Grade 10

Advanced

9 min

Lesson 10 of 12

Limiting Reactants and Excess

Not Started

Identifying which reactant runs out first and how it dictates the total amount of product possible.

If you have 100 burger patties but only 10 buns, how many burgers can you actually serve? In chemistry, just like in a kitchen, the ingredient that runs out first dictates the entire outcome of your work.

Learning Objectives

1.Identify the limiting reactant in a chemical reaction given starting masses using mole ratios.
2.Calculate the precise mass of excess reactant remaining after a reaction reaches completion.
3.Evaluate the efficiency of a reaction using the concept of 'atom economy' to promote sustainable chemistry.

The Concept of the Limiting Reactant

In a chemical reaction, reactants are rarely present in the exact stoichiometric proportions required by the balanced equation. The limiting reactant (or reagent) is the substance that is completely consumed first, bringing the reaction to a halt. Any reactant that remains after the limiting reactant is exhausted is called the excess reactant. The amount of product formed—the theoretical yield—is strictly determined by the amount of the limiting reactant available. To find it, we cannot simply look at the masses; we must convert to moles and compare them using the coefficients from the balanced chemical equation.

Consider the formation of water:

2H_2(g) + O_2(g) \rightarrow 2H_2O(l)

1. Suppose you have

4.0

mol of

H_2

and

3.0

mol of

O_2

. 2. According to the ratio,

2

moles of

H_2

require

1

mole of

O_2

. 3. Therefore,

4.0

mol of

H_2

would need

\frac{4.0}{2} = 2.0

mol of

O_2

. 4. Since we have

3.0

mol of

O_2

(which is more than

2.0

O_2

is in excess and

H_2

is the limiting reactant.

Quick Check

If a reaction requires a 1:1 ratio of Reactant A to Reactant B, and you have 5 moles of A and 7 moles of B, which one is the limiting reactant?

Compare the available moles to the required ratio.

Answer

Reactant A is the limiting reactant because it will run out first.

Calculating Excess Mass

Once the limiting reactant is identified, we can determine exactly how much of the excess reactant was used and how much remains. This is a three-step process: 1. Use the moles of the limiting reactant to calculate the moles of the excess reactant consumed. 2. Subtract the consumed moles from the initial moles to find the moles remaining. 3. Convert the remaining moles back into grams using the molar mass ( $M$ ). This is vital for industrial efficiency and reducing waste in 'Green Chemistry' applications.

Reaction: $Mg + 2HCl \rightarrow MgCl_2 + H_2$ . You start with $12.15g$ of $Mg$ ( $0.5$ mol) and $1.0$ mol of $HCl$ . 1. The ratio is $1:2$ . To use $0.5$ mol of $Mg$ , you need $0.5 \times 2 = 1.0$ mol of $HCl$ . 2. Since you have exactly $1.0$ mol of $HCl$ , both are consumed entirely (stoichiometric amounts). 3. If you had $1.5$ mol of $HCl$ instead, you would have $1.5 - 1.0 = 0.5$ mol of $HCl$ left over. 4. Mass remaining: $0.5 \text{ mol} \times 36.5 \text{ g/mol} = 18.25g$ of $HCl$ .

Quick Check

Why must we convert grams to moles before identifying the limiting reactant?

Think about the weight of a dozen eggs vs. a dozen bricks.

Answer

Because chemical equations react in particle ratios (moles), not mass ratios. Different substances have different molar masses.

Atom Economy: The Efficiency Metric

In advanced chemistry, we don't just care about yield; we care about Atom Economy. This measures how many atoms from the reactants actually end up in the desired product versus being wasted in side-products. A reaction with high atom economy is more 'green' because it minimizes waste. Formula:

\text{Atom Economy} = \left( \frac{\text{Molar Mass of Desired Product}}{\text{Sum of Molar Masses of All Reactants}} \right) \times 100\%

Even if a reaction has a

100\%

yield, its atom economy might be low if it produces many useless by-products.

Reaction: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$ . You start with $160g$ $Fe_2O_3$ ( $1$ mol) and $112g$ $CO$ ( $4$ mol). 1. Ratio is $1:3$ . $1$ mol $Fe_2O_3$ needs $3$ mol $CO$ . 2. We have $4$ mol $CO$ , so $CO$ is in excess. 3. Moles of $CO$ remaining: $4 - 3 = 1$ mol. 4. Mass of $CO$ remaining: $1 \text{ mol} \times 28.01 \text{ g/mol} = 28.01g$ .

Key Takeaways

1The limiting reactant is the reagent that is completely consumed and determines the maximum amount of product possible.
2To identify the limiting reactant, always convert masses to moles and divide by the stoichiometric coefficients.
3Excess reactant remaining = (Initial Moles - Moles Consumed) × Molar Mass.
4Atom Economy evaluates the percentage of reactant atoms that become part of the desired product.

Quick Check

Multiple Choice

In the reaction $A + 2B \rightarrow C$ , if you have 2 moles of A and 2 moles of B, which is the limiting reactant?

Multiple Choice

Which factor does NOT affect the identification of the limiting reactant?

True/False

A reaction with 100% percentage yield can still have a low atom economy.

What's Next?

Review Tomorrow

In 24 hours, try to explain the 'Have vs. Need' method for finding limiting reactants to a friend or write it down from memory.

Practice Activity

Find a balanced equation for the combustion of propane ( $C_3H_8$ ) and calculate the limiting reactant if you have 10g of propane and 10g of oxygen.

Browse

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Limiting Reactants and Excess

Learning Objectives

The Concept of the Limiting Reactant

Calculating Excess Mass

Atom Economy: The Efficiency Metric

Key Takeaways

Quick Check

What's Next?

Limiting Reactants and Excess

Learning Objectives

The Concept of the Limiting Reactant

Calculating Excess Mass

Atom Economy: The Efficiency Metric

Key Takeaways

Quick Check

What's Next?