Limiting Reactants

Understanding Limiting Reactants

When two reactants are combined, if they are in the correct stoichiometric ratio, both will be completely consumed, resulting in no leftover reactants.
However, in many practical scenarios, reactants are given in specific masses, and it's not immediately clear which one will run out first. The reactant that is completely consumed first is called the limiting reactant (or limiting reagent), as it limits the amount of product that can be formed. The other reactant(s) will be in excess.

To solve problems involving limiting reactants, a specific method is recommended to consistently determine the limiting reactant and the maximum possible product formation.

Method for Solving Limiting Reactant Problems

When faced with a limiting reactant problem, follow these steps:

Make two (or more) assumptions: Assume each reactant, one at a time, is the limiting reactant.
Calculate the potential product amount for each assumption: For each assumption, perform stoichiometric calculations to determine the maximum amount of a specific product that could be produced if that reactant were indeed limiting.
Identify the true limiting reactant: The actual maximum amount of product that can be formed is the smaller value among all the calculated product amounts. The reactant that leads to this smaller product amount is the true limiting reactant.
Confirm the product yield: The smallest calculated product amount is the correct answer for the maximum product yield.

Analogy: If you have a maximum of $10 in your wallet, and also know the maximum possible for the wallet itself is $18,000,000,000, you can never go above $10. The smaller limit always prevails.

Step-by-Step Example: Reaction of Hydrogen and Oxygen

Problem: Combine $5.0 ext{ grams of } H2$ with $35 ext{ grams of } O2$ , and add a spark to initiate a reaction. How many grams of $H_2O$ will be produced?

This problem gives two extensive starting points (masses of reactants) and asks for an extensive quantity (mass of product). This indicates a limiting reactant scenario.

Balanced Chemical Equation

The reaction between hydrogen ( $H2$ ) and oxygen ( $O2$ ) to produce water ( $H2O$ ) is: $2H2 + O2 \rightarrow 2H2O$

Step 1: Assume $H_2$ is Limiting

If $H2$ is the limiting reactant, we will use all $5.0 ext{ grams}$ of it to determine the maximum $H2O$ that can be produced.

Convert grams of $H2$ to moles of $H2$ :
Molar mass of $H2 = 2 \times 1.008 \text{ g/mol} = 2.016 \text{ g/mol}$ $\frac{5.0 \text{ g } H2}{1} \times \frac{1 \text{ mol } H2}{2.016 \text{ g } H2}$
Convert moles of $H2$ to moles of $H2O$ (using mole ratio from balanced equation):
From the balanced equation, $2 \text{ moles of } H2$ produce $2 \text{ moles of } H2O$ .
$\dots \times \frac{2 \text{ mol } H2O}{2 \text{ mol } H2}$
Convert moles of $H2O$ to grams of $H2O$ :
Molar mass of $H2O = (2 \times 1.008) + 16.00 = 18.016 \approx 18.02 \text{ g/mol}$ $\dots \times \frac{18.02 \text{ g } H2O}{1 \text{ mol } H_2O}$

Calculation:
$\frac{5.0 \text{ g } H2}{1} \times \frac{1 \text{ mol } H2}{2.016 \text{ g } H2} \times \frac{2 \text{ mol } H2O}{2 \text{ mol } H2} \times \frac{18.02 \text{ g } H2O}{1 \text{ mol } H2O} \approx 44.74 \text{ g } H2O$

Rounding to two significant figures (due to $5.0 ext{ g } H2$ ), if $H2$ is limiting, $45 \text{ grams of } H_2O$ would be produced.

Step 2: Assume $O_2$ is Limiting

If $O2$ is the limiting reactant, we will use all $35 ext{ grams}$ of it to determine the maximum $H2O$ that can be produced.

Convert grams of $O2$ to moles of $O2$ :
Molar mass of $O2 = 2 \times 16.00 \text{ g/mol} = 32.00 \text{ g/mol}$ $\frac{35 \text{ g } O2}{1} \times \frac{1 \text{ mol } O2}{32.00 \text{ g } O2}$
Convert moles of $O2$ to moles of $H2O$ (using mole ratio from balanced equation):
From the balanced equation, $1 \text{ mole of } O2$ produces $2 \text{ moles of } H2O$ .
$\dots \times \frac{2 \text{ mol } H2O}{1 \text{ mol } O2}$
Convert moles of $H2O$ to grams of $H2O$ :
Molar mass of $H2O = 18.02 \text{ g/mol}$ (as calculated previously). $\dots \times \frac{18.02 \text{ g } H2O}{1 \text{ mol } H_2O}$

Calculation:
$\frac{35 \text{ g } O2}{1} \times \frac{1 \text{ mol } O2}{32.00 \text{ g } O2} \times \frac{2 \text{ mol } H2O}{1 \text{ mol } O2} \times \frac{18.02 \text{ g } H2O}{1 \text{ mol } H2O} \approx 39.42 \text{ g } H2O$

Rounding to two significant figures (due to $35 ext{ g } O2$ ), if $O2$ is limiting, $39 \text{ grams of } H_2O$ would be produced.

Identifying the Limiting Reactant

Assuming $H2$ means a maximum of $45 \text{ g } H2O$ can be produced.
Assuming $O2$ means a maximum of $39 \text{ g } H2O$ can be produced.

The smaller of these two values is $39 \text{ g } H2O$ . This means that no more than $39 \text{ grams of } H2O$ can be produced from the given amounts of reactants, because the amount of oxygen supplied will run out first. Therefore:

The actual amount of $H_2O$ produced is $39 \text{ grams}$ .
$O_2$ is the limiting reactant.
$H_2$ is the excess reactant.

Calculating Leftover Reactant

Since $H2$ is the excess reactant, there will be some $H2$ left over after the reaction. To find out how much is left, we first need to calculate how much $H2$ was actually consumed to produce the $39 \text{ grams}$ of $H2O$ .

Convert grams of $H2O$ (actual product amount) to moles of $H2O$ :
$\frac{39 \text{ g } H2O}{1} \times \frac{1 \text{ mol } H2O}{18.02 \text{ g } H_2O}$
Convert moles of $H2O$ to moles of $H2$ (using mole ratio from balanced equation):
From the balanced equation, $2 \text{ moles of } H2O$ require $2 \text{ moles of } H2$ .
$\dots \times \frac{2 \text{ mol } H2}{2 \text{ mol } H2O}$
Convert moles of $H2$ to grams of $H2$ :
Molar mass of $H2 = 2.016 \text{ g/mol}$ . $\dots \times \frac{2.016 \text{ g } H2}{1 \text{ mol } H_2}$

Calculation for $H2$ consumed:
$\frac{39 \text{ g } H2O}{1} \times \frac{1 \text{ mol } H2O}{18.02 \text{ g } H2O} \times \frac{2 \text{ mol } H2}{2 \text{ mol } H2O} \times \frac{2.016 \text{ g } H2}{1 \text{ mol } H2} \approx 4.36 \text{ g } H_2$

Rounding to two significant figures, $4.4 \text{ grams of } H_2$ were consumed.

Calculate leftover $H2$ :
Initial amount of $H2$ supplied: $5.0 \text{ g}$
Amount of $H2$ consumed: $4.4 \text{ g}$ Leftover $H2 = 5.0 \text{ g} - 4.4 \text{ g} = 0.6 \text{ grams of } H_2$

This systematic approach ensures accurate determination of the limiting reactant, product yield, and leftover reactant amounts, which is crucial for exam problems.