A sample containing 6.40 g of O2 gas has a volume of 27.0 L at constant pressure

Question

A sample containing 6.40 g of O2 gas has a volume of 27.0 L at constant pressure and temperature.

(1) A sample containing 6.40 g of O2 gas has a volume of 27.0 L at constant pressure and temperature. What is the new volume, in liters, after 0.600 mole of O2 gas is added to the initial sample 6.40 g of O2?

(2) A sample containing 6.40 g of O2 gas has a volume of 27.0 L at constant pressure and temperature. If oxygen is released until the volume is 10.0 L how many moles of O2 are removed?

(3)A sample containing 6.40 g of O2 gas has a volume of 27.0 L at constant pressure and temperature. What is the volume, in liters, after 6.00 g of He is added to the O2 gas in the container?

Explanation / Answer

At constant pressure and temperature V1 / n1 = V2 / n2

number of moles of O2 = 6.4/32 = 0.2

27 / 0.2 = V2 / (0.2 +0.6)

V2 = 108 L

V1 / n1 = V2 / n2

27 / 0.2 = 10 / n2

n2 = 0.0741 moles

number of moles removed = 0.2 - 0.0741 = 0.1259 moles

number of moles of He added = 6.4 / 4 = 1.6 moles

using same above formula,

27 / 0.2 = V2 / (0.2+1.6)

V2 = 243 L

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