Answer:
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
