Question

Under standard conditions what volume of dihydrogen will be evolved from treatment of a mass of `0.08*g` magnesium metal with excess acid?

Answer 1

You have got the stoichiometric reaction......

Explanation:

`Mg(s) + 2HCl(aq) rarr MgCl_2(aq)+H_2(g)uarr`

And you have a molar quantity of `(0.08*g)/(24.3*g*mol^-1)=3.29xx10^-3*mol` with respect to the metal.

And thus get an equivalent molar quantity of dihydrogen....

i.e. `"moles of dihydrogen gas"=3.29xx10^-3*mol`.

We know that 1 mol of Ideal gas occupies a volume of `24.8*dm^3` under standard conditions. I do not what the standard pertains in your syllabus so you will have to do some digging.....

And so volume....

`=3.29xx10^-3*molxx24.8*dm^3*mol^-1-=0.082*dm^3` approx. `82*mL`.