When a student doesn't polish a sample of Mg ribbon before massing it, what happens when calculating the molar volume? Is it that the Mg ribbon doesn't completely react therefore less volumes of gas are produced?

Thanks :) so just to be clear, does it matter what the Mg is reacting with, or is it always going to have the coating of the MgO on it?

In air the coating will be MgO. In some other atmosphere it may be something else.

When a student doesn't polish the sample of Mg ribbon before massing it, the presence of impurities on the surface of the ribbon can affect the accuracy of the experiment and result in an incorrect calculation of the molar volume.

By not polishing the Mg ribbon, impurities like oxide layers or other contaminants can be present on the surface. These impurities may act as a barrier, hindering the reaction between the Mg ribbon and the acid. As a result, the reaction may not proceed to completion, leading to less volume of gas being produced.

The molar volume is calculated by dividing the volume of gas produced by the number of moles of reacting substances. If the reaction is incomplete due to the presence of impurities, the calculated molar volume will be lower than the actual value.

To obtain accurate results for calculating the molar volume, it is important to ensure a clean and polished surface of the Mg ribbon. This can be achieved by using fine sandpaper or a metal file to remove any impurities or oxide layers before weighing the ribbon. By doing so, the entire surface area of the ribbon is exposed and available for reaction, leading to a more accurate measurement of the molar volume.

"Old" magnesium has a thin coating of MgO on the surface. The purpose of polishing is to remove the oxide coating. If you don't remove the oxide coating then the mass of "real" Mg is too high (due to the oxide coating being weighed along with the Mg) and less gas is evolved than should be.