14 g of iron was fused with 6.4 g of sulfur. When products are processed with hydrochloric acid, gases are generated in volume.

1) We have a reaction of fusion of sulfur with iron: Fe + S = FeS.
According to it, we have that the amount of iron is equal to the amount of sulfur in the reaction.
2) Calculate the amount of substance: n (Fe) = m (Fe) / M (Fe) = 14 g / 56 g / mol = 0.25 mol, n (S) = m (S) / M (S) = 6 , 4 g / 32 g / mol = 0.2 mol.
3) Since n (S) <n (Fe), we consider it as a deficiency, and 0.2 mol of iron will enter into the reaction and 0.2 mol of FeS is formed.
4) We have the reaction of interaction of iron (II) sulfide with hydrochloric acid: FeS + 2HCl = FeCl2 + H2S, according to it n (FeS) = n (H2S) = 0.2 mol.
5) Then V (H2S) = n (H2S) * Vm = 0.2 mol * 22.4 L / mol = 4.48 L.
ANSWER: with a volume of 4.48 liters.