Question

Air is 21% oxygen by volume. Oxygen has a density of 1.31 g/L. What is the volume in liters of a room that holds enough air to contain 55kg of oxygen?

Answer 1

First divide the mass of oxygen by its density,
55 kg / 1.31 L = 42 liters
Now if V is the volume of the room, then 21% of this volume is oxygen.
This can be written as:
0.21 V = 42 liters
Divide both sides of the equation by 0.21 to get V
V = 42/0.21 liters
V = 200 liter

Answer 2

Answer:

199,927.29 L is the volume air that holds enough air to contain 55 kg of oxygen.

Explanation:

Density of oxygen = 1.31 g/L

Mass of oxygen gas,m = 55 kg = 55,000 g

Volume of the oxygen gas = V

Volume of the air = $V_a$

$Density=\frac{Mass}{Volume}$

$1.31 g/L=\frac{55,000 g}{V}$

V = 41.984.732 L

Air is 21% oxygen by volume.That is in 100 L of air 21 L of oxygen  is present.

So, when oxygen gas is 41.984.732 L the volume of the air will be:

$21\%=\frac{41.984.732 L}{V_a}\times 100$

$V_a=199,927.29 L$

199,927.29 L is the volume air that holds enough air to contain 55 kg of oxygen.