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

Jul 18, 2014

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

#### Explanation:

In looking at the balanced equation, we see:

$6 H C l + 2 A l \to 2 A l C {l}_{3} + 3 {H}_{2}$

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