CHEM problem

[匿名]

2009-10-19 1:15 am

3g of ammonium sulphate is boiled with 40 cm^3 of 1.25M potassium hydroxide until no more gas bubbles evolve.the solution is then neutralized with 25cm^3
0.5M hydroxide acid.calculate the percentage by mass of ammonia inthe ammonium sulphate.

回答 (3)

老爺子

2009-10-19 2:22 am

✔ 最佳答案

The "hydroxide acid" should be the mistype of "hydrochloric acid".

Consider the reaction between excess KOH and HCl.
KOH + HCl → KCl + H2O
Mole ratio KOH : HCl = 1 : 1
No. of moles of HCl used = 0.5 x 25/1000 = 0.0125 mol
No. of moles of excess KOH = 0.0125 mol

Consider the reaction between (NH4)2SO4 and KOH.
(NH4)2SO4 + 2KOH → K2SO4 + 2NH3 + 2H2O
Mole ratio KOH : NH3 = 2 : 2 = 1 : 1
Total no. of moles of KOH added = 1.25 x40/1000 = 0.05 mol
No. of moles of KOH reacted = 0.05 - 0.0125 mol = 0.0375 mol
No. of moles of NH3 formed = 0.0375 mol
Molar mass of NH3 = 14 + 1x3 = 17 g/mol
Mass of NH3 formed = 0.0375 x17 = 0.6375 g
% by mass of NH3 in (NH4)2SO4 = (0.6375/3) x 100% = 21.25%

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[匿名]

2009-10-19 1:38 am

你條題目系米有問題??
根本無hydroxide acid呢個chemical

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Albert

2009-10-19 1:23 am

u can go to yahoo!

參考: me

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