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28 June, 21:17

The reaction between ethyne (acetylene, C 2 H 2) and hydrogen. The product is ethane (C 2 H 6). Which is the limiting reactant? Which is the excess reactant? Explain

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  1. 28 June, 21:46
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    Three possible cases:

    - If amount are equal for each reactant (for example 1 mol each), the limiting is the hydrogen and the excess reagent is the acetylene.

    - When moles of H₂ are greater than C₂H₂

    The acetylene is the limiting reagent so the H₂ is the excess

    - When moles of C₂H₂ are greater than H₂

    For this case, H₂ is the limiting reactant and the excess is the C₂H₂

    Explanation:

    First of all we determine the reaction:

    Reactants, acetylene and hydrogen

    Products are ethane

    Then, the balanced reaction is: C₂H₂ + 2H₂ → C₂H₆

    1 mol of acetylene reacts with 2 moles of hydrogen ir order to produce 1 mol of ethane.

    If amount are equal for each reactant, the limiting is the hydrogen,

    For example, 1 mol each

    For 1 mol of acetylene I need 2 moles of H₂. I've only got 1 mol, so I do not have enough H₂. The excess reagent is the acetylene.

    - When moles of H₂ are greater than C₂H₂

    For example, 3 moles of H₂ and 0.5 mol of C₂H₂

    2moles of H₂ need 1 mol of C₂H₂ for the reaction

    Then 3 moles of H₂ will need (3. 1) / 2 = 1.5 moles

    We have 0.5 moles, so the acetylene is the limiting reagent, again.

    - When moles of C₂H₂ are greater than H₂

    For example 1 mol of C₂H₂ and 0.001 moles of H₂

    If I have 1 mol of C₂H₂, I definetly need the double of moles of hydrogen, so in this case, H₂ is the limiting reactant and the excess is the C₂H₂

    If we have 1 mol of H₂ and 0.5 mol of C₂H₂, notice that moles of acetylene are lower than hydrogen

    1 mol of C₂H₂ needs 2 moles of H₂

    So 0.5 moles of C₂H₂ will need 1 mol of H₂ (it's ok because we have 1 mol)

    2 moles of H₂ need 1 mol of C₂H₂ for reaction

    Then, 1 mol of H₂ will need 0.5 moles of C₂H₂ (it's ok because we have that amount)

    In this case, there is no excess neither limiting. That's why we can choose any of them to determine the moles (or mass) for the product
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