Question

The H 2 produced in a chemical reaction is collected through water in a eudiometer. If the pressure in the eudiometer is 760.0 torr and the vapor pressure of water under the experimental conditions is 23.8 torr, what is the pressure (torr) of the H 2 gas

Answer 1

Answer:

the pressure (torr) of theH₂ gas is 736.2 torr

Explanation:

Given that

The H₂ gas  produced in a chemical reaction is collected through water in a eudiometer; during this process, the gas collected contains some droplets of water vapor along with these gas.

So; the total pressure in the eudiometer = Pressure in the H₂ gas  - Pressure of the water vapor

Where;

$P_{Totsl}$ =  total pressure in the eudiometer = 760.0 torr

$P_{H_2}$ = Pressure in the H₂ gas  = ???

$P_{H_2O}$ = Pressure in the water vapor  = 23.8 torr

Now:

$P_{Totsl}$ = $P_{H_2}$ + $P_{H_2O}$

- $P_{H_2}$ = + $P_{H_2O}$ - $P_{Totsl}$

$P_{H_2}$  = - $P_{H_2O}$ + $P_{Totsl}$

$P_{H_2}$  =  (- 23.8 + 760) torr

$P_{H_2}$  = 736.2 torr

Thus; the pressure (torr) of theH₂ gas is 736.2 torr

Answer 2

If the pressure in a eudiometer is 760.0 torr and the vapor pressure of water is 23.8 torr, the pressure of the H₂ gas is 736.2 torr.

A eudiometer is a laboratory device that measures the change in volume of a gas mixture following a physical or chemical change. Usually, the gas is collected over water.

According to Dalton's law of partial pressures, the sum of the sum partial pressure of the hydrogen and the partial pressure of the water is equal to the total pressure in the eudiometer.

$P = pH_2 + pH_2O\\\\pH_2 = P - pH_2O = 760.0 torr - 23.8 torr = 736.2 torr$

If the pressure in a eudiometer is 760.0 torr and the vapor pressure of water is 23.8 torr, the pressure of the H₂ gas is 736.2 torr.

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