#3 What size container would 67.3mol CO2 need at STP?

The volume of the container would be 1507 L

Explanation:

These question can be solved in two different ways:

Let's do the short road.

We know that 1 mol of anything at STP conditions occupies 22.4 L

By a rule of three:

1 mol contains 22.4L at STP

Then, 67.3 mol of gas would be contained in (67.3 . 22.4) /1 = 1507.5 L

We assume CO₂ as an Ideal Gas, so we can apply P . V = n . R . T

At STP, T is 273.15K and P = 1 atm. We replace:

1 atm . V = 67.3 mol . 0.082 . 273.15K

V = (67.3 mol . 0.082 . 273.15K) / 1 atm → 1507 L

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.