Question

500 dm3 of moist O2 gas was collected over water at 27°C and 726 torr pressure. Find the mass in gm. Of dry O2 gas at S.T.P. When the vapor pressure of water at 27°C is 26 torr

Answer 1

Answer:

(a) Moles of H2 = mass/molar mass = 1.02/2.02 = 0.500 mol

Moles of Cl2 = mass/molar mass = 17.73/70.90 = 0.250 mol

Total moles = 0.500 + 0.250 = 0.750 mol.

Total pressure = 98.8 kPa.

Partial pressure of each gas is proportional to its mole fraction in the mixture.

Therefore partial pressure of H2 = (0.500/0.750) x 98.8 = 65.9 kPa.

(b) The number of molecules does not change, only the volume (reduced)

and therefore the partial pressure of each gas (increased).

Total moles of gas = 0.750 mol

Therefore number of molecules = moles x NA

= 0.750 x 6.022 x 1023 = 4.52 x 1023 molecules.

(c) The reaction equation is

H2(g) + Cl2(g) ® 2HCl(g)

1 mol 1 mol 2 mol

From the stoichiometry, 2 moles of reactants produce 2 moles of product,

so the total moles of gas remains unchanged and also the temperature is unchanged.

Thus Boyle's Law can be used to calculate the new pressure.

P1V1 = P2V2

As V2 = (2/3)/V1,

98.8 x V1 = P2 x (2/3)V1

P2 = 148 kPa.

2

Each gas is undergoing an expansion when introduced into the 10.00 L container, so each will exert a partial pressure which is less than its original pressure. As temperature and number of moles of each gas in unchanged, Boyle's Law can be used to calculate the new (partial) pressure of each gas in the mixture.

P1V1 = P2V2

Oxygen:

202.6 x 2.00 = P(O2) x 10.00

P(O2) = 40.5 kPa

Neon:

303.9 x 3.00 = P(Ne) x 10.00

P(Ne) = 91.2 kPa

Total pressure = sum of partial pressures

= 40.5 + 91.2 = 131.7 kPa

3

Gases collected over water are saturated with water vapour (see preamble).

Therefore Ptotal = Pgas + Pwater

Explanation:

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