Consider the balanced equation for the following reaction:
6HCl(aq) + 2Al(s) → 3H2(g) + 2AlCl3(s)
Which species is the limiting reagent in the reaction if 69.0 grams of HCl reacts with 78.0 grams of Al?

In this reaction HCl is the limiting reactant and aluminium is the reactant in excess.

Explanation:

Step 1: Data given

Mass of HCl = 69.0 grams

Mass of Al = 78.0 grams

Molar mass HCl = 36.46 g/mol

Atomic mass Al = 26.99 g/mol

Step 2: The balanced equation

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

Step 3: calculate moles HCl

Moles HCl = mass HCl / molar mass HCl

Moles HCl = 69.0 grams / 36.46 g/mol

Moles HCl = 1.89 moles

Step 4: Calculate moles Al

Moles Al = 78.0 grams / 26.99 g/mol

Moles Al = 2.89 moles

Step 5: Calculate the limiting reactant

For 6 moles HCl we need 2 moles Al  to produce 3 moles H2 and 2 moles AlCl3

HCl is the limiting reactant. It will completely be consumed (1.89 moles). Aluminium is the reactant in excess. There will react 1.89 / 3 = 0.63 moles

There will remain 2.89 - 0.63 = 2.26 moles aluminium

In this reaction HCl is the limiting reactant and aluminium is the reactant in excess.

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