Question

A personal flotation device is being blown up through exhalation of breath, which is made up of roughly 78.0% N, 16.04% O₂, 5.00% CO₂, and .960% Ar. At STP, 10 breaths will cause the volume of the system to increase by 600mL. If the system is inflated with 65 breaths, how many moles of Oxygen will it contain? What volume in mL does the O₂ occupy?

Answer 1

Solution:

1. Moles of O₂ in 10 breaths:

10 breaths = 600 mL volume

O₂ volume in 600 mL = 0.600 L x (16.04 / 100.00) = 0.0962 L

22.4 L is occupied by 1 mole O₂. Therefore, 0.0962 L = (0.0962 L x 1 mol) / 22.2 L = 0.0042946 mole

2. Moles of O₂ in 65 breaths:

10 breath = 0.0042946 mole O₂

65 breath = x

x = (65 breath x 0.0042946 mole O₂) / 10 breaths = 0.027924 mole O₂.

Hence, 65 breaths will contain 0.027924 moles O₂.

3. Volume of O₂:

1 mole O₂ = 22.4 L

0.027924 mole O₂ = y

y= (22.4 L x 0.027924 mole O₂) / 1 mole O₂ = 0.625 L

1 L = 1000 mL

0.625 L = 0.625 L x (1000 mL / 1 L) = 625 mL

Hence, 625 mL is the volume of O₂.