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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

Respuesta :

Yna9141
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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