Aluminum reacts with hydrochloric acid to produce aluminum chloride and hydrogen gas. What mass of hydrochloric acid reacts when 87.7 grams of aluminum dissolves in excess hydrochloric acid

355.5 g.

Explanation:

The balanced equation for the mentioned reaction is:

2Al + 6HCl → 2AlCl₃ + 3H₂,

It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.

Firstly, we need to calculate the no. of moles of (87.7 g) of Al:

no. of moles of Al = mass/atomic mass = (87.7 g) / (26.98 g/mol) = 3.251 mol.

Using cross multiplication:

2.0 mol of Al need to react → 6.0 mol of HCl, from stichiometry.

3.251 mol of Al need to react →? mol of HCl.

∴ the no. of moles of HCl needed to react with (87.7 g) 3.251 mol of Al = (3.251 mol) (6.0 mol) / (2.0 mol) = 9.752 mol.

Now, we can get the mass of HCl needed to react with (87.7 g) 3.251 mol of Al:

mass of HCl = (no. of moles) (molar mass) = (9.752 mol) (36.46 g/mol) = 355.5 g.