Question

What is the volume of 58.9 grams of oxygen gas (O2) at STP?

Answer 1

41.2 L

Step-by-step explanation

Given that:

The mass of oxygen gas (O₂) = 58.9 grams

What to find:

The volume of 58.9 grams of oxygen gas (O₂) at STP.

Step-by-step solution:

The first step is to convert 58.9 grams of oxygen gas (O₂) given to moles using the mole formula:

`Moles=\frac{Mass}{Molar\text{ }mass}`

The molar mass of oxygen gas (O₂) is 31.998 g/mol.

`Moles\text{ }of\text{ }O_2=\frac{58.9g}{31.998g\text{/}mol}=1.8407\text{ }mol`

The final step is to convert the 1.8407 moles of oxygen gas (O₂) to volume using the conversion factor below.

Conversion factor: The molar volume of any gas at STP is 22.4 L. That is, one mole of any gas has a volume of 22.4 L or 22,400 mL.

So,

`\begin{gathered} 1\text{ }mol\text{ }O_2=22.4\text{ }L \\ \\ 1.8407\text{ }mol\text{ }O_2=x \\ \\ x=\frac{1.8407\text{ }mol}{1\text{ }mol}*22.4\text{ }L \\ \\ x=41.2\text{ }L \end{gathered}`