When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed? 2Al + 6HCl → 2AlCl3 + 3H2

What is the limiting reactant?: HCl is the limitng reactant.

How many moles of H₂ are formed?: 6.5 moles of H₂ are formed.

Explanation:

Part A: what is the limiting reactant?

1) Balanced chemical equation: given

2Al + 6HCl → 2AlCl₃ + 3H₂

2) Stoichiometric mole ratio:

Use the coefficients of the balanced equation:

2 mol Al : 6 mol HCl : 2 mol AlCl₃ : 3H₂

3) Compare the stoichiometric mole ratio of the reactants with their actual ratio:

Theoretical ratio: 2 mol Al / 6 mol HCl ≈ 0.33 mol Al / mol HCl

Actual ratio: 6.0 mol Al / 13 mol Cl ≈ 0.46 mol Al / mol Cl

Since the actual ratio indicates that there is a greater number of moles of Al (0.46) per mol of Cl than what is required by the stoichiometric ratio (0.33), Al is in excess and HCl is the limiting reactant.

Answer: the limiting reactant is HCl.

Part B. How many moles of H₂ are formed?

3. Determine how many moles of H₂ can be formed

Theoretical ratio using limiting reactant:

6 mol HCl / 3 mol H₂ = 13 mol HCl / x

⇒ x = 13 mol HCl * 3 mol H₂ / 6 mol HCl = 6.5 mol H₂.

The answer must be reported with two significant digits, such as the data are given.

Answer: 6.5 moles of H₂ are formed