Question

In an experiment, hydrogen is collected in a 455-milliliter jar above water at a pressure of 101,3 kilopascals and a

Answer 1

Answer:

n = 1.9 ×10⁻⁵ mol

Explanation:

Given data:

Volume of hydrogen gas = 455 mm

Pressure of hydrogen gas = 101.3 kpa

Temperature = 29.1°C

Partial pressure of water vapor = 4.0 kpa

Number of moles of hydrogen gas = ?

Solution:

First of all we will convert the units.

Pressure of hydrogen gas = 101.3 kpa= 101.3/101 =  1 atm

Partial pressure of water vapor = 4.0 kpa = 4.0/ 101 = 0.04 atm

Temperature = 29.1 + 273 = 302.1 K

Volume of hydrogen gas = 455 / 1×10⁶ = 0.000455 L

Now we will calculate the total pressure.

Total pressure = Partial pressure of hydrogen gas + partial pressure of water vapors

Total pressure = 1 atm + 0.04 atm

Total pressure = 1.04 atm

Now we will calculate the number of moles;

PV = nRT

n = PV/RT

n =  1.04 atm × 0.000455 L / 0.0821 atm.L / mol.K × 302.1 K

n = 0.00047 /24.80/mol

n = 1.9 ×10⁻⁵ mol

Answer 2

Answer:

First, calculate the partial pressure of hydrogen by subtracting the total pressure of the hydrogen–water vapor mixture from the partial pressure of water vapor:

101.3 kPa − 4.0 kPa = 97.3 kPa.

Convert the temperature to kelvins:

29.1°C + 273.15 = 302.25 K.

Convert milliliters to liters by dividing by 1,000 to get 0.455 L.

Since the pressure is in kilopascals, use the R value 8.314 .

Now substitute the known values into the ideal gas equation:

n =  

n =  

n =  

n = 0.0176176 mol

n = 0.0176 mol

Explanation:

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