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When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?


2 Al + 6HCl _ 2 AlCl3 + 3 H2


Al is the limiting reactant, 9.0 mol H2 can be formed


HCl is the limiting reactant, 6.5 mol H2 can be formed


Al is the limiting reactant, 6.0 mol H2 can be formed


HCl is the limiting reactant, 4.3 mol H2 can be formed

Verified Answer :

The best and most correct answer among the choices provided by your question is the second choice or letter B " Al is the limiting reactant, 6.0 mol H2 can be formed." This  answer was based from my calculations and computations of the equation.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

martinbrusconi

Answer:

HCl is the limiting reactant, 6.5 mol H2 can be formed

Explanation:

If you check the estequiometric equation for 2 mol of Al you will need 6 mols of HCl. Having 6 mols of Al it is necessary to have 18 mols of HCl for a complete reaction, you have 13 mols. Therefore, there is less HCl than needed; it is the limiting reactant.

On the other hand, in a hypothetic complete reaction with 6 mol of Al and 18 mols of HCl nine mols of H2 would be produced. Here you can calculate taking the direct proportion as follows:

[tex] 13/18*9=6.5 mols of H2[\tex]

So it is possible to produce 6.5 mol of H2.

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