How many liters of O2 gas at STP is equivalent to 14g of O2?
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To answer this question, you must use the ideal gas equation:
P V = n R T
"P" stands for pressure, "V" stands for volume, "n" stands for number of moles, "R" stands for gas constant, and "T" stands for temperature.
The problem mentions that this scenario occurs at STP, which stands for standard temperature and pressure. This automatically gives us two important pieces of information: (1) the temperature in the scenario is 273 K & (2) the pressure in the scenario is 1 atm.
It's also given to us that there are 14 g of O2 present in the scenario, which can help us to calculate the number of moles! By looking at the periodic table, we can see that 1 mole of oxygen has a mass of 15.999 g. Since 1 mole of O2 is composed of 2 moles of oxygen, we can find that 1 mole of O2 has a mass of 31.998 g.
`2 xx (15.999 g) = 31.998 g`
We then use the molar mass of O2 to find the number of moles:
`(14 g)/(1) xx (1 mol)/(31.998 g) = 0.44 mol`
The gas constant -- "R" -- is a constant, and we choose which ones to use based on the other units in our scenario. Since this scenario deals with atm, K, and mol, we will use the following for "R":
Now we can put everything we know into the ideal gas equation & solve for volume:
`(1 atm) xx V = (0.44 mol) xx (0.08206(L atm)/(mol K)) xx (273 K)`
`V = 9.9 L`
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