Answer:
[tex]\boxed{\text{20 g}}[/tex]
Explanation:
We can use the Ideal Gas Law to solve this problem
pV = nRT
Data:
p = 1.02 atm
V = 15 L
T = 28 °C
Calculations:
(a) Convert temperature to kelvins
T = (28 + 273.15) K = 301.15 K
(b) Calculate the number of moles
[tex]\text{1.02 atm} \times\text{15 L} = n \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1}\times \text{301.15 K}\\\\15.3 = n \times \text{24.71 mol}^{-1}\\\\n = \dfrac{15.3}{\text{24.71 mol}^{-1}} = \text{0.619 mol}[/tex]
(c) Calculate the mass
[tex]\text{Molar mass} = \dfrac{\text{mass}}{\text{moles}}\\\\M = \dfrac{m}{n}\\\\\text{32.00 g}\cdot \text{mol}^{-1} = \dfrac{m}{\text{0.619 mol}}\\\\m = 32.00 \times 0.619 \text{ g} = \textbf{20 g}\\\\\text{The mass of oxygen is } \boxed{\textbf{20 g}}[/tex]
