12-3
Limiting Reactants
 Balanced equations give exact mole ratios
 When the reactions is carried out, the reactants are
usually not available in those same ratio amounts
 One reactant will be in excess, while one will be
limited.
 The reaction will continue until the limiting reactant
runs out.
 Limiting reactant- limits the extent of the reactants and
determines the amount of product produced
 Excess reactants- reactants left over after the reaction
has stopped
 Steps for calculating the product when limiting
reactant is present: Ex. S8(l) + 4Cl2  4S2Cl2(l)
200.0gS reacts with 100.0gCl: mass of product = ?
1. find # of moles of each reactant (mult. by inverse
of mol. mass: 1/mol. mass)
100.0g Cl2 X 1mol Cl2 =1.410 mol Cl2
70.91g Cl2
200.0g S8 X 1mol S8 = 0.7797 mol S8
256.5g S8
2. determine whether reactants are in correct mole
ratio (divide available moles of chlorine by
available moles of sulfur)
1.410 mol Cl2 available = 1.808 mol Cl2 available
0.7797 mol S8 available 1 mol S8 available
need a 4:1 ratio, but have a 1.808:1 ratio
An alternate method would be to calculate the
moles of product produced by both reactants. The
reactant producing the least amount is the limiting
reactant!
Therefore chlorine will be used up first and is
limiting reactant!
3. change moles of limiting reactant to moles of
product (multiply moles of limiting reactant by
the mole ratio that relates the product to the
limiting reactant)
1.410 mol Cl2 X 4 mol S2Cl2 = 1.410 mol S2Cl2
4 mol Cl2
4. change moles of product to grams of product
(multiply moles of product by its molar mass in
grams)
1.410 mol S2Cl2 X 135.0g S2Cl2 = 190.4g S2Cl2
1 mol S2Cl2
To find out how much of the excess reactant is
used:
5. change moles of limiting reactant to moles of
excess reactant (multiply moles of limiting
reactant by the mole ratio that relates the excess
reactant to the limiting reactant)
1.410 mol Cl2 X 1 mol S8 = 0.3525 mol S8
4 mol Cl2
6. change actual used moles of excess reactant to
grams (multiply moles of excess reactant by its
molar mass in grams)
0.3525 mol S8 X 256.5g S8 = 90.42g S8
1 mol S8
7. amount of sulfur left un-reacted may be
calculated by subtracting amount used from the
amount available
200.0g S8 available – 90.42g S8 needed = 109.6g S8 excess
Complete Practice Problems Page 368 #20 &21