F.6 Chem Mole (15分!超趕!!!)

waterlongszeto

2010-10-08 3:04 am

On combusion, 300cm³ of a gaseous hydrocarbon form 900cm³ of carbon dioxide and 0.7232g of water. Assuming all volumes to be measuring at s.t.p., find the molecular formula of the hydrocarbon.

A mixture of gases has the composition of 40% ammonia, 25% hydrogen and 35% nitrogen by volume. The pressure is 1.5 x 10⁵ Nm⁻2. What will the partial pressure of the other gases be if all ammonia is removed is removed by the addition of phosphorus(V) oxide?

A sample of ethanedioic acid crystals [(HCO₂)₂ •nH₂O] weighting 1.26g was dissolved in water and titrate with molar sodium hydroxide. If 20.0cm3 of the hydroxide was required for the titration, find the number of molecules of water in each molecules of the acid.

An ammonia plant produces 10⁶ kg of ammonia per day by the following process:
N₂ + 3H₂ ⇀ 2NH₃
The plants operates at 200 atm. pressure and 525'C.
What volume of ammonia is produces at that condition each day?
What volume does that correspond to at 1.0 atm pressure pressure and 25'C?

回答 (1)

用戶26240

2010-10-08 4:19 am

✔ 最佳答案

"On combustion ……"

Let CxHy be the molecular formula.
CxHy + (x + y/4)H2O → xCO2 + (y/2)H2O
Mole ratio CxHy : CO2 : H2O = 1 : x : y/2

No. of moles of CxHy = 300/22400 = 0.0134 mol
No. of moles of CO2 = 900/22400 = 0.0402 mol
No. of moles of H2O = 0.7232/(1x2 + 16) = 0.0402 mol

x = 0.0402/0.0134 = 3
y/2 = 0.0402/0.0134, Hence x = 6

Molecular formula = C3H6

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"A mixture ……"

If the total volume and temperature are constant, the partial pressures of all other gases will be unchanged when all ammonia is removed.

Partial pressure of H2 = (1.5 x 10⁵) x 25% = 3.75 x 10⁴ N m¯²
Partial pressure of N2 = (1.5 x 10⁵) x 35% = 5.25 x 10⁴ N m¯²

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"A sample ……"

(HCO2)2 + 2NaOH → (NaCO)2 + 2H2O
Mole ratio (HCO2)2 : 2NaOH = 1 : 2

No. of moles of NaOH = 1 x (20/1000) = 0.02 mol
No. of moles of (HCO2)2 used = 0.02 x (1/2) = 0.01 mol
No. of moles of (HCO2)2•nH₂O = 0.01 mol

Molar mass of (HCO2)2•nH₂O:
2 x (1 + 12 + 16x2) + n x (1x2 + 16) = 1.26/0.01
90 + 18n = 126
n = 2

No. of molecules of water in each molecule of the acid = 2

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"An ammonia ……"
PV = (n/M)RT
200 x V = [(10⁶ x 10³)/(14 + 1x3)] x 0.082 x (273 + 525)
V = 1.925 x 10⁷ L
Volume at 200 atm and 525°C = 1.925 x 10⁷ L

P1V1/T1 = P2V2/T2
200 x (1.925 x 10⁷) / (273 + 525) = 1 x V2 / (273 + 25)
V2 = 1.438 x 10⁹ L
Volume at 1 atm and 25°C = 1.438 x 10⁹ L

參考: 土扁

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