Question

You have a 6.75 L container of pure gaseous oxygen at a constant pressure of 1.25 atm. If the temperature is 298 K, calculate the number of grams of oxygen (O2) gas in the container. (R = 0.0821 L∙atm/mol∙K) A. 11.1 g B. 92.8 g C. 2.90 g D. 0.346 g

Answer 1

There is 11.1 grams of O2 in the container ( option A)

Step-by-step explanation:

Step 1: Data given

Volume of the container = 6.75 L

Pressure = 1.25 atm

Temperature = 298 K

Step 2: Calculate number of mol of oxygen

p*V = n*R*T

⇒ with p = the total pressure of oxygen = 1.25 atm

⇒ with V = the volume of the oxygen gas = 6.75 L

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

⇒ with R = the gas constant = 0.0821 L*atm/K*mol

⇒ with T = the temperature in the container = 298 K

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

n = (1.25*6.75)/ (0.0821*298)

n = 0.345 moles O2

Step 3: Calculate mass of O2

Mass O2 = moles O2 * molar mass O2

Mass O2 = 0.345 moles O2 = 32 g/mol

Mass O2 ≈ 11.1 grams O2

There is 11.1 grams of O2 in the container ( option A)