Stoichiometry Chemistry Example 2

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Example 2

hard
How many grams of NH3\text{NH}_3 can be produced from 28.028.0 g of N2\text{N}_2 and excess H2\text{H}_2? (N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3)

Solution

  1. 1
    Moles of N2=28.028.02=1.00 mol\text{N}_2 = \frac{28.0}{28.02} = 1.00\,\text{mol}.
  2. 2
    Mole ratio: 1 mol N2\text{N}_2 → 2 mol NH3\text{NH}_3. So 1.001.00 mol N2\text{N}_2 produces 2.002.00 mol NH3\text{NH}_3.
  3. 3
    Mass of NH3=2.00×17.03=34.1 g\text{NH}_3 = 2.00 \times 17.03 = 34.1\,\text{g}.

Answer

34.1 g of NH334.1\,\text{g of NH}_3
When one reactant is in excess, the other (limiting reactant) determines the maximum product. Here N2\text{N}_2 is the focus since H2\text{H}_2 is stated to be in excess.

About Stoichiometry

The branch of chemistry that uses balanced chemical equations and mole ratios to calculate the precise quantities of reactants consumed and products formed in chemical.

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