Consider the following unbalanced equation: HCl(aq) + Al(s) → H2(g) + AlCl3(s) If 38.0 moles of HCl(aq) reacts with an excess of Al(s), what is the theoretical yield of H2(g) in moles?

Answer 1

The first thing you want to do for most problems that involve a chemical equation is to balance it.

6HCl (aq) + 2Al (s) → 3H₂ (g) + 2AlCl₃ (s)

I will follow this question to answer any questions that may arise on how to balance this equation.

We're told that the Al is in excess, so in terms of looking for theoretical yield, we don't need to worry about it, as it won't affect the amount that can be produced. What we DO need to worry about is the HCl, which is our limiting reactant. Luckily, we have the mole value given.

When we're trying to use the amount of one compound to find the amount of another, we use stoichiometry, and the mole to mole conversion. We want the theoretical yield of hydrogen, and we're given the moles of HCl, so those are the two compounds we'll be plugging into our dimensional analysis.

First, let's go over theoretical yield, since it'll come up in more advanced versions of a problem like this. Theoretical yield is the maximum amount of a compound that can be produced from your limiting reactant, assuming all experimental conditions are perfect and there is absolutely no error. It is very rare for your actual yield to be the same as your theoretical, but it's important to find your theoretical yield so that, if you're doing an actual experiment, you know what percentage of the theoretical yield you actually produced (actual yield and percent yield).

In this problem, we're only concerned about the theoretical yield, which is a simple mole to mole conversion. We start with the mole amount we're given, and because we want moles of HCl to cancel itself out, we put it on the bottom of our dimensional analysis.

38.0 mol HCl • 3 mol H₂

———— = 19.0 mol H₂

6 mol HCl

We use the coefficients in our balanced equation as our conversion values for hydrogen and HCl.