Theoretical Yield Chemistry Example 2

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Example 2

medium
Calculate the theoretical yield of water when 8.08.0 g of hydrogen reacts with excess oxygen. (2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}; H = 1.0081.008, O = 16.00 g/mol16.00\,\text{g/mol})

Solution

  1. 1
    Moles of H2=8.02.016=3.97 mol\text{H}_2 = \frac{8.0}{2.016} = 3.97\,\text{mol}.
  2. 2
    From the equation: 2 mol H2\text{H}_2 → 2 mol H2O\text{H}_2\text{O} (1:1 ratio). So moles of H2O=3.97 mol\text{H}_2\text{O} = 3.97\,\text{mol}.
  3. 3
    Theoretical yield =3.97×18.02=71.5 g= 3.97 \times 18.02 = 71.5\,\text{g}.

Answer

Theoretical yield=71.5 g of H2O\text{Theoretical yield} = 71.5\,\text{g of H}_2\text{O}
The theoretical yield calculation follows the standard stoichiometric pathway: convert mass to moles, apply the mole ratio from the balanced equation, then convert back to mass. With excess oxygen, hydrogen is the limiting reactant.

About Theoretical Yield

The maximum amount of product that could be formed in a chemical reaction, calculated from the stoichiometry of the balanced equation using the limiting reactant.

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More Theoretical Yield Examples

Example 1 easy

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

Example 3 medium

In the reaction [formula], calculate the theoretical yield of NaCl from [formula] g of sodium with e

Example 4 hard

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