When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. 2 Al ( s ) + 6 HCl ( aq ) ⟶ 2AlCl3 ( aq ) + 3 H 2 ( g ) What volume of H 2 ( g ) is produced when 8.60 g Al ( s ) reacts at STP

The answer to your question is  V = 19.9 L

Explanation:

Data

mass of Al = 8.60 g

volume of H₂ = ?

Balanced chemical reaction

                 2Al  +  6HCl  ⇒   2AlCl₃  +  3H₂

1.- Use proportions to calculate the mass of H₂ produced

Atomic mass of Al = 2 x 27 = 54 g

Atomic mass of H₂ = 6 x 1 = 6 g

                54 g of Al 6 g of H₂

                  8 g  of Al  x

                  x = (8 x 6) / 54

                  x = 48 / 54

                  x = 0.89 g of H₂

2.- Calculate the volume of H₂

- Convert the H₂ to moles

                    1 g of H₂ 1 mol

                   0.89 g     x

                    x = 0.89 moles

                    1 mol of H₂ 22.4 l

                   0.89 moles   x

                    x = (0.89 x 22.4) / 1

                   x = 19.9 L