When 5.0 mol al react with 13 mol hcl, what is the limiting reactant, and how many moles of alcl3 can be formed? 2 al + 6hcl → 2 alcl3 + 3 h2 a. al is the limiting reactant, 5.0 mol alcl3 can be formed b. hcl is the limiting reactant, 4.3 mol alcl3 can be formed c. al is the limiting reactant, 7.5 mol alcl3 can be formed d. hcl is the limiting reactant, 6.5 mol alcl3 can be formed

2 moles of Al will react will 6 moles of HCl 
5 moles of Al will react with ((6÷2)×5 )=15 moles of HCl
Thus, HCl is the limiting reactant(you have only 13 moles, you need 15 moles).
Limiting reagent determines amount of product.
6 moles of HCl will produce 2 moles AlCl3
13 moles of HCl will produce (2÷6)(13)=4.3 moles of AlCl3
⇒B IS THE ANSWER

From the balanced chemical reaction, 
                                2Al + 6HCl ---> 2AlCl3 + 3H2
we based the limiting and excess reactant on the product AlCl3. Each 2 moles of aluminum will form 2 moles of aluminum chloride and 6 moles of HCl will be needed for the same production. From the given,
                         (5 mol Al) x (2 mol AlCl3 / 2 mol Al) = 5 mole AlCl3
                         (13 mol HCl) x (2 mol AlCl3 / 6 mol AlCl3) = 4.3 mol AlCl3
Thus, the answer to this question is letter B. 

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