Question

What is the maximum volume of hydrogen that can be liberated when 1.2 moles of zinc are added to excess dilute hydrochloride acid at room temperature and pressure?

Answer 1

Would it not be approx. `30.5*L`?

Explanation:

We address the stoichiometric equation...

`Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) +H_2(g)uarr`

And thus one equivalent of dihydrogen gas is produced by one equiv of zinc metal....

Here, we used `1.2*mol` of zinc metal reductant...and therefore we gets `1.2*mol` dihydrogen gas. And depending on your syllabus (and these days they are all over the place), one mole of gas occupies `22.4*dm^3` at `0` `""^@C`, or `25.4*dm^3` at `298` `K`...and so with the second definition, we take the product...

`25.4*dm^3*mol^-1xx1.20*mol=??L`

Answer 2

26.8 litres

Explanation:

The balanced chemical reaction would be

`"Zn"+color(red)(2)"HCl"->"ZnCl"_2+"H"_2`

`=>color(blue)(1"mol")"Zn"+color(blue)(2"mol")"HCl"->color(blue)(1"mol")"ZnCl"_2+color(blue)(1"mol")"H"_2`

That means 1 mol Zn will yield 1 mol `"H"_2`. So, 1.2 mol Zn will yield 1.2 mol `"H"_2`.

1.2 × molar volume = 1.2 × 22.4
= 26.88 litres