2Li(s) + Br2(l) → 2LiBr(s)

Given:

Mass of lithium (Li) = 25.0 g

Mass of bromine (Br2) = 100 g

To determine the limiting reactant and calculate the mass of lithium bromide (LiBr) formed, we need to compare the mole ratio of the reactants with their stoichiometric coefficients in the balanced chemical equation.

Convert the masses of Li and Br2 to moles:

Moles of Li = 25.0 g / 6.941 g/mol = 3.605 mol

Moles of Br2 = 100 g / 159.81 g/mol = 0.626 mol

Calculate the mole ratio of reactants:

Li : Br2 = 3.605 mol / 2 : 0.626 mol / 1

Li : Br2 = 1.8025 : 0.626

Li : Br2 ≈ 3 : 1

Comparing the mole ratio with the stoichiometric ratio of 2:1 in the balanced equation, we can observe that Lithium is the limiting reactant as it is present in a lesser mole ratio compared to Bromine.

Therefore, all of the lithium will react, and the amount of LiBr formed will be determined by the amount of lithium.

Now, calculate the moles of LiBr formed using the stoichiometry from the balanced equation:

Moles of LiBr = Moles of Li × (2 mol LiBr / 2 mol Li)

Moles of LiBr = 3.605 mol × 1

Moles of LiBr = 3.605 mol

Finally, convert the moles of LiBr to grams:

Mass of LiBr = Moles of LiBr × Molar mass of LiBr

Mass of LiBr = 3.605 mol × 86.85 g/mol

Mass of LiBr = 312.35 g

Therefore, the mass of lithium bromide (LiBr) formed when 25.0 g of lithium and 100 g of bromine react is 312.35 g.