Question

What is the volume of 7.25 g of oxygen gas at STP?

Answer 1

Answer:

5.15 L

Step-by-step explanation:

This looks like a case where we can use the Ideal Gas Law to calculate the volume.

pV = nRT       Divide both sides by p

 V = (nRT)/p

Data:

m = 7.25 g

R = 0.083 14 L·atm·K⁻¹mol⁻¹

T = (0 +273.15) K = 273.15K

p = 1 bar (Note: STP is 1 bar and 0°C)

Calculations:

(a) Moles of O₂

n = m/M

n = 7.25/32.00

n = 0.2266 mol

(b) Volume

V = (0.2266 × 0.083 14 × 273.15)/1

V = 161.3/2.00

V = 5.15 L  

Answer 2

The volume of 7.25 g oxygen gas at STP has been 5.075 L.

The oxygen has been assumed to be an ideal gas. The volume of 1 mole of an ideal gas at STP has been 22.4 L.

Volume of Oxygen at STP

The moles of a gas has been given as:

$\rm Moles=\dfrac{Mass}{Molar\;mass}$

The given mass of oxygen has been 7.25 g.

The molar mass of oxygen has been 32 g.

The moles of oxygen gas have been given as:

$\rm Moles\;O_2=\dfrac{7.25}{32}\\Moles\;O_2= 0.226\;mol$

The moles of oxygen available has been 0.226 mol.

The volume of 0.226 mole oxygen at STP has been:

$\rm 1\;mol=22.4\;L\\0.226\;mol=0.226\;\times\;22.4\;L\\0.226\;mol=5.075\;L$

The volume of 7.25 g oxygen gas at STP has been 5.075 L.

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