Experiment 5, Molar Volume of Gases

Experiment 5, Molar Volume of Gases - Part 1 2H 2 O 2(l...

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Unformatted text preview: Part 1: 2H 2 O 2 (l) catalyst > 2H 2 O (l) + O 2 (g) In this experiment we used the decomposition of hydrogen peroxide, H 2 O 2 , to determine the molar volume of oxygen. Since the reaction is slow, a catalyst, FeCl 3 , was used to speed up the reaction without being consumed in the reaction itself. Because of the law of conservation of mass, if the weight of the products after the reaction is finished is subtracted from the weight of reactants before the reaction, the mass of oxygen gas produced can be found. The volume of O 2 produced is how much water was displaced. To then calculate the molar volume, the volume of oxygen gas was divided by the moles of O 2 present. The molar volume we calculated for oxygen, 22.74 L/mol was too low compared to the true value of O 2 at STP, 22.39 mol/L. Some sources of error include not equalizing the pressure inside and outside the flask and some oxygen gas dissolving in water. If the pressure inside of the flask was more than outside, the pressure would have been less than it should be...
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This note was uploaded on 03/14/2010 for the course CHEM 109A 109A taught by Professor Aue during the Spring '09 term at UCSB.

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