Answer:
0.56L
Explanation:
This question requires the Ideal Gas Law:
where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.
Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:![R=0.0821\frac{L\cdot atm}{mol\cdot K}](https://tex.z-dn.net/?f=R%3D0.0821%5Cfrac%7BL%5Ccdot%20atm%7D%7Bmol%5Ccdot%20K%7D)
Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means
and ![P=1atm](https://tex.z-dn.net/?f=P%3D1atm)
Lastly, we must calculate the number of moles of
there are. Given 0.80g of
, we will need to convert with the molar mass of
. Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2: ![32g\text{ }O_2=1mol\text{ }O_2](https://tex.z-dn.net/?f=32g%5Ctext%7B%20%7DO_2%3D1mol%5Ctext%7B%20%7DO_2)
Thus, ![\frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n](https://tex.z-dn.net/?f=%5Cfrac%7B0.80g%20%5Ctext%7B%20%7DO_2%7D%7B1%7D%20%5Cfrac%7B1mol%5Ctext%7B%20%7DO_2%7D%7B32g%5Ctext%7B%20%7DO_2%7D%3D0.25mol%5Ctext%7B%20%7DO_2%3Dn)
Isolating V in the Ideal Gas Law:
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
![V=\frac{nRT}{P}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BnRT%7D%7BP%7D)
...substituting the known values, and simplifying...
![V=\frac{(0.025 mol \text{ }O_2)(0.0821\frac{L\cdot atm}{mol \cdot K} )(273.15K)}{(1atm)}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B%280.025%20mol%20%5Ctext%7B%20%7DO_2%29%280.0821%5Cfrac%7BL%5Ccdot%20atm%7D%7Bmol%20%5Ccdot%20K%7D%20%29%28273.15K%29%7D%7B%281atm%29%7D)
![V=0.56L \text{ } O_2](https://tex.z-dn.net/?f=V%3D0.56L%20%5Ctext%7B%20%7D%20O_2)
So, 0.80g of
would occupy 0.56L at STP.