The volume of hydrogen released by the interaction of 130 g of zinc with 36.5 g of hydrochloric acid ..

The reaction equation for the interaction of zinc with acid looks like this:

Zn + 2HCl = ZnCl2 + H2.

It shows that the number of moles of released hydrogen is equal to the number of moles of zinc reacted. It remains to calculate the latter.

Zinc in the reaction zone 130 g: 65 g / mol = 2 mol, HCl – 36.5 g: 36.5 g / mol = 1 mol, but it takes 2 times more. Thus, we have an excess of Zn and it reacts in an amount of 1/2 mol. From here we get the amount and volume of H2: 1/2 mol * 22.4 l / mol = 11.2 l.