How many grams of chlorine gas are in a 17.8 liter sample at 1.1 atmospheres and 29 degrees celsius

For this problem, we can use the Ideal Gas Law to solve for the number of moles of chlorine gas present in this container, and then we can convert this molar measurement into grams.

The Ideal Gas Law is presented as:

PV=nRT

where P is pressure, V is volume, in liters, n is the number of moles of substance, R is the gas constant associated with the units for pressure, and T is temperature, in Kelvins.

Starting with the IGL, we can substitute the information given in the problem, and then we can simplify to a number of moles of Chlorine gas.

PV=nRT
(1.1 atm)(17.8 L) = n(0.0821 L•atm/mol•K)(302.15 K)
19.58=24.807n
n=0.7893 moles

Chlorine is a gas in this problem, which means it is the di-Chlorine form most commonly found as a gas (Cl2). This molecule weighs twice as much as elemental chlorine (2 mol x 35.453 g/mol). So, we can multiply this new weight by the number of moles of chlorine gas we calculated:

70.9 g/mol • 0.7893 mol = 55.96 g Cl2

55.96 grams of diChlorine gas is present in the sample.

Answer:

5.6⋅10−^19  j

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