Limiting Reactants
A limiting reactant is the compound or molecule which is used up the soonest in a reaction.
To find the limiting reactant, we need to compare the moles of the reactants together and
determine which has the smallest amount.
Remember how to convert from grams to moles:
𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑍 = (𝑔𝑟𝑎𝑚𝑠 𝑍 ∗
1 𝑚𝑜𝑙
)
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑍 (𝑔)
Example, if we have 5.2 grams of Na:
(5.2 𝑔 𝑁𝑎 ∗
1 𝑚𝑜𝑙 𝑁𝑎
) = 0.226 𝑚𝑜𝑙 𝑁𝑎
22.99 𝑔 𝑁𝑎
We will repeat this process for each reactant in a balanced chemical equation, if moles are not
already given in the problem.
For example:
We have 2.29g of H2SO4 and 5.77g of NaOH. Find the limiting reactant.
H2SO4 + 2 NaOH → Na2SO4 + 2 H2O
Find the molecular weight of both reactants:
-
H2SO4 = (2 x 1.008 g/mol H) + (1 x 32.06 g/mol S) + (4 x 15.999 g/mol O) = 98.072 g/mol
-
NaOH = (1 x 22.99 g/mol Na) + (1 x 15.999 g/mol O) + (1 x 1.008 g/mol H) = 39.997 g/mol
Convert these to moles, starting with the grams given in the problem.
-
2.29 𝑔 H2SO4 ∗
-
5.77 𝑔 NaOH ∗
1 𝑚𝑜𝑙 H2SO4
98.072 𝑔 H2SO4
1 𝑚𝑜𝑙 NaOH
39.997 𝑔 NaOH
= 0.0233 𝑚𝑜𝑙 H2SO4
= 0.144 𝑚𝑜𝑙 NaOH
Compare the two values by dividing them by their respective coefficients in the balanced
chemical equation.
-
0.0233 𝑚𝑜𝑙 𝐻2𝑆𝑂4
1
0.144 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
2
= 0.0233 mol H2SO4
= 0.072 mol NaOH
The lower value is 0.0233 mol of H2SO4, meaning that it is the limiting reactant, and NaOH is the
excess reactant.
To find the amount of excess, follow this process:
Take the limiting reactant’s initial mole value (before dividing by coefficients) and convert it to
the excess reactant using the coefficient ratio found in the balanced chemical equation.
-
2 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
0.0233 𝑚𝑜𝑙 H2SO4 ∗ (1 𝑚𝑜𝑙 𝐻2𝑆𝑂4) = 0.0466 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
Then, convert this to grams of the excess reactant.
-
39.997 𝑔 𝑁𝑎𝑂𝐻
0.0466 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 ∗ (
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
) = 1.864 𝑔 𝑁𝑎𝑂𝐻
This is how much is used up by the reaction, but it is not the amount of excess. We started with
5.77 g of NaOH, and we used 1.864 g. The amount of excess is the difference of these two
values.
-
5.77 𝑔 𝑁𝑎𝑂𝐻 − 1.864 𝑔 𝑁𝑎𝑂𝐻 = 3.906 𝑔 𝑁𝑎𝑂𝐻
This is how much NaOH is left over, or is in excess.
Refer to the practice worksheet for more practice over these problems!
Theoretical vs Actual Yield, and Percent Yield
Theoretical yield is how much product the reaction should make, while the actual yield is how
much the reaction actually made when the reaction was run. The percent yield compares these
two and gives feedback on how efficient the methods were during experimentation.
To find theoretical yield of one of our products, we need to convert from moles of our limiting
reactant to moles of the product, and then to grams of the product.
Using the example from above, we can find theoretical yield:
-
MW of Na2SO4 = (2 x 29.990 g/mol Na) + (1 x 32.06 g/mol S) + (4 x 15.999 g/mol O) =
156.036 g Na2SO4
-
1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂4
156.036 𝑔 𝑁𝑎2𝑆𝑂4
0.0233 𝑚𝑜𝑙 H2SO4 ∗ ( 1 𝑚𝑜𝑙 𝐻2𝑆𝑂4 ) ∗ (
1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂4
) = 3.636 𝑔 𝑁𝑎2𝑆𝑂4
A student ran this reaction and found that their actual yield was 2.78 g Na2SO4. The formula for
percent yield is as follows:
% 𝑦𝑖𝑒𝑙𝑑 = (
𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
) ∗ 100
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
For our example, this would be:
2.78 𝑔 𝑁𝑎2𝑆𝑂4
% 𝑦𝑖𝑒𝑙𝑑 = (
) ∗ 100 = 76.46%
3.636 𝑔 𝑁𝑎2𝑆𝑂4
Percent yield
Evaluation
Close to 100%
Quantitative
Above 90%
Excellent
Above 80%
Very good
Above 70%
Good
Above 50%
Fair
Below 40%
Poor