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March 30, 2017

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Find the limiting reactant for the initial quantities of reactants
2k(s) + Cl2(g)--> 2KCl(s)

1.8 mol K; 1 mol Cl2

I'm lost

  • Stoichiometry - ,

    Very simple. You just run the usual stoichiometry twice.
    First you have 1.8 moles K.
    moles KCl produced = 1.8 moles K x (2 moles KCl/2 moles K) = 1.8 moles KCl

    Then you have 1 mole Cl2.
    moles KCl produced = 1 mole Cl2 x (2 moles KCl/1 mole Cl2) = 2 moles KCl.

    Obviously both answers can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller value AND the reagent producing the smaller value is the limiting reagent.

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