1) Aluminum metal reacts with chlorine gas to form solid aluminum trichloride, AlCl3. What mass of chlorine gas (Cl2) is needed to react completely with 163 g of aluminum?

Question

2) Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron.

3Mg(s) + 2FeCl3(s) → 3MgCl2(s) + 2Fe(s)

A mixture of 41.0 g of magnesium ( Picture = 24.31 g/mol) and 175 g of iron(III) chloride ( Picture = 162.2 g/mol) is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.

Answer 1

1.
mols Al = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Al to mols Cl2.
Now convert mols Cl2 to g Cl2. g = mols x molar mass.

2.
Do what I did above several times; ie. as follows:
mols Mg = ?
mols Fe = ?
Using the coefficients in the balanced equation, convert mols Mg to mols Fe.
Do the same for mols FeCl3 to mols Fe.
It is likely the two values will not be the same. The correct answer in limiting reagent problems is ALWAYS the smaller number and the reagent producing that value is the limiting reagent (LR).
To determine how much of the LR was used use the coefficients as above to convert mols Mg to mols FeCl3. Then subtract amount used from the initial amount to find the amount remaining after reaction is complete.