posted by Monique on .
Find the limiting reactant for the initial quantities of reactants
2k(s) + Cl2(g)--> 2KCl(s)
1.8 mol K; 1 mol Cl2
Very simple. You just run the usual stoichiometry twice.
First you have 1.8 moles K.
moles KCl produced = 1.8 moles K x (2 moles KCl/2 moles K) = 1.8 moles KCl
Then you have 1 mole Cl2.
moles KCl produced = 1 mole Cl2 x (2 moles KCl/1 mole Cl2) = 2 moles KCl.
Obviously both answers can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller value AND the reagent producing the smaller value is the limiting reagent.