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The hydrogen gas formed in a chemical reaction is collected over water at 30.0 °C at a total pressure of 732 mmHg. What is the partial pressure of the hydrogen gas collected in this way? If the total volume of gas collected is 722 mL, what mass of hydrogen gas is collected?

Respuesta :

zubam2002

Answer:

Explanation: i) P2 = 732 mmHg; T1 = 273 K; T2 = (273 + 30) K = 300 K;

∴ P1/T1 = P2/T2

P1/273 = 732/300

P1 = 732 X 273/300 = 666.12 mmHg

∴ Partial pressure = (732 - 666.12) =  65.88 mmHg

ii) Pressure ( at atmosphere) = 732/760 = 0.96 atm; R= 0.0821 atm/mole.K; T = 300 K; V = 722/1000 = 0.722 L

∴ Number of moles, n = PV/RT ( from nRT = PV)

n = 0.96 X 0.722/(0.0821 X 300) = 0.6954/24.63 = 0.028 mol

∴ mass of hydrogen gas = Molar mass X n ( where molar mass of hydrogen is 2g) = 0.028 X 2 = 0.056g

pstnonsonjoku

The mass of the gas collected is 0.054 g.

From Dalton's law of partial pressure, the total pressure of a gas is the sum of all the partial pressure at that temperature.

a) At a total pressure of 732 mmHg, the partial pressure of water at 30.0 °C is 31.8 mmHg

Hence;

Pressure of the gas =  732 mmHg - 31.8 mmHg = 700.2 mmHg

b) Using the formula;

PV = nRT

P = 700.2 mmHg or 0.92 atm

V = 722 mL or 0.722 L

n = ?

R = 0.082 atmL K-1mol-1

T = 30.0 °C + 273 = 303 K

n = PV/RT

n = 0.92 atm × 0.722 L/ 0.082 atmL K-1mol-1 × 303 K

n = 0.664/24.85

n = 0.027 moles

Mass of hydrogen collected = 0.027 moles × 2 g/mol = 0.054 g

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