Question

calculate the grams of oxygen gas present in a 2.50 L sample kept at 1.66 atm pressure and a temperature of 10.0 C

Answer 1

Answer:

5.72 grams of O₂ are present in the sample

Explanation:

Let's calculate the moles of O₂ by the Ideal Gases Law.

P . V = n . R . T

1.66 atm . 2.50L = n . 0.082 . 283K

(1.66 atm . 2.50L) / 0.082 . 283K = n

0.179 mol = n

Molar mass O₂ = 32 g/m

Mol . molar mass = mass → 0.179 m . 32g/m = 5.72 g

Answer 2

Answer:

The mass of oxygen gas is 5.72 grams

Explanation:

Step 1: Data given

Volume of oxygen gas (O2) = 2.50L

Pressure of O2 gas = 1.66 atm

Temperature = 10.0 °C = 283 Kelvin

Step 2: Calculate moles of O2 gas

p*V = n*R*T

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

⇒ with n = the moles of O2 gas = TO BE DETERMINED

⇒ with p = the pressure of the O2 gas =1.66 atm

⇒ with V = the volume of the gas = 2.50 L

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

⇒ with T= the temperature = 10.0°C = 283 K

n = (1.66 * 2.50)/(0.08206*283)

n = 0.1787 moles O2

Step 3: Calculate mass of O2 grams

Mass O2 = moles O2 = Molar mass O2

Mass O2 = 0.1787 moles * 32.0 g/mol

Mass O2 =  5.72 grams

The mass of oxygen gas is 5.72 grams