If 22.5 g of lithium and 33.0 g of bromine are present at the beginning of the reaction find the excess reactant and the mass in excess?

Roger the Mole

(22.5 g Li) / (6.941 g Li/mol) = 3.2416 mol Li (33.0 g Br2) / (159.808 g Br2/mol) = 0.20650 mol Br2 0.20650 mole of Br2 would react completely with 0.20650 x (2/1) = 0.41300 mole of Li, but there is more Li present than that, so Li is in excess, and Br2 is the limiting reactant. ((3.2416 mol Li initially) - (0.41300 mol Li reacted)) x (6.941 g Li/mol) = 19.6 g Li left over (0.20650 mol Br2) x (2 mol LiBr / 1 mol Br2) x (86.845 g LiBr/mol) = 35.9 g LiBr produced