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22 February, 21:42

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants. 2li (s) + f2 (g) → 2lif (s) 1.0 g li; 1.0 g f2 10.5 g li; 37.2 g f2 2.85*103 g li; 6.79*103 g f2

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  1. 22 February, 22:12
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    Since, we have the reaction as,

    2Li (s) + F2 (g) - - > 2LiF (s)

    we are only concerned with the limiting reactants. We calculate for the amount of product that can be produced with the given amount of reactants.

    a. 1 g Li (1 mol / 6.941 g of Li) (2 mol LiF/2 mol Li) = 0.144 mol LiF2

    1 g F2 (1 mol/38 g) (2 mol LiF2/1 mol F2) = 0.052 mol LiF2

    Answer: 1 g of F2

    b. 10.5 g Li (1 mol/6.941 g of Li) (2 mol LiF/2 mol Li) = 1.512 mol LiF2

    37.2 g F2 (1 mol/38 g) (2 mol LiF2/1 mol F2) = 1.958 mol LiF2

    Answer: 10.5 g of Li

    c. (2.85 x 10^3 g Li) (1 mol/6.941 g of Li) (2 mol LiF/2 mol Li) = 410.60 mol LiF2

    (6.79 x 10^3 g F2) (1 mol/38 g) (2 mol LiF2/1 mol F2) = 357.368 mol of LiF2

    Answer: 6.79 x 10^3 g F2
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