if 315mL of 1.25 M iron(ii) sulfate is added to 405mL of.95 M ammonium phosphate, what is the limiting reactant

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if 315mL of 1.25 M iron(ii) sulfate is added to 405mL of.95 M ammonium phosphate, what is the limiting reactant

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Edana Edana 7 months 2021-07-11T14:49:34+00:00 1 Answers 24 views 0

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    2021-07-11T14:51:13+00:00

    Answer:

    FeSO₄ is the Limiting Reactant

    Explanation:

    3FeSO₄(aq) + 2(NH₄)₃PO₄ (aq) => 3(NH₄)₃PO₄(aq) + Fe₃(PO₄)(s)

    315ml(1.25M)FeSO₄ + 405ml(0.95M)(NH₄)₃PO₄

    => 0.315L(1.25M)FeSO₄       +      0.405ml(0.95M)(NH₄)PO₄

    => 0.394mole FeSO₄           +      0.385mole (NH₄)PO₄

    Divide by respective coefficient => the smaller value is the Limiting Reactant.

    => 0.394/3 = 0.131                      => 0.385/2 = 0.193

    Smaller value is associated with FeSO₄

    ∴the limiting reactant is FeSO₄.

    Summery:

    1. convert reactant quantities to moles

    2. divide reactants by respective coefficients of balanced equation

    3. note smaller value after dividing by respective coefficient => Limiting Reactant.

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