The Reactivity of Magnesium Metal with Hydrocholric Acid

The Reactivity of Magnesium Metal with Hydrochloric Acid
Adam Capriola
CHM 1122 Section 155
February 7, 2007
The ideal gas law is used to define how gasses typically act. It is not precise, as gasses
do not usually act ideally, but it works for most laboratory conditions. The ideal gas law is
defined by PV = nRT (pressure in atm * volume in L = moles * constant * temperature in K).
There is also a rule that states the total pressure of a system is equal to the sum of its partial
pressures. Using this information, a small piece of metal magnesium is reacted with
hydrochloric acid. The reaction is represented by the equation Mg (s) + 2HCl (aq) --> MgCl2
(aq) + H2 (g). Using the sum of partial pressures rule and the ideal gas law to find moles, the
molar mass of Mg (s) can be calculated only knowing the pressure of H2 (g).
First, a strip of magnesium metal was obtained and weighed on an analytical balance.
This weight was recorded to three significant figures. Next, the magnesium strip was wrapped
around the tip of a copper wire, and was then encaged by the same copper wire by wrapping the
wire around the magnesium. Following this, the tip of a buret was calibrated and this volume
was recorded. Next, about 3 mL of concentrated HCl was put in the empty buret with the
stopcock closed, and distilled water was then added until the buret was almost full. The copper
wire with the magnesium was then lowered into the buret and held in place with a one-holed
rubber stopper to make sure the wire would stay in place and would not fall to the bottom.
Distilled water was squirted into the one-holed rubber stopped to make sure the buret was
completely filled. The buret was then quickly inverted and clamped into a 600 mL beaker half
filled with tap water. The water level on the buret was recorded after production of gas ceased.
Finally, the distance between the height of water in the beaker and water in the buret was also
recorded. The process was repeated for one more trial.
Mass of Mg sample, g
Volume of uncalibrated
portion of buret, mL
Final buret reading, mL
Volume of hydrogen, mL
Temperature of hydrogen, ºC
Barometric pressure, mm Hg
Differences in water levels
between buret and beaker,
mm H2O
Pressure difference of water
levels, mm Hg
Aqueous vapor pressure at
temperature of hydrogen, mm
Pressure of hydrogen after
correction for difference in
water levels and vapor
pressure, mm Hg
Pressure converted to atm,
Absolute temperature, K
Volume of hydrogen, L
Moles of hydrogen gas
produced, moles
Calculated molar mass of Mg,
Actual molar mass of Mg,
Percent error
Trial 1
Trial 2
Average percent error: 0.4525
In order to find the volume of hydrogen, I took the total volume of the buret (50 mL plus
the calibrated part) and subtracted my final reading. Using the results from my first trial, 50.0
mL + 1.20 mL – 9.40 mL = 41.80 mL. To find the pressure difference of water levels, I took the
distance in difference of water levels, multiplied by the density of water, and divided by the
density of Hg. From the first trial, 135 mm (1.0 g/mL) / 13.5 g/mL = 10.0 mm. The aqueous
vapor pressure was found using a chart. At 23 ºC, the aqueous vapor pressure is 21.2 torr (or
21.1 mm Hg). Using the equation Patm = PH2O (l) + PH20 (g) + PH2 (g), I could solve for the pressure
of H2 (g). Subbing in the numbers from the first trial, 762.76 torr = 21.2 torr + 10.0 torr + PH2 (g).
PH2 (g) = 731.6 torr. To find atmospheres, I multiplied by 1 atm/760 torr. 731.6 torr (1 atm/760
torr) = 0.9626 atm. I then rearranged the ideal gas law to find moles of hydrogen produced. n =
PV/RT, n = 0.9626 atm * 0.04180 L / (0.0821 L atm mole-1 K-1 * 296 K), n = 0.001656 moles.
In the equation for the reaction, there are an equal number of moles of hydrogen and magnesium
used. To find the molar mass of magnesium, I took the mass of the magnesium used divided by
the number of moles used. From trial 1, 0.0406 g / 0.001656 moles = 24.52 g/mol. To find
percent error, I took the absolute value of the true value minus my calculated value divided by
the true value and multiplied by 100%. Using trial 1, ( | 24.31 - 24.52 | ) / 24.31 * 100% =
0.8638%. Finally, to find the average percent error, I added the two percents and divided by
two: (0.8638% + 0.04113%) / 2 = 0.4525%.
My results are very good. An average percent error of 0.4525% is extremely small, so I
feel that I performed the experiment well and had some good luck. I felt that when performing
the experiment, I could have gotten some error from my temperature readings. The temperature
seemed to fluctuate a couple degrees during the experiment and I thought that would negatively
affect my final results. I also thought I could have gotten error from my difference in water
levels measurement because it was hard to hold the ruler steady and read the ruler at the same
time. Lastly, since the ideal gas law is not precise, error could have come from that. Other than
those questionable factors, the experiment went smoothly.
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