The hydrogen gas formed in a chemical reaction is collected over water at 30.0 degrees c at a total pressure of 732 mmhg what is the partial pressure of the hydrogen gas collected

when a gas is collected over water at 30.0 deg C , the H2O vapor pressure is 32 Torr Part A:What is the partial pressure of the hydrogen gas collected in this way? at a total pressure of 732 mm Hg. - 32 Torr H2O vapor = 700. Torr H2 your answer is answer is 700 mm Hg. Part B: If the total volume of gas collected is 724 mL, what mass of hydrogen gas is collected? find moles PV = nRT (700.Torr)(0.724 Litres) = n (62.36 Torr-Litres/mol-K) (303 Kelvin) n = 0.02682 moles of H2 using molar mass, find moles: ( 0.02682 moles of H2) (2.016 grams H2 / mol) = 0.05407 grams of H2 your answer rounded to 3 sig figs is 0.0541 grams of H2

The partial pressure of the hydrogen gas collected : 700.4974 mmHg

Further explanation

Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases

Can be formulated:

P tot = P1 + P2 + P3 ....

The partial pressure is the pressure of each gas in a mixture

The hydrogen gas is formed in a chemical reaction evaporate and mixed with water vapor

Hydrogen gas pressure and water-vapor pressure produce total pressure of the gas

So

Ptotal = P water vapor + P Hydrogen gas

At 30.0 o C the vapor pressure of water is 4.2 kPa. (a vapor pressure table is in the attached picture)

Because 1kPa = 7,50062 mmHg, then 4.2 kPa = 31,5026 mmHg

So the partial pressure of the Hydrogen gas is

732 mmHg (P total) = 31.5026 mmHg (P water vapor) + P Hydrogen gas

 P Hydrogen gas = 732 - 31.5026

 P Hydrogen gas = 700.4974 mmHg

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