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

Answer:

0.56L

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:$R=0.0821\frac{L\cdot atm}{mol\cdot K}$

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:

$PV=nRT$

$V=\frac{nRT}{P}$

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

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

$V=0.56L \text{ } O_2$

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

Answer 2

Answer:

0.56 L

Explanation: