Theoretical Yield Examples in Chemistry

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Theoretical Yield.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Chemistry.

Concept Recap

The maximum amount of product that could form based on stoichiometry and the limiting reactant.

The perfect-world result — the most product you could possibly get if nothing is lost or wasted.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Theoretical yield is calculated from the limiting reactant using balanced equation mole ratios.

Common stuck point: Theoretical yield assumes 100% conversion—real yields are always less.

Worked Examples

Example 1

easy
Define theoretical yield and explain how it differs from actual yield.

Solution

  1. 1
    Theoretical yield is the maximum amount of product that can be formed from a given amount of reactants, assuming the reaction goes to completion with no losses.
  2. 2
    It is calculated using stoichiometry from the balanced equation and the limiting reactant.
  3. 3
    Actual yield is the amount of product actually obtained in the laboratory, which is always less than or equal to the theoretical yield due to side reactions, incomplete reactions, and losses during purification.

Answer

\text{Theoretical yield = maximum possible product; actual yield} \leq \text{theoretical yield}
The theoretical yield represents an ideal scenario. In practice, reactions rarely achieve 100% yield because of practical limitations such as side reactions, equilibrium constraints, and mechanical losses during product isolation.

Example 2

medium
Calculate the theoretical yield of water when 8.0 g of hydrogen reacts with excess oxygen. (2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}; H = 1.008, O = 16.00\,\text{g/mol})

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

medium
In the reaction 2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}, calculate the theoretical yield of NaCl from 11.5 g of sodium with excess chlorine. (Na = 23.0, Cl = 35.45\,\text{g/mol})

Example 2

hard
A student reacts 6.50 g of zinc with 20.0 mL of 3.00\,\text{M HCl}. The reaction is \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2. Determine the limiting reactant and the theoretical yield of \text{ZnCl}_2. (Zn = 65.38, \text{ZnCl}_2 = 136.28\,\text{g/mol})
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Background Knowledge

These ideas may be useful before you work through the harder examples.

stoichiometrylimiting reactant