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24 October, 03:44

Consider the balanced equation for the following reaction:

6HCl (aq) + 2Al (s) → 3H2 (g) + 2AlCl3 (s)

Which species is the limiting reagent in the reaction if 69.0 grams of HCl reacts with 78.0 grams of Al?

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  1. 24 October, 05:12
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    In this reaction HCl is the limiting reactant and aluminium is the reactant in excess.

    Explanation:

    Step 1: Data given

    Mass of HCl = 69.0 grams

    Mass of Al = 78.0 grams

    Molar mass HCl = 36.46 g/mol

    Atomic mass Al = 26.99 g/mol

    Step 2: The balanced equation

    6HCl (aq) + 2Al (s) → 3H2 (g) + 2AlCl3 (s)

    Step 3: calculate moles HCl

    Moles HCl = mass HCl / molar mass HCl

    Moles HCl = 69.0 grams / 36.46 g/mol

    Moles HCl = 1.89 moles

    Step 4: Calculate moles Al

    Moles Al = 78.0 grams / 26.99 g/mol

    Moles Al = 2.89 moles

    Step 5: Calculate the limiting reactant

    For 6 moles HCl we need 2 moles Al to produce 3 moles H2 and 2 moles AlCl3

    HCl is the limiting reactant. It will completely be consumed (1.89 moles). Aluminium is the reactant in excess. There will react 1.89 / 3 = 0.63 moles

    There will remain 2.89 - 0.63 = 2.26 moles aluminium

    In this reaction HCl is the limiting reactant and aluminium is the reactant in excess.
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