MHS Chemistry
Stoichiometry Lab
(Magnesium version)
Name: _________________________
Section: A B C D E F G

Magnesium metal reacts with sulfuric acid to form hydrogen gas and a magnesium sulfate solution. In this investigation, you will predict the mass of hydrogen gas produced by a given amount of reactants, then measure the amount, and compare them.

Magnesium metal is very reactive in air, and usually covered with a rough layer of corrosion. This layer should be polished with steel wool before beginning the reaction.

Sulfuric acid is an aqueous solution of hydrogen sulfate. The concentration is usually described in terms of “moles per liter” which is abbreviated “mol/L” or “molar” or “M.” The most concentrated sulfuric acid is 15 M, but for safety reasons we will be using a solution that is about 0.75 M.


  1. Put on a pair of safety goggles, and open your lab notebook to a new page.
  2. Obtain approximately 0.25 grams of magnesium ribbon. Polish it until it is shiny, and record the exact mass of the ribbon you actually use.
  3. Put 20.0 mL of sulfuric acid into a small beaker. Record the mass of the beaker with solution in it.
  4. Place the ribbon on the balance tray next to the beaker of acid, and record the total mass. [If this total mass does not equal the sum of number 2 and 3, you should remeasure.]
  5. With the beaker still on the balance, drop the ribbon into the acid. What happens to the overall mass?
  6. When the magnesium is gone, record the final mass of the beaker with the solution.
  7. Dispose of the materials in the beaker according to your teacher’s instructions.


  1. Write the chemical equation for the reaction that took place. Don’t forget to include the states of matter (s, l, g, or aq). Balance it.
  2. Calculate the number of moles of (clean) magnesium used.
  3. Predict the number of moles of hydrogen that should be produced by this much magnesium reacting.
  4. Calculate the mass of hydrogen gas produced during your reaction.
  5. Calculate the number of moles of hydrogen gas actually produced.
  6. Calculate the “percent yield” for your reaction. If your prediction (#3) and calculation (#5) were the same, your percent yield was 100%.
  7. Why was your percent yield what it was? In other words, if you got 100%, why was it so perfect? If you got less than 100%, what happened?

[Moles Notes][MHS Chem page]