i love math but sadly i am dumb
Answer:
You would Multiply the mass of one mole of each element by the ratio of that element to one mole of the compound.
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
Group 2 because the outer electrons ( also known as valence electrons) are 2 and that’s how we know which group it is in
