Question

A mixture of hydrogen and argon gases is maintained in a 6.47 L flask at a pressure of 3.43 atm and a temperature of 85 °C. If the gas mixture contains 1.10 grams of hydrogen, the number of grams of argon in the mixture is _______ g.

Answer 1

Answer:

The mixture contains 8.23 g of Ar

Explanation:

Let's solve this with the Ideal Gases Law

Total pressure of a mixture = (Total moles . R . T) / V

We convert T° from °C to K → 85°C + 273 = 358K

3.43 atm = (Moles . 0.082 L.atm/mol.K . 358K) / 6.47L

(3.43 atm . 6.47L) / (0.082 L.atm/mol.K . 358K) = Moles

0.756= Total moles from the mixture

Moles of Ar + Moles of H₂ = 0.756 moles

Moles of Ar + 1.10 g / 2g/mol = 0.756 moles

Moles of Ar = 0.756 moles - 0.55 moles H₂ → 0.206

We convert the moles to g → 0.206 mol . 39.95 g / 1 mol = 8.23 g

Answer 2

Answer:

The mass of argon is 8.42 grams

Explanation:

Step 1: Data given

Volume = 6.47 L

Pressure = 3.43 atm

Temperature = 85 °C = 358 K

Mass of hydrgen = 1.0 grams

Step 2: Calculate the number of moles

p*V = n*R*T

n = (p*V)/(R*T)

⇒with n = the number of moles = TO BE DETERMINED

⇒with p = the pressure = 3.43 atm

⇒with V = the volume = 6.47 L

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T =the temperature = 85 °C  = 358 K

n = (3.43*6.47) / (0.08206 * 358)

n = 0.7554 moles

Step 3: Calculate moles Ar

Thus,  0.7554 = mol of Ar + mol of H2

 

There is 1.10 grams ofH2 gas

Moles H2 =  1.10 grams / 2.02 g/mol

Moles H2 = 0.5446 moles H2

Moles Ar = 0.7554 - 0.5446 = 0.2108 moles

 

Step 4: Calculate mass Ar

Mass Ar = moles Ar * molar mass Ar

Mass Ar = 0.2108 moles *39.95 g/mol

MAss Ar = 8.42 grams

The mass of argon is 8.42 grams