Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixture of 41.0 g

Question

Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixture of 41.0 g of magnesium and 175.0 g of iron(III) chloride is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete. Select one: a. Limiting reactant is Mg; 7.4 g of FeCl3 remain. b. Limiting reactant is Mg; 46.5 g of FeCl3 remain. c. Limiting reactant is FeCl3; 1.7 g of Mg remain. d. Limiting reactant is FeCl3; 37.8 g of Mg remain. e. Limiting reactant is Mg; 134.0 g of FeCl3 remain.

in progress 0
Calantha 5 years 2021-08-16T11:58:30+00:00 1 Answers 469 views 0

Answers ( )

  1. trucchi
    0
    2021-08-16T12:00:22+00:00
    Reply

    Answer:


    Limiting reactant is FeCl3; 1.7 g of Mg remain.


    Explanation:


    From the question;

    The equation is;

    3Mg(s) + 2FeCl₃(s) → 3MgCl₂(s) + 2Fe(s)


    Amount of Mg = 41 g/24.31 g/mol = 1.687 moles

    The limiting reactant yields the least amount of MgCl2

    3 moles of Mg yields 3 moles of MgCl2

    Hence 1.687 moles of Mg yields yields 1.687 moles of MgCl2


    FeCl₃ = 175 g/162.2 g/mol = 1.0789 moles

    2 moles of FeCl3 yields 3 moles of MgCl2

    1.0789  moles of FeCl3 yields 1.0789  * 3/2 = 1.61835 moles of MgCl2

    Hence FeCl3 is the limiting reactant

    3 moles of Mg reacts with 2 moles of FeCl3

    x moles of Mg reacts with 1.0789  of FeCl3

    x = 3 * 1.0789  /2 = 1.61835 moles of Mg

    Mass of Mg reacted = 1.61835 moles * 24.31 = 39.342 g

    Mass of excess Mg = 41 g – 39.342 g = 1.7 g

     


                       


                       



     





                               


                             



           


Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )