Question

A mixture of xenon and nitrogen gases, at a total pressure of 783 mm Hg, contains 9.00 grams of xenon and 6.65 grams of nitrogen. What is the partial pressure of each gas in the mixture? PXe = mm Hg PN2 = mm Hg

Answer 1

Partial pressure Xe = 175.4 mm Hg

Partial pressure N2 = 607.6 mm Hg

Step-by-step explanation:

Step 1: Data given

Total pressure of the mixture = 783 mm Hg

Mass of Xenon = 9.00 grams

Atomic mass Xenon = 131.29 g/mol

Mass of N2 gas = 6.65 grams

Molar mass of 28.0 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles xenon = 9.00 grams / 131.29 g/mol

Moles Xenon = 0.06855 moles

Moles nitrogen gas = 6.65 grams / 28.0 g/mol

Moles nitrogen gas = 0.2375 moles

Step 3: Calculate mol ratio

Mol ratio = number of moles / total moles

Mol ratio xenon = 0.06855 / (0.06855+0.2375)

Mol ratio xenon = 0.224

Mol ratio nitrogen gas = 0.2375 / (0.06855+0.2375)

Mol ratio nitrogen gas = 0.776

Step 4: Calculate the partial pressures

Partial pressure = mol ratio * total pressure

Partial pressure Xe = 0.224 * 783 mmHg

Partial pressure Xe = 175.4 mm Hg

Partial pressure N2 = 0.776 * 783 mmHg

Partial pressure N2 = 607.6 mm Hg

175.4 + 607.6 = 783 mmHg

Answer 2

The answers are:

Partial pressure Xenon = 174.3 mmHg

Partial pressure Nitrogen = 608.7 mmHg

Step-by-step explanation:

According to Dalton's law, the partial pressure of a gas in a mixture is equal to the total pressure multiplied by the molar fraction of the gas:

P= X Ptotal

In this case we hace a mixture of xenon (Xe) and nitrogen (N₂) gases. In order to calculate the partial pressures of each gas, we have to first calculate the molar fraction of each one:

X= number of moles of gas/total moles

Xenon is a monoatomic gas (molecular weight= molar mass= 131.3 g/mol) and nitrogen is a diatomic gas (molecular weight= molar mass N x 2= 14 g/mol x 2= 28 g/mol)

moles Xe= mass Xe/molecular weight Xe= 9 g/131.3 g/mol= 0.068 mol

moles N₂= mass N₂/molecular weight N₂= 6.65 g/28 g= 0.2375 mol

Total moles= moles Xe + moles N₂ = 0.068 mol + 0.2375 mol = 0.3055 mol

⇒X(Xe)= moles Xe/total moles= 0.068 mol/0.3055= 0.2226

P(Xe)= X(Xe) x Ptotal = 0.2225 x 783 mmHg = 174.3 mmHg

⇒X(N₂)= moles N₂/total moles= 0.2375 mol/0.3055= 0.7774

P(Xe)= X(Xe) x Ptotal = 0.7774 x 783 mmHg = 608.7 mmHg