Consider the following reaction: 2Al(s) + 3Cl2(g) 2AlCl3(s)

If you begin with 0.552 mol of aluminum and 0.887 mol of chlorine, what is the "Limiting Reactant" and "Theoretical Yield" of AlCl3 in moles?

The limiting reagent is Al and the theoretical yield is 0.552 mol AlCl₃

The first step is to establish the balanced chemical equation for the reaction:

?2Al(s)+3Cl2?(g)→2AlCl3?(s)?

LIMITING REAGENT:

Based on the balanced chemical equation, the stoichiometric mole ratio is ?2 mol Al3 mol Cl2??=1 mol Al1.5 mol Cl2??? . Where the actual mole ratio is determined below.

?0.552 mol Al0.887 mol Cl2??=1 mol Al1.608 mol Cl2???

Notice that the actual mole ratio is greater than the stoichiometric mole ratio. Thus, we have an excess of Cl₂ and a limited amount of Al.

THEORETICAL YIELD:

The limiting reagent determines the maximum amount of product. Thus, we base the amount of product formed on the moles of Al present initially.

?0.552 mol Al×2 mol Al2 mol AlCl3??=0.552 mol AlCl3???