Question

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

Answer 1

Answer:

• 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