question archive Consider the following reaction: 2Al(s) + 3Cl2(g) 2AlCl3(s) If you begin with 0

Consider the following reaction: 2Al(s) + 3Cl2(g) 2AlCl3(s) If you begin with 0

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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?

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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???