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yuradex [85]
3 years ago
10

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
Chemistry
1 answer:
kodGreya [7K]3 years ago
5 0
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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Answer:

8. the answer is B.

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Explanation:

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8. In this case, by bearing to mind that the limiting reactant is always completely consumed and the excess one remain as a leftover at the end of the reaction, we can also infer that as all the limiting reactant is consumed, it must determine the maximum amount of product as the excess reactant will hypothetically produce a greater mass than expected; thus, the answer to this question is B.

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