What is the volume of hydrogen gas released when 25cm3 of 0.2 mol dm-3 hydrochloric acid reacts with 0.2 g of zinc powder?

Answer 1

First, let's write down the correctly balanced equation. We will need this in order to look at the stoichiometry of the reaction (mol ratio of hydrogen gas to hydrochloric acid and zinc).

Zn(s) + 2HCl(aq) ==> ZnCl₂(aq) + H₂(g) ... balanced equation

Next, we will find which reactant, if any, is limiting:

moles Zn present = 0.2 g Zn x 1 mol Zn / 65 g = 0.0031 moles Zn

moles HCl present = 25 cm³ x 1 dm³ / 1000 cm³ x 0.2 mol/dm³ = 0.005 moles HCl

Since it takes TWO moles HCl per one mole Zn, HCl is LIMITING as we don't have 2x as much.

Now that we know HCl is limiting, we can use the 0.005 mols HCl to determine moles and volume of H₂ gas.

0.005 moles HCl x 1 mol H₂ / 2 moles HCl = 0.0025 moles H₂ gas formed.

If we assume this reaction takes place at STP (standard temperature and pressure), 1 mole of any gas is equal to 22.4 dm³ in volume.

Volume H₂ gas formed = 0.0025 mols H₂ x 22.4 dm³ / mole = 0.056 dm³ = 0.06 dm³ (1 sig. fig.)