Answer:
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.