in dm^3) = 27.3 / 1000 = 0.0273 dm^3
Molarity = no. of moles of solute / volume
Therefore, No. of moles of HCl = 0.6 x 0.273 = 0.0164 moles (3 sig. fig.)
c) Deduce the number of moles of the unreacted NaOH neutralized by the hydrochloric acid.
NaOH + HCl --> NaCl + H2O
No. of moles of NaOH : No. of moles of HCl = 1:1
No. of moles of NaOH = No. of moles of HCl = 0.0164 moles
d) Use your answers from parts(a) and (c) to calculate the number of moles of NaOH which reacted with the ammonium sulphate.
No. of moles of unreacted NaOH = 0.0164 moles
No. of moles of original NaOH = 0.05 moles
Therefore, no. of moles of NaOH reacted = 0.05 - 0.0164 = 0.0336 moles
Molarity of NaOH = 0.5M
Volume of NaOH (in dm^3) = 100 / 1000 = 0.1dm^3
Molarity = no. of moles of solute / volume
Therefore, No. of moles of NaOH = 0.5 x 0.1 = 0.05 moles
b) Calculate the number of moles of HCl in 27.3cm^3 of 0.600mol dm^-3 hydrochloric acid.
Molarity of HCl = 0.6M
Volume of NaOH (in dm^3) = 27.3 / 1000 = 0.0273 dm^3
Molarity = no. of moles of solute / volume
Therefore, No. of moles of HCl = 0.6 x 0.273 = 0.0164 moles (3 sig. fig.)
c) Deduce the number of moles of the unreacted NaOH neutralized by the hydrochloric acid.
NaOH + HCl --> NaCl + H2O
No. of moles of NaOH : No. of moles of HCl = 1:1
No. of moles of NaOH = No. of moles of HCl = 0.0164 moles
d) Use your answers from parts(a) and (c) to calculate the number of moles of NaOH which reacted with the ammonium sulphate.
No. of moles of unreacted NaOH = 0.0164 moles
No. of moles of original NaOH = 0.05 moles
Therefore, no. of moles of NaOH reacted = 0.05 - 0.0164 = 0.0336 moles
2011-02-06 9:06 am
功課就留他們自己做罷.
收錄日期: 2021-04-13 17:48:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110205000051KK00872