Question

A sample containing 5.60 g O 2 gas has a volume of 21.0 L . Pressure and temperature remain constant.Part A What is the new volume if 0.400 mole O 2 gas is added? Part B Oxygen is released until the volume is 7.50 L . How many moles of O 2 are removed? Part C What is the volume after 5.50 g He is added to the O 2 gas already in the container?

Answer 1

Part A : new volume = 69.0 L

Part B : moles O₂ removed = 0.113 mol   (if wrong then try 0.513 mol)

Part C : volume = 186 L (if wrong then try 172 L)

Explanation

Part A

initial mass O₂ = 5.60 g

initial moles O₂ = (initial mass O₂) / (moles O₂)

initial moles O₂ = (5.60 g) / (32.0 g/mol)

initial moles O₂ = 0.175 mol

According to Avogadro's law,

V₁ / n₁ = V₂ / n₂

where V₁ = initial volume = 21.0 L

n₁ = initial moles O₂ = 0.175 mol

V₂ = final volume

n₂ = final moles O₂ = (0.175 mol) + (0.400 mol) = 0.575 mol

V₂ = V₁ * (n₂ / n₁)

V₂ = (21.0 L) * (0.575 mol / 0.175 mol)

V₂ = (21.0 L) * (3.28)

V₂ = 69 L