Question

What is the volume of 0.80 grams of o2 gas at stp? (5 points) group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 liters

Answer 1

0.56 L

Step-by-step explanation:

Answer 2

0.56L

Step-by-step explanation:

This question requires the Ideal Gas Law:
`PV=nRT` 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(L\cdot atm)/(mol\cdot K)`

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means
`T=273.15K` and
`P=1atm`

Lastly, we must calculate the number of moles of
`O_2(g)` there are. Given 0.80g of
`O_2(g)`, we will need to convert with the molar mass of
`O_2(g)`. 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`

Thus,
`\frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n`

Isolating V in the Ideal Gas Law:

`PV=nRT`

`V=(nRT)/(P)`

...substituting the known values, and simplifying...

`V=\frac{(0.025 mol \text{ }O_2)(0.0821(L\cdot atm)/(mol \cdot K) )(273.15K)}{(1atm)}`

`V=0.56L \text{ } O_2`

So, 0.80g of
`O_2(g)` would occupy 0.56L at STP.