What mass of potassium would react with excess chlorine to produce 156 grams of potassium chloride

The balanced equation for the above reaction is as follows;

2K + Cl₂ - - > 2KCl

stoichiometry of K to KCl is 2:2

the number of KCl moles produced - 156 g / 74.5 g/mol = 2.09 mol

we have been given that Cl₂ is in excess. therefore K is the limiting reactant.

amount of product formed depends on number of limiting reactant present,

2 mol of K reacts to give 2 mol of KCl

therefore 2.09 mol of KCl being formed means - 2.09 mol of K has reacted

mass of K reacted is - 2.09 mol x 39 g/mol = 81.51 g of K

mass of K reacted is 81.51 g