Anonymous
Anonymous asked in Science & MathematicsAlternativeParapsychology · 2 months ago

If 35 g of Mg reacted with 15 g N2. Determine the yield-limiting reagent, mass of Mg3N2 formed, mass of the exces reagent remaining reaction?

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  • 2 months ago
    Favourite answer

    Balanced equation:

    3 Mg + N2 --> Mg3N2

    moles Mg = 35 g / 24.305 g/mol = 1.44 mol Mg

    moles N2 = 15 g / 28.0 g/mol = 0.536 mol N2

    To consume all of the Mg available would require:

    1.44 mol Mg X (1 mol N2/3 mol Mg) = 0.48 mol N2

    Because more N2 is available than this, Mg is the limiting reactant

    Mass excess N2 = (0.536 mol - 0.480 mol) X 28.0 g/mol = 1.6 g excess N2

    Mass Mg3N2 = 1.44 mol Mg X (1 mol Mg3N2 / 3 mol Mg) X 101 g/mol = 48 g Mg3N2

  • Bobby
    Lv 7
    2 months ago

    3 Mg + N2 = Mg3N2

    moles of Mg = 35g /  24.30 g/ mol  = 1.440 moles

    these need 1.440 / 3 moles of N2 for a complete reaction = 0.480 moles 

    but we have 15g / 28.0134 g/ moles of N2 = 0.535 moles 

    so the N2 is in excess and the yield limiting reagent is Mg 

    according to the equation we will get  0.480 moles of Mg3N2

    Mass of Mg3N2 =   0.480 moles * 100.9284 g/mol = 48 g to 2 significant figures

    moles of N2 in excess = 0.535 - 0.480 moles = 0.0553 moles of N2 

    mass of excess N2 = 0.0553 * 28.0134 g/ moles = 1.5 g of N2 excess

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