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9 November, 11:32

what is the limiting reactant when 1.50g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advanced batteries, according to the following unbalanced equation?

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  1. 9 November, 12:34
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    Limiting reactant is Lithium

    Explanation:

    We are given;

    Mass of Lithium as 1.50 g Mass of nitrogen is 1.50 g

    We are required to determine the rate limiting reagent.

    First, we write the balanced equation for the reaction

    6Li (s) + N₂ (g) → 2Li₃N

    From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.

    Second, we determine moles of Lithium and nitrogen given.

    Moles = Mass : Molar mass

    Moles of Lithium

    Molar mass of Li = 6.941 g/mol

    Moles of Li = 1.50 g : 6.941 g/mol

    = 0.216 moles

    Moles of nitrogen gas

    Molar mass of Nitrogen gas is 28.0 g/mol

    Moles of nitrogen gas = 1.50 g : 28.0 g/mol

    = 0.054 moles

    According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen. Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles : 6) of nitrogen gas. On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.

    Thus, Lithium is the limiting reagent while nitrogen is in excess.
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