Limiting Reactant Chemistry Example 2

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Example 2

hard
Given 10.010.0 g of N2\text{N}_2 and 10.010.0 g of H2\text{H}_2, how many grams of NH3\text{NH}_3 can be produced? (N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3)

Solution

  1. 1
    Moles: N2=10.028.02=0.357 mol\text{N}_2 = \frac{10.0}{28.02} = 0.357\,\text{mol}. H2=10.02.016=4.96 mol\text{H}_2 = \frac{10.0}{2.016} = 4.96\,\text{mol}.
  2. 2
    Required H2\text{H}_2 for 0.3570.357 mol N2\text{N}_2: 3×0.357=1.071 mol3 \times 0.357 = 1.071\,\text{mol}. Available: 4.964.96 mol. N2\text{N}_2 is limiting.
  3. 3
    Moles of NH3=2×0.357=0.714 mol\text{NH}_3 = 2 \times 0.357 = 0.714\,\text{mol}. Mass =0.714×17.03=12.2 g= 0.714 \times 17.03 = 12.2\,\text{g}.

Answer

12.2 g of NH312.2\,\text{g of NH}_3
Always check both reactants to find the limiting one. The limiting reactant produces the lesser amount of product and determines the theoretical yield.

About Limiting Reactant

The reactant that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product that can be formed.

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