Question

If `80*g` of magnesium chloride salt were isolated from the treatment of magnesium metal by excess hydrochloric acid, how much metal was present?

Answer 1

We interrogate the stoichiometric equation..........and find that approx. `20*g` of magnesium was present.

Explanation:

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

If there are `80*g` salt produced, this represents a molar quantity of `(80*g)/(95.21*g*mol^-1)=0.840*mol`.

And thus there MUST have been an `0.840*mol` quantity of metal present, given the 1:1 stoichiometry of the reaction. And so we gots a mass with respect to the metal of...

`0.840*gxx24.31*g*mol^-1=??*g`

And thus there were also a `0.840*mol` quantity of dihydrogen gas evolved. If the gas is (reasonably!) assumed to behave ideally, then a volume of `0.840*molxx24.5*L*mol^-1=20.6*L` under standard condtions of `1*atm`, and `298*K` is evolved.