Reactant Chemistry Example 4

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Example 4

hard
A student mixes 5.05.0 g of sodium (Na\text{Na}) with excess water. The reaction is: 2Na+2H2O2NaOH+H22\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2. Calculate the mass of the reactant sodium consumed and the moles of H2\text{H}_2 produced. (Na = 23.0g/mol23.0\,\text{g/mol})

Solution

  1. 1
    All 5.05.0 g of sodium is consumed since water is in excess. Moles of Na =5.023.0=0.217mol= \frac{5.0}{23.0} = 0.217\,\text{mol}.
  2. 2
    From the equation, 2 moles of Na produce 1 mole of H2\text{H}_2. So moles of H2=0.2172=0.109mol\text{H}_2 = \frac{0.217}{2} = 0.109\,\text{mol}.
  3. 3
    The reactant Na is the limiting reactant here, determining the amount of product formed.

Answer

5.0g Na consumed,  0.109mol H2 produced5.0\,\text{g Na consumed},\; 0.109\,\text{mol H}_2\text{ produced}
When one reactant is in excess, the other reactant (sodium in this case) is completely consumed and determines the maximum amount of product. This is the concept of the limiting reactant.

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A starting substance that is consumed and chemically transformed during a chemical reaction, appearing on the left side of a chemical equation before the reaction.

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