Question

How many grams of oxygen gas are contained in a 15 L sample at 1.02 atm and 28oC?

Answer 1

`\boxed{\text{20 g}}`

Step-by-step 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

`\text{1.02 atm} *\text{15 L} = n * \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^(-1)\text{mol}^(-1)* \text{301.15 K}\\\\15.3 = n * \text{24.71 mol}^(-1)\\\\n = \frac{15.3}{\text{24.71 mol}^(-1)} = \text{0.619 mol}`

(c) Calculate the mass

`\text{Molar mass} = \frac{\text{mass}}{\text{moles}}\\\\M = (m)/(n)\\\\\text{32.00 g}\cdot \text{mol}^(-1) = \frac{m}{\text{0.619 mol}}\\\\m = 32.00 * 0.619 \text{ g} = \textbf{20 g}\\\\\text{The mass of oxygen is } \boxed{\textbf{20 g}}`