How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl? __ HCl + __ Al __ AlCl3 + __ H2
.63 g Al is required to react with 35 mL of 2.0 M hydrochloric acid.
Thanks for your stoichiometry question.
In looking at the balanced equation, we see:
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.
= .63 g Al