Procedure:
1. Measure and cut a piece of magnesium ribbon 3.5 to 4.5 cm long. Do not exceed 4.5 cm.
2. Measure the mass of the magnesium ribbon to the nearest milligram (0.001 g) and record
the mass.
3. Fold up the magnesium ribbon into a small, tight bundle. Tie it with a piece of thread (10
to 15 cm long).
4. Add about 10 mL of 3.0M HCl to a gas collection tube.
5. Hold the gas collection tube at an angle as you fill it completely with tap water, until it is
nearly overflowing. Holding the tube at an angle allows the air bubbles to escape.
6. Place the magnesium in the mouth of the gas collection tube so that it is about three cm
below the surface of the water. Fold the thread extension over the side of the tube. Insert
a one-hole stopper into the opening so that the magnesium ribbon is held firmly in place.
7. Fill a 400-mL beaker half full with water.
8. Holding your finger over the hole in the rubber stopper, invert the tube into the 400-mL
beaker of water. Clamp the gas collection tube in place on the ring stand so that the
stoppered end of the tube is under the water but not touching the bottom of the beaker.
Warning: Even though the acid has been diluted by the water, be sure to wash
your hands after handling the gas collection tube.
9. Observe the reaction. When no more hydrogen bubbles are visible, the reaction is
complete. Tap the tube gently to release any hydrogen gas bubbles on the side of the tube.
10. Wait an additional five minutes so that the hydrogen gas reaches room temperature.
11. Cover the hole in the stopper with your finger and transfer the tube to a large cylinder
filled with water. Lower or raise the tube until the liquid level on the inside of the tube is
the same as the outside. Record the volume of the hydrogen gas inside the gas collection
tube. (Be sure to read the liquid level at eye level.)
12. Measure and record the temperature and pressure of the lab room.
13. Obtain the appropriate vapor pressure of water from the table below and record.
Data and Observations:
Present all relevant data in a data table below. Include an observations section for any
observations that you made during the lab.
Mass of magnesium strip (grams)
Volume of gas collected (mL)
Barometric pressure (atm)
Room temperature (C)
Vapor pressure of water (torr)
0.032g
31mL
1.1atm
22C
19.8torr
Calculations:
Calculations
2 - show your work, I don't think this is right but I can't tell without work
3 - don't you want the moles of H2, so shouldn't you use the partial pressure of H2?
5 - correct once you've corrected 2 and 3
1. Write the balanced equation for the reaction conducted in this lab, including appropriate
phase symbols.
Mg(s) + 2HCl(aq)  MgCL2(aq) + H2 (g)
2. Determine the partial pressure of the hydrogen gas collected in the gas collection tube.
1.1atm = P1 + 0.0395atm
1.1atm – 0.0395atm = 1.0605atm
3. Calculate the moles of hydrogen gas collected.
𝑛(𝐻2 ) =
(1.0605𝑎𝑡𝑚)(0.31𝐿)
= 0.0136𝑚𝑜𝑙 𝐻2
(0.0821)(295.15𝐾)
Moles of H2 gas collected = 0.0136mol H2
4. If magnesium was the limiting reactant in this lab, calculate the theoretical yield of the
gaseous product. Show all steps of your calculation.
0.03232g Mg x (1mol Mg / 24.305g Mg) x (1mol H2 / 1mol Mg) x (2.014g H2 / 1mol
H2) = 0.267g H2  theoretical yield
5. Determine the percent yield of this reaction, showing all steps of your calculation.
0.0136𝑚𝑜𝑙 𝐻2 ×
1𝑚𝑜𝑙 𝐻2
2.014𝑔 𝐻2
×
= 0.0274𝑔 𝐻2
1𝑚𝑜𝑙 𝐻2
1𝑚𝑜𝑙 𝐻2
0.0274𝐻2
× 100 = 10.26%
0.267𝑔 𝐻2
10.26% = percent yield
Conclusion:
1. Would the following errors increase, decrease, or have no effect on the calculated moles
of gas collected in the experiment? Explain your answers in complete sentences.
a. The measured mass of the magnesium was smaller than the true mass.
If the measured mass of the magnesium was smaller than the true mass, this
would cause an increase in the calculated moles. This is because the smaller the
nuumber that you divide by, the bigger the answer.
If the measured mass of the magnesium was smaller than the true mass, this
would have no effect because this measurment is not used in the calculation nor
is it used in the ideal gas law (PV=nRT)
b. The actual temperature of the hydrogen gas is lower than room temperature.
This would cause an increase in the answer that I get because more temperature
means more volume because the particles are moving faster as the temperature
goes up.
This would cause the answer to become larger beause I would be dividing by
a smaller number. For example, 100/50 = 2, but 100/2 = 50.
2. Explain in terms of particle collisions and Dalton’s law why it can be assumed that the
total pressure inside the gas collection tube is equal to the atmospheric pressure outside of
the tube.
If the particles outside the tube are creating more exertion than the gas particles on
the inside of the tube, then the tube ould collapse. If the particles on the inside of the
tube are creating more exertion than the gas particles on the outside of the tube,
then the tube would explode.
3. If an undetected air bubble was trapped inside the gas collection tube, how would this
affect your calculated percent yield? Explain your answer.
When you measure your percent yield, you would be counting that extra air bubble, thus
interruping your calculation making it wrong. This is because that extra air bubble is not
part of that particluar gas making it unpure.
The air bubble would make the actual yield larger, but the theoretical yield isn’t
affected because you don’t need this amount to calculate the theoretical yield. But,
because the actual yield is larger, this would affect the percent yield because the
answer would not be correct.