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A mixture of neon and xenon gases, at a total pressure of 739 mm Hg, contains 0.919 grams of neon and 19.1 grams of xenon. What is the partial pressure of each gas in the mixture?_______g Xe

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Answer:

Partial pressure of neon = 175 mmHg

Partial pressure of xenon = 564 mmHg

Step-by-step explanation:

The partial pressure of a gas in a mixture can be calculated as the product of the mole fraction of the gas (Xi) and the total pressure (Pt), as follows:

Pi = Xi Pt

The total pressure is 739 mmHg ⇒ Pt = 739 mmHg

In order to calculate the mole fraction of each gas, we have to first calculate the number of moles of each gas (n) by dividing the mass of the gas into the molar mass (MM):

For neon gas (Ne):

MM(Ne) = 20.18 g/mol

n(Ne)= mass/MM = 0.919 g x 1 mol/20.18 g = 0.045 mol Ne

For xenon gas (Xe):

MM(Xe) = 131.3 g/mol

n(Xe)= mass/MM = 19.1 g x 1 mol/131.3 g = 0.145 mol Xe

Now, we calculate the mole fraction (X) by dividing the number of moles of the gas into the total number of moles (nt):

nt= moles Ne + moles Xe = 0.045 mol + 0.145 mol = 0.190 mol

X(Ne) = moles Ne/nt = 0.045 mol/0.190 mol = 0.237

X(Xe) = moles Xe/nt = 0.145/0.190 mol = 0.763

Finally, we calculate the partial pressures of Ne and Xe as follows:

P(Ne) = X(Ne) x Pt = 0.237 x 739 mmHg = 175 mmHg

P(Xe) = X(Xe) x Pt = 0.763 x 739 mmHg = 564 mmHg

User Johannes Metzner
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