# LM 04 Avagadro's Law-solutions - liquid water Ignoring the...

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nanni (arn437) – LM 04 Avagadro’s Law – vanden bout – (50985) 1 This print-out should have 2 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points 16 . 8 lbs oF a gas occupies 358 Ft 3 at a certain pressure and temperature. The amount oF gas is then increased to 39 . 2 lbs. What is the new volume For this amount assuming the pressure and temperature haven’t changed? Correct answer: 835 . 3 Ft 3 . Explanation: The volume oF a gas is proportional to the amount (Avogadro’s Law). Even though we don’t know the identity oF the gas we do know the amount in lbs. Volume will scale with lbs in the same way it scales with moles. ANY amount you pick will scale accordingly. Avogadro’s Law: V 1 amount 1 = V 2 amount 2 V 2 = V 1 p amount 2 amount 1 P = 358 p 39 . 2 16 . 8 P V 2 = 835 . 3 Ft 3 002 10.0 points Suppose that 2 L oF propane gas (C 3 H 8 ) at 1 atm and 298 K is mixed with 7 . 2 L oF oxy- gen gas at the same pressure and temperature and burned to Form carbon dioxide gas and
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Unformatted text preview: liquid water. Ignoring the volume oF water Forward, fnd the fnal volume oF the reaction mixture (including products and excess reac-tant) at 1 atm and 298 K iF the reaction goes to completion. Correct answer: 4 . 88 L. Explanation: The balanced equation is C 3 H 8 (g) + 5 O 2 (g)-→ 3 CO 2 (g) + 4 H 2 O( ℓ ) . Because the reaction remains at 1 atm pressure and 298 K throughout, we can use the volumes directly to solve the problem ( V ∝ n ). 2 L oF propane requires 10 L oF O 2 to react in order to go to completion; O 2 is thus the limiting reactant. Use the volume oF oxygen to fnd the volume oF propane reacted: V propane = (7 . 2 L O 2 ) 1 L propane 5 L O 2 = 1 . 44 L propane This leaves 0 . 56 L oF unreacted propane and produces ? L CO 2 = (7 . 2 L O 2 consumed) × 3 L CO 2 produced 5 L O 2 consumed = 4 . 32 L CO 2 . The total volume is V = 0 . 56 L + 4 . 32 L = 4 . 88 L ....
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