Answer:
28 liters of O2
Step-by-step explanation:
A marvelous byproduct of establishing a STP (Standard Temperature and Pressure) is a conversion factor that 1) seems impossible, and 2) saves a lot of time. That factor is 22.4 Liters/mole gas. At STP ALL gases occupy 22.4 liters per mole of that gas. Hydrogen, argon, methane, etc. all will occupy 22.4 liters at 1 atm pressure and 0°C.
That made no sense to me, but after realizing the amount of time it can save answering question as this, I'm a firm believer.
Without this conversion factor, the way to answer the question is to use the ideal gas law: PV=nRT, where n is the number of moles and R is the universal gas constant. Not a super difficult equation, but looking up the gas constant and making sure all the units match and cancel can take time and requires focus.
But the conversion factor is straightforward. We are given 40 g of O2. Let's coivert the 40g into moles O2. Divide the mass by the molar mass of O2 (32 g/mole):
(40g O2)/(32 g/mole O2) = 1.25 moles of O2
Now is when the ideal gas law might be employed. We have P, T, n, and R, so plug them is and calculate volume. This is done below, just as an illustration.
But since we are at STP, we can use the handy coversion factor to determine the volume of O2:
(1.25 moles O2)*(22.4L/mole O2) = 28.0 liters of O2
In contrast, the ideal gas law calculation would look like this:
Rearrange to isolate V, the unknown:
V = (nRT)/P
Enter the data, ensuring the correct gas constant is used for the units provided:
V = [(1.25 moles)*(0.0820575L⋅atm⋅K^1⋅mol^1)(273.15K)]/(1atm)
Ouch
V = 28.0 liters
Same answer, more effort.