Lab 5 Discussion

Lab 5 Discussion - of stoichiometry We found the mass of...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Ying (Alan) Sun 11-15-07 Lab #5: The Molar Volume of Gases Discussion The purpose of this lab was to learn to determine molar volume in an experimental environment utilizing two different methods. The first method was by determining the weight of the flask containing H2O2 before and after the catalyst FeCl3 was introduced, thereby determining the mass of oxygen. We found the mass to be 0.256g, which is 0.008 mol. From there we were able to determine the volume of oxygen by looking up the water vapor pressure at the recorded temperature, and subtracting that from the current barometric pressure to give the pressure in torr for oxygen (741.367 torr), then utilized the ideal gas law V2 = V1 (P1 / P2) * (T2 / T1). We found V2 to be 0.2L. Simply dividing the volume of O2 by moles of O2 (hence volume/mole = Molar Volume), we found the molar volume of oxygen to be 25L/mol. This is only off by approximately 2.6 L/mol from the ideal gas molar volume, so our experiment yielded fairly accurate results. The next method to determine molar volume involved a step
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: of stoichiometry. We found the mass of the reactant Mg to be 0.085g. Since the mole ratio of Mg to H2 is 1:1, moles of hydrogen was equal to moles of magnesium. Dividing 0.085g by 24g (atomic weight of Mg), we found we had 0.0033 mol Mg, which equaled 0.0033 mol H2. Once again, using the same ideal gas law, we determined the volume of hydrogen at STP (V2 = 0.085L (743.523 torr / 762.762 torr) * (294K / 2965K)) to be 0.083L. Dividing 0.083L by 0.0033mol, we found the molar volume of hydrogen to be 25.2 L/mol. The main source of error for this experiment is within the reactions themselves. Hydrogen and oxygen gas most likely partially dissolved in the water, or may have not reacted in the original solution. In both cases, the results would be thrown off, resulting in a lower observed molar volume. Another source of error is caused by equipment accuracy but this is to be expected....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online