Question

How many grams of oxygen gas are there in a 2.3 L tank at 7.5 atm and 24 C

Answer 1

Answer:

22.6 g

Explanation:

First we use the PV=nRT equation:

• P = 7.5 atm

• V = 2.3 L

• n = ?

• R = 0.082 atm·L·mol⁻¹·K⁻¹

• T = 24 °C ⇒ 24 + 273.16 = 297.16 K

We input the data:

• 7.5 atm * 2.3 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 297.16 K

And solve for n:

• n = 0.708 mol

Then we convert 0.708 moles of oxygen gas (O₂) to grams, using its molar mass:

• 0.708 mol * 32 g/mol = 22.6 g