Question

If the mixture in question 1 is in a 3.0 Liter container at 34 °C, what mass (in grams)of oxygen is present?If oxygen was 0.472 atm

Answer 1

The mass (in grams) of oxygen present = 1.79 grams.

Step-by-step explanation

Given

Volume, V = 3.0 L

Temperature, T = 34 °C = (34 + 273.15 K) = 307.15 K

Pressure, P = 0.472 atm

What to find:

The mass (in grams) of oxygen present.

Step-by-step solution:

Step 1: Calculate the moles of oxygen present.

Using the ideal gas law:

`PV=nRT`

R is the molar gas constant = 0.0820574 L•atm/mol•K.

`\begin{gathered} 0.472\text{ }atm*3.0\text{ }L=n(0.0820574\text{ }L•atm/mol•K*307.15\text{ }K) \\ \\ n=\frac{0.472\text{ }atm*3.0\text{ }L}{0.0820574\text{ }L•atm/mol•K*307.15\text{ }K} \\ \\ n=0.056\text{ }mol \end{gathered}`

The moles of oxygen present is 0.056 mol.

Step 2: Convert 0.056 mol oxygen to mass in grams.

The molar mass of oxygen gas = 31.998 g/mol

Using the mole formula below, the mass of oxygen can be calculated as follows:

`\begin{gathered} Moles=\frac{Mass}{Molar\text{ }mass} \\ \\ \Rightarrow Mass=Moles* Molar\text{ }mass \\ \\ Mass=0.056\text{ }mol*31.998\text{ }g\text{/}mol \\ \\ Mass=1.791888\text{ }g\approx1.79\text{ }grams \end{gathered}`

The mass (in grams) of oxygen present = 1.79 grams.