Theoretical Yield Chemistry Example 3

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Example 3

medium
In the reaction 2Na+Cl2→2NaCl2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}, calculate the theoretical yield of NaCl from 11.511.5 g of sodium with excess chlorine. (Na = 23.023.0, Cl = 35.45 g/mol35.45\,\text{g/mol})

Solution

  1. 1
    Moles of Na =11.523.0=0.500 mol= \frac{11.5}{23.0} = 0.500\,\text{mol}. Mole ratio: 2 mol Na → 2 mol NaCl (1:1). So moles of NaCl =0.500 mol= 0.500\,\text{mol}.
  2. 2
    Theoretical yield =0.500×58.44=29.2 g= 0.500 \times 58.44 = 29.2\,\text{g}.

Answer

29.2 g of NaCl29.2\,\text{g of NaCl}
With excess chlorine, sodium is the limiting reactant that determines the theoretical yield. The 1:1 mole ratio makes this calculation straightforward.

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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