Limiting reactant Knowledge-Based Study

Study notes

1. What is a Limiting Reactant?

   In chemical reactions, reactants are rarely combined in exact stoichiometric proportions. The limiting reactant (or limiting reagent) is the substance that is completely consumed first in a reaction. Because the reaction stops once this substance is exhausted, the limiting reactant dictates the maximum quantity of product that can form.

   The reactant that remains after the reaction has ceased is called the excess reactant.

   • Limiting Reactant: Fully consumed; determines the theoretical yield.
   • Excess Reactant: Partially consumed; a portion remains unreacted.

   How to Determine the Limiting Reactant

   To identify the limiting reactant from given quantities of starting materials, follow these systematic steps:

   1. Write and balance the chemical equation.
   2. Convert all starting masses or volumes into moles.
   3. Divide the available moles of each reactant by its corresponding stoichiometric coefficient from the balanced equation.
   4. The reactant yielding the smallest mole-to-coefficient ratio is the limiting reactant.

   Consider the synthesis of ammonia: $$\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$$ If 2.0 moles of $\text{N}_2$ and 4.5 moles of $\text{H}_2$ are mixed:

   • For $\text{N}_2$: $\frac{2.0\text{ mol}}{1} = 2.0$
   • For $\text{H}_2$: $\frac{4.5\text{ mol}}{3} = 1.5$

   Because 1.5 is smaller than 2.0, $\text{H}_2$ is the limiting reactant, and $\text{N}_2$ is in excess.

   Determining Product Yield

   The theoretical yield is the maximum amount of product that can be formed from the given amounts of reactants, calculated entirely from the moles of the limiting reactant.

   Using the previous example where $\text{H}_2$ is the limiting reactant (4.5 moles):

   $$4.5\text{ mol }\text{H}_2 \times \frac{2\text{ mol }\text{NH}_3}{3\text{ mol }\text{H}_2} = 3.0\text{ mol }\text{NH}_3$$

   To express this yield in grams, multiply the calculated moles of product by its molar mass ($17.03\text{ g/mol}$):

   $$3.0\text{ mol }\text{NH}_3 \times 17.03\text{ g/mol} = 51.09\text{ g }\text{NH}_3$$

   Quantifying the Unreacted Substance

   To determine how much of the excess reactant remains unreacted at the end of the reaction:

   1. Use the amount of the limiting reactant to calculate the moles of excess reactant consumed.
   2. Subtract the consumed moles from the initial moles supplied.
   3. Convert the remaining moles back to mass if required.

   Continuing with the ammonia synthesis:

   • Moles of $\text{N}_2$ consumed: $$4.5\text{ mol }\text{H}_2 \times \frac{1\text{ mol }\text{N}_2}{3\text{ mol }\text{H}_2} = 1.5\text{ mol }\text{N}_2$$
   • Moles of $\text{N}_2$ remaining: $$2.0\text{ mol }\text{initial} - 1.5\text{ mol }\text{consumed} = 0.5\text{ mol }\text{N}_2$$
   • Mass of remaining $\text{N}_2$: $$0.5\text{ mol} \times 28.02\text{ g/mol} = 14.01\text{ g}$$