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

May 5, 2017

We interrogate the stoichiometric equation..........and find that approx. $20 \cdot g$ of magnesium was present.

#### Explanation:

$M g \left(s\right) + 2 H C l \left(a q\right) \rightarrow M g C {l}_{2} \left(a q\right) + {H}_{2} \left(g\right) \uparrow$

If there are $80 \cdot g$ salt produced, this represents a molar quantity of $\frac{80 \cdot g}{95.21 \cdot g \cdot m o {l}^{-} 1} = 0.840 \cdot m o l$.

And thus there MUST have been an $0.840 \cdot m o l$ 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 $0.840 \cdot m o l$ quantity of dihydrogen gas evolved. If the gas is (reasonably!) assumed to behave ideally, then a volume of $0.840 \cdot m o l \times 24.5 \cdot L \cdot m o {l}^{-} 1 = 20.6 \cdot L$ under standard condtions of $1 \cdot a t m$, and $298 \cdot K$ is evolved.