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?

2 Answers
May 6, 2018

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

May 6, 2018

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