# You have a 40 oz. canister of compressed CO2 powering your paintball gun. You calculate that each time you pull the trigger you release 3.69 grams of CO2. What will the volume of the released gas be immediately after the shot, assuming STP? To solve this problem, we need to first convert 3.69 grams of CO2 to moles:

Molar mass of CO2 = 44.1 grams/mole

3.69 grams CO2 x 1 mol/44.1 g = 0.084 moles CO2

There are two ways to calculate the volume. One is to use the ideal gas law, PV=nRT:

P = standard pressure = 1.00 atm

V = unknown

n = 0.084 moles

R = ideal gas constant = 0.0821 L-atm/mol-K

T = standard temperature = 273 K

V = nRT/P = [(0.084 mol)(0.0821 L-atom/mol-K)(273 K)]/1.00 atm = 1.88 L

A second method, which is faster, is to use the molar volume of gas. The volume of one mole of an ideal gas at STP is 22.4 L. Since the conditions are those of STP you can multiply the number of moles by the molar volume:

0.084 mol x (22.4L/1 mol) = 1.88 L

This is the volume that the CO2 gas will expand to as soon as it's released from the canister.

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