Question

how do each of the following affect the calculated volume of hydrogen gas at stp?

A: The strip of magnesium used in the experiment was partly oxidized and it was not cleaned before the strip was bent and placed in the buret.

B: The bent strip of magnesium dropped out of the end of the buret and some bubbles of H2 from the reaction were not trapped in the buret.

C: you did not correct the bariometric pressure for the vapor pressure of water.

Answer 1

A: The strip of magnesium used in the experiment was partly oxidized and it was not cleaned before the strip was bent and placed in the buret.

B: The bent strip of magnesium dropped out of the end of the buret and some bubbles of H2 from the reaction were not trapped in the buret.

C: you did not correct the bariometric pressure for the vapor pressure of water.

The reaction between Mg and HCl is as follows:

Mg+ 2HCl -- > MgCl2 + H2

A: if strip of magnesium used in the experiment was partly oxidized and it was not cleaned before the strip was bent and placed in the buret then the mass of Mg is increased which will increased the Number of observed H2.

PV= nRT

V= nRT / P

Which also increase the volume of H2 because volume is directly related to the number of moles.

B: If the bent strip of magnesium dropped out of the end of the buret and some bubbles of H2 from the reaction were not trapped in the buret means here the number of moles of H2 is less as compared to form.

PV= nRT

V= nRT / P

Which also decrease the volume of H2 because volume is directly related to the number of moles.

C: you did not correct the bariometric pressure for the vapor pressure of water.

If we did not correct the bariometric pressure for the vapor pressure of water then we read more pressure values than the actual. Hence the volume of H2 will show less values then the actual .