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

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

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)

your answer would be “dry”