How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl? __ HCl + __ Al  __ AlCl3 + __ H2

1 Answer

Answer:

.63 g Al is required to react with 35 mL of 2.0 M hydrochloric acid.

Explanation:

Thanks for your stoichiometry question.

In looking at the balanced equation, we see:

#6 HCl + 2 Al -> 2 AlCl_3 +3H_2#

The first thing we need to do is to determine the number of moles of HCl that are reacting.

Molarity = moles/liters of solution

2.0 M = X/.035 L X = .070 moles of HCl

Now, we can determine the grams of Al that will be needed to react with .070 moles of HCl by using the balanced equation.

#".070 moles HCl" * ( "2 mol Al"/"6 mol HCl")("26.98g Al"/"1 mol Al")#

= .63 g Al