A mixture of hydrogen and nitrogen gases at a total pressure of

620 mm Hg contains hydrogen at a partial pressure of 326 mm Hg.

If the gas mixture contains 0.279 grams of hydrogen, how many

grams of nitrogen are present?

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A mixture of hydrogen and nitrogen gases at a total pressure of

620 mm Hg contains hydrogen at a partial pressure of 326 mm Hg.

If the gas mixture contains 0.279 grams of hydrogen, how many

grams of nitrogen are present?

Given:

2 moles HNO3

Required:

Grams of nitrogen

Solution:

3HNO2 -> HNO3 + 2NO + H2O

Moles of N2 = (2 moles HNO3)(2 moles NO /1 mole HNO3) = 4 Moles of NO

Molar mass of N2 = 28 g/mol

Mass of N2 = (4 Moles of NO)(1/2 mol N2/1 mol NO)( 28g/mol)

Mass of N2 = 56g H2O

Answer : The mass of nitrogen present are, 4.965 grams.

Explanation :

According to the Raoult's law,

[tex]p_{He}=X_{He}\times p_T[/tex]

where,

[tex]p_{He}[/tex] = partial pressure of gas = 231 mmHg

[tex]p_T[/tex] = total pressure of gas = 641 mmHg

[tex]X_{He}[/tex] = mole fraction of helium gas = ?

Now put all the given values in this formula, we get the mole fraction of helium gas.

[tex]231mmHg=X_{He}\times 641mmHg[/tex]

[tex]X_{He}=0.36[/tex]

Now we have to calculate the mole fraction of nitrogen gas.

[tex]X_{He}+X_{N_2}=1[/tex]

[tex]X_{N_2}=1=0.36=0.64[/tex]

Now we have to calculate the mass nitrogen gas.

[tex]\frac{X_{He}}{X_{N_2}}=\frac{n_{He}}{n_{N_2}}[/tex]

[tex]\frac{X_{He}}{X_{N_2}}=\frac{\frac{w_{He}}{M_{He}}}{\frac{w_{N_2}}{M_{N_2}}}[/tex]

where,

n = moles, w = mass, M = molar mass

Now put all the given values in this expression, we get:

[tex]\frac{0.36}{0.64}=\frac{\frac{0.399}{4}}{\frac{w_{N_2}}{28}}[/tex]

[tex]w_{N_2}=4.965g[/tex]

Therefore, the mass of nitrogen present are, 4.965 grams.

0.641 g of Nitrogen are present in the mixture.

Explanation:

We use the Ideal Gases Law, to solve this question.

For the mixture:

P mixture . V mixture = mol mixture . R . T

We convert the T° to K → 23°C + 273 = 296 K

R = Ideal gases constant → 0.082 L.atm/mol.K

1 atm . 2L = mol mixture . 0.082 L.atm/mol.K . 296K

2 atm.L / ( 0.082 mol /L.atm) . 296 = 0.0824 moles

We know that sum of partial pressure = 1

Partial pressure N₂ + Partial pressure O₂ = 1

1 - 0.722 atm = Partial pressure N₂ → 0.278 atm

We apply the mole fraction concept:

Partial pressure N₂ / Total pressure = Moles N₂ / Total moles

Moles N₂ = (Partial pressure N₂ / Total pressure) . Total moles

Moles N₂ = (0.278 atm / 1 atm) . 0.0824 mol → 0.0229 moles

We convert the moles to mass → 0.0229 mol . 28 g/mol = 0.641 g

641 mg

6.50 grams of the compound X is nitrogen

Explanation:

Step 1: Data given

Mass of compound X = 37.8 grams

Compound contains 17.2 % N

Step 2: Calculate mass of nitrogen

Mass nitrogen = % Nitrogen * total mass

Mass nitrogen = 17.2 % * 37.8 grams = 0.172 * 37.8 grams

Mass nitrogen = 6.50 grams

6.50 grams of the compound X is nitrogen

About 7.109g

Explanation:

25000mL is the same as 25 liters.

[tex]PV=nRT \\\\0.5 (25)=n (0.0821) (300) \\\\n\approx 0.508[/tex]

Multiplying this by the molar mass of nitrogen, you get about 7.109 grams of nitrogen. Hope this helps!