# A 55*mL volume of hydrochloric acid was poured on to an excess of magnesium turnings. What volume of dihydrogen gas evolved?

Jul 17, 2017

Well, one mole of gas at $\text{STP}$ occupies $22.4 \cdot L$.......We do not know the concentration of hydrochloric acid.......

#### Explanation:

And thus we interrogate the redox reaction........

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

And thus the moles of magnesium is equivalent to the moles of dihydrogen gas if there is molar equivalence.

Now I happen to know that if ${\rho}_{\text{HCl}} = 1.18 \cdot g \cdot m o {l}^{-} 1$, then the concentration of $H C l$ is $10.17 \cdot m o l \cdot {L}^{-} 1$ (and such data SHOULD have been supplied with the question).

And so we have a molar quantity of $55 \times {10}^{-} 3 \cdot L \times 10.17 \cdot m o l \cdot {L}^{-} 1 = 0.559 \cdot m o l$ with respect to $H C l$.

$\text{Moles of magnesium} = \frac{0.865 \cdot g}{24.305 \cdot g \cdot m o {l}^{-} 1} = 0.0356 \cdot m o l$.

Magnesium metal is in excess, and we do not have a handle on the volume of dihydrogen evolved.