F4 chem~~
2009-04-30 1:40 am
27.82 g of hydrated potassium carbonate (K2CO3 . nH2O) was dissolved and made up to 1.00 dm^3. 25.00cm^3 of the solution is completely reacted with 24.4 cm^3 of 0.20M nitric acid. Determine the value of n.
回答 (1)
2009-04-30 2:39 am
✔ 最佳答案
Consider titration of 25 cm3 of the solution with nitric acid:
K2CO3 + 2HNO3 → 2KNO3 + H2O + CO2
Mole ratio K2CO3 : HNO3 = 1 : 2
No. of moles of HNO3 used = 0.2 x (24.4/1000) = 0.00488 mol
No. of moles of K2CO3 = 0.00488 x (1/2) = 0.00244 mol
Consider the preparation of the 1 dm3 of the solution:
No. of moles of K2CO3 = 0.00244 x (1000/25) = 0.0976 mol
No. of moles of K2CO3•nH2O used = 0.0976 mol
Molar mass of K2CO3•nH2O = 27.82/0.0976 = 285 g mol-1
Consider the molar mass of K2CO3•nH2O:
39x2 + 12 + 16x3 + n(1x2 + 16) = 285
138 + 18n = 285
18n = 147
n = 8.17
n ≈ 8
=
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